Trigonometric Identities

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This section covers:

Before we get started, here is a table of some common trig identities for future reference:

Reciprocal and Quotient Identities

\(\displaystyle \sin \theta =\frac{1}{{\csc \theta }}\,\,\,\,\,\,\,\,\,\,\,\csc \theta =\frac{1}{{\sin \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac{1}{{\sec \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \theta =\frac{1}{{\cos \theta }}\,\)

\(\displaystyle \tan \theta =\frac{1}{{\cot \theta }}=\frac{{\sin \theta }}{{\cos \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cot \theta =\frac{1}{{\tan \theta }}=\frac{{\cos \theta }}{{\sin \theta }}\)

Pythagorean Identities

\(\begin{array}{l}\sin \left( {A+B} \right)=\sin A\cos B+\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {A+B} \right)=\cos A\cos B-\sin A\sin B\\\sin \left( {A-B} \right)=\sin A\cos B-\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {A-B} \right)=\cos A\cos B+\sin A\sin B\end{array}\)

 

\(\displaystyle {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\cot }^{2}}\theta +1={{\csc }^{2}}\theta \)

Sum and Difference Identities

\(\displaystyle \tan \left( {A+B} \right)=\frac{{\tan A+\tan B}}{{1-\tan A\tan B}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {A-B} \right)=\frac{{\tan A-\tan B}}{{1+\tan A\tan B}}\)

Double Angle Identities

\(\sin \left( {2A} \right)=2\sin A\cos A\)          \(\begin{align}\cos \left( {2A} \right)&={{\cos }^{2}}A-{{\sin }^{2}}A\\&=1-2{{\sin }^{2}}A\\&=2{{\cos }^{2}}A-1\end{align}\)               \(\displaystyle \tan \left( {2A} \right)=\frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}\)

Half Angle Identities

    \(\displaystyle \sin \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1-\cos A}}{2}}}\,\,\,\,\,\,\,\,\,\,\cos \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1+\cos A}}{2}}}\,\,\,\,\,\,\,\,\,\,\,\tan \left( {\frac{A}{2}} \right)=\frac{{\sin A}}{{1+\cos A}}=\frac{{1-\cos A}}{{\sin A}}\)

 

 

An “identity” is something that is always true, so you are typically either substituting or trying to get two sides of an equation to equal each other. Think of it as a reflection; like looking in a mirror. An example of a trig identity is \(\displaystyle \csc (x)=\frac{1}{\sin (x)}\); for any value of \(x\), this equation is true.

Trig identities are sort of like puzzles since you have to “play” with them to get what you want. You will also have to do some memorizing for these, since most of them aren’t really obvious. You may not like Trig Identity problems, since they can resemble the proofs that you had to in Geometry. I actually love them, since I love to do puzzles!

There are typically two types of problems you’ll have with trig identities: working on one side of an equation to “prove” it equals the other side, and also solving trig problems by substituting identities to make the problem solvable.

We’ll start out with the simpler identities that you’ve seen before.

Reciprocal and Quotient Identities

You’ve already seen the reciprocal and quotient identities. You can also write these as “\(\sin x\)”, and so on.

Trig Function

Reciprocal Function

Hints

\(\displaystyle \sin \theta =\frac{1}{{\csc \theta }}\)

 \(\displaystyle \csc \theta =\frac{1}{{\sin \theta }}\) I remember that sin is the reciprocal of cosecant (csc) since the “s” in the middle of csc matches the “s” at the beginning of sin.

\(\displaystyle \cos \theta =\frac{1}{{\sec \theta }}\)

\(\displaystyle \sec \theta =\frac{1}{{\cos \theta }}\)

I remember that secant (sec) is the reciprocal of cos, since cos contains “c s” and sec contains “s c”.

\(\displaystyle \cos \theta =\frac{1}{{\sec \theta }}\)

\(\displaystyle \cot \theta =\frac{1}{{\tan \theta }}\)

It’s easy to remember that tan is the reciprocal of cotangent (cot), since they both have “tangent” in them.

\(\displaystyle \tan \theta =\frac{{\sin \theta }}{{\cos \theta }}\)

\(\displaystyle \cot \theta =\frac{{\cos \theta }}{{\sin \theta }}\)

We’ve already seen these, but the way I remember the order is that cot has the cos on the top (in the numerator).

 

Here are some examples of simple identity proofs with reciprocal and quotient identities. Typically, to do these proofs, you must always start with one side (either side, but usually take the more complicated side) and manipulate the side until you end up with the other side. (Some teachers will let you go down both sides until the two sides are equal).

The best way to solve these is to turn everything into sin and cos. Note how we work on one side only and pull down the other side when it matches. It doesn’t matter which side we start on, but typically, it’s the most complicated.

Sometimes we have to find common denominators, like in the last example. We didn’t need to turn it into sin and cos, since we only had tan and cot in the identity (although it still would have worked).

And note that there may be more than one way to do these! Several answers may be “right”.

Reciprocal and Quotient Trig Identities

     \(\displaystyle \sin x\cot x\,=\,\cos x\)

 

\(\require{cancel} \displaystyle \cancel{{\sin x}}\left( {\frac{{\cos x}}{{\cancel{{\sin x}}}}} \right)=\cos x\,\surd \)

\(\sin \theta (\cot \theta +\sec \theta )=\cos \theta +\tan \theta \)

 

\(\begin{align}\sin \theta \cot \theta +\sin \theta \sec \theta &=\\\cancel{{\sin \theta }}\frac{{\cos \theta }}{{\cancel{{\sin \theta }}}}+\sin \theta \frac{1}{{\cos \theta }}&=\\\cos \theta +\frac{{\sin \theta }}{{\cos \theta }}&=\cos \theta +\tan \theta \,\surd \end{align}\)

\(\displaystyle \frac{{1+\cot \theta }}{{1+\tan \theta }}=\,\cot \theta \)

 

\(\displaystyle \begin{align}\frac{{1+\frac{1}{{\tan \theta }}}}{{1+\tan \theta }}&=\\\frac{{\frac{{\tan \theta }}{{\tan \theta }}+\frac{1}{{\tan \theta }}}}{{1+\tan \theta }}&=\\\frac{{\frac{{\tan \theta +1}}{{\tan \theta }}}}{{1+\tan \theta }}&=\\\frac{{\cancel{{\tan \theta +1}}}}{{\tan \theta }}\times \frac{1}{{\cancel{{1+\tan \theta }}}}&=\\\frac{1}{{\tan \theta }}&=\cot \theta \,\surd \end{align}\)

 

Pythagorean Identities

The Pythagorean identities are derived from (you guessed it!) the Pythagorean Theorem. Going back to the unit circle, notice that \({{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\):

Pythagorean Identities

           “Proof“:    \(\displaystyle \begin{align}\sin \theta =\frac{y}{r}\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac{x}{r}\,\,\,\,\,\,\,\,\,\,\,\,{{y}^{2}}+{{x}^{2}}=1\,\,\,\,\left( {\text{Pythagorean}} \right)\\{{\sin }^{2}}\theta +{{\cos }^{2}}\theta ={{\left( {\frac{y}{r}} \right)}^{2}}+{{\left( {\frac{x}{r}} \right)}^{2}}={{\left( {\frac{y}{1}} \right)}^{2}}+{{\left( {\frac{x}{1}} \right)}^{2}}={{y}^{2}}+{{x}^{2}}=1\end{align}\)

                                  (Note that \({{\left( {\sin x} \right)}^{2}}\) is written as \({{\sin }^{2}}x\).)

 

From the first Pythagorean identity, we can derive the other two:

\(\begin{align}{{\sin }^{2}}\theta +{{\cos }^{2}}\theta &=1\\\frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}&=\frac{1}{{{{{\cos }}^{2}}\theta }}\\{{\tan }^{2}}\theta +1&={{\sec }^{2}}\theta \end{align}\)                                \(\begin{align}{{\sin }^{2}}\theta +{{\cos }^{2}}\theta &=1\\\frac{{{{{\sin }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }}&=\frac{1}{{{{{\sin }}^{2}}\theta }}\\1+{{\cot }^{2}}\theta &={{\csc }^{2}}\theta \end{align}\)

 

The three Pythagorean Identities are:

\(\displaystyle \begin{array}{c}{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\\{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \\{{\cot }^{2}}\theta +1={{\csc }^{2}}\theta \end{array}\)

Here are some examples of solving Pythagorean Identities. To “prove” the identities, we use the following “tricks” if we can:

  1. Match trig functions; for example, if you have a \(\cos \text{, }{{\cos }^{2}}\), and \({{\sin }^{2}}\), turn the \({{\sin }^{2}}\) into \(\left( {1-{{{\cos }}^{2}}} \right)\).
  2. Use common denominators to combine terms.
  3. Use conjugates (the first term and then change sign and then second term) to multiply numerator or denominator with two terms (binomials). This will create a difference of two squares to work with.
  4. Turn all trig functions into sin and cos if you have other trig functions, such as tan.
Pythagorean Identify Proofs Hints
                 \(\displaystyle \frac{{\csc x}}{{\cos x}}-\frac{{\cos x}}{{\sin x}}\,=\,\tan x\)

 

\(\displaystyle \require{cancel} \begin{align}\frac{1}{{\sin x}}\cdot \frac{1}{{\cos x}}-\frac{{\cos x}}{{\sin x}}&=\\\frac{1}{{\sin x\cdot \cos x}}-\frac{{\cos x}}{{\sin x}}\cdot \frac{{\cos x}}{{\cos x}}&=\\\frac{{1-{{{\cos }}^{2}}x}}{{\sin x\cos x}}&=\\\frac{{{}^{{\sin x}}\cancel{{{{{\sin }}^{2}}x}}}}{{\cancel{{\sin x}}\cos x}}&=\,\tan x\,\,\,\,\,\,\surd \end{align}\)

  • Convert everything to sin and cos
  • Find common denominator to combine fractions
  • Use Pythagorean identity to turn two terms in the numerator into one
  • Simplify

               \(\displaystyle \frac{{{{{\cos }}^{2}}\theta }}{{1+\sin \theta }}\,=\,1-\sin \theta \)

 

\(\displaystyle \begin{align}\frac{{1-{{{\sin }}^{2}}\theta }}{{1+\sin \theta }}\,\,\,&=\\\frac{{\left( {1-\sin \theta } \right)\cancel{{(1+\sin \theta )}}}}{{\cancel{{1+\sin \theta }}}}\,&=\,1-\sin \theta \,\,\,\,\,\,\,\,\surd \end{align}\)

  • Convert everything to sin by using Pythagorean identity
  • Factor numerator (difference of squares)
  • Simplify

And here are a few more that are more complicated:

Pythagorean Identity Proofs Hints

  \(\displaystyle \frac{2}{{\sin x}}\,=\,\frac{{\tan x}}{{\sec x+1}}+\frac{{\sec x+1}}{{\tan x}}\)

 

\(\require{cancel} \displaystyle \begin{align}&=\frac{{\tan x}}{{\sec x+1}}\left( {\frac{{\tan x}}{{\tan x}}} \right)+\frac{{\sec x+1}}{{\tan x}}\left( {\frac{{\sec x+1}}{{\sec x+1}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac{{{{{\tan }}^{2}}x}}{{\tan x\left( {\sec x+1} \right)}}+\frac{{{{{\left( {\sec x+1} \right)}}^{2}}}}{{\tan x\left( {\sec x+1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac{{{{{\tan }}^{2}}x}}{{\tan x\left( {\sec x+1} \right)}}+\frac{{{{{\sec }}^{2}}x+2\sec x+1}}{{\tan x\left( {\sec x+1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac{{\left( {{{{\tan }}^{2}}x+1} \right)+{{{\sec }}^{2}}x+2\sec x}}{{\tan x\left( {\sec x+1} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac{{{{{\sec }}^{2}}x+{{{\sec }}^{2}}x+2\sec x}}{{\tan x\left( {\sec x+1} \right)}}\,\,=\,\frac{{2\sec x\cancel{{\left( {\sec x+1} \right)}}}}{{\tan x\cancel{{\left( {\sec x+1} \right)}}}}\\\frac{2}{{\sin x}}\,\,&=\frac{2}{{\cancel{{\cos x}}}}\cdot \frac{{\cancel{{\cos x}}}}{{\sin x}}\,\,\,\surd \end{align}\)

  • Start with the right-hand side, since it’s more complicated
  • Find common denominator to add fractions, since we have to turn two terms into one
  • Convert \(\left( {{{{\tan }}^{2}}x+1} \right)\) into \({{\sec }^{2}}x\), since we have more sec than tan in the problem at this point
  • Factor out a \(2\sec x\) in numerator
  • Turn everything into sin and cos to match left-hand side
  • Simplify
                           \(\displaystyle \frac{{1-{{{\tan }}^{2}}y}}{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}\,=\,{{\sec }^{2}}y\)

 

\(\displaystyle \begin{align}\frac{{1-\frac{{{{{\sin }}^{2}}y}}{{{{{\cos }}^{2}}y}}}}{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}\,\,&=\\\frac{{\frac{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}{{{{{\cos }}^{2}}y}}}}{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}\,\,&=\\\frac{{\cancel{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}}}{{{{{\cos }}^{2}}y}}\cdot \frac{1}{{\cancel{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}}}\,\,&=\\\frac{1}{{{{{\cos }}^{2}}y}}\,&=\,\,{{\sec }^{2}}y\,\,\,\surd \end{align}\)

  • The trick here is that it looks like you should use a Pythagorean identity because of all the squares, but you don’t (you can try, but it won’t get you anywhere)
  • Convert everything to sin and cos
  • Find common denominator in the numerator to combine fractions
  • To divide the fractions, flip the bottom (put 1 over it) and multiply
  • Simplify
                              \(\displaystyle \frac{{{{{\cos }}^{4}}\theta -{{{\sin }}^{4}}\theta }}{{{{{\left( {2{{{\cos }}^{2}}\theta -1} \right)}}^{2}}}}=\frac{1}{{1-2{{{\sin }}^{2}}\theta }}\)

 

\(\displaystyle \begin{align}\frac{{\left( {{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta } \right)\cancel{{\left( {{{{\cos }}^{2}}\theta +{{{\sin }}^{2}}\theta } \right)}}}}{{{{{\left( {2{{{\cos }}^{2}}\theta -1} \right)}}^{2}}}}&=\\\frac{{{{{\cos }}^{2}}\theta -\left( {1-{{{\cos }}^{2}}\theta } \right)}}{{{{{\left( {2{{{\cos }}^{2}}\theta -1} \right)}}^{2}}}}&=\\\frac{{2{{{\cos }}^{2}}\theta -1}}{{{{{\left( {2{{{\cos }}^{2}}\theta -1} \right)}}^{2}}}}&=\\\frac{1}{{2{{{\cos }}^{2}}\theta -1}}&=\\\frac{1}{{2\left( {1-{{{\sin }}^{2}}\theta } \right)-1}}&=\frac{1}{{1-2{{{\sin }}^{2}}\theta }}\,\,\,\surd \end{align}\)

·

  • We need to get rid of the fourth powers by using the difference of squares. Let’s leave the denominator alone, since there would be a fourth root if we multiplied out
  • Aha! We can get rid of the \(({{\sin }^{2}}x+{{\cos }^{2}}x)\) since it is 1, and we are multiplying
  • Turn everything we can into cos’s to see if we can simplify; we can!
  • Now turn back to sin since that’s on the right-hand side.
  • That was difficult! And remember that there are other ways to do this!

                                                                                                                                                  \(\displaystyle \frac{{1-\sin x+\cos x}}{{1+\sin x+\cos x}}\,=\,\frac{{\cos x}}{{1+\sin x}}\)

 

\(\displaystyle \begin{align}\frac{{1-\left( {\sin x-\cos x} \right)}}{{1+\left( {\sin x+\cos x} \right)}}\,\cdot \frac{{1+\left( {\sin x-\cos x} \right)}}{{1+\left( {\sin x-\cos x} \right)}}&=\\\frac{{1-{{{\left( {\sin x-\cos x} \right)}}^{2}}}}{{\left( {1+\sin x+\cos x} \right)\left( {1+\sin x-\cos x} \right)}}&=\\\frac{{1-\left( {{{{\sin }}^{2}}x-2\sin x\cos x+{{{\cos }}^{2}}x} \right)}}{{1+2\sin x+{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}&=\\\frac{{1-\left( {1-2\sin x\cos x} \right)}}{{1+2\sin x+{{{\sin }}^{2}}x-{{{\cos }}^{2}}x}}&=\\\frac{{2\sin x\cos x}}{{\cancel{1}+2\sin x+{{{\sin }}^{2}}x-\left( {\cancel{1}-{{{\sin }}^{2}}x} \right)}}&=\\\frac{{2\sin x\cos x}}{{2\sin x+2{{{\sin }}^{2}}x}}&=\\\frac{{\cancel{{2\sin x}}\cos x}}{{\cancel{{2\sin x}}\left( {1+\sin x} \right)}}&=\frac{{\cos x}}{{1+\sin x}}\,\,\,\surd \end{align}\)

  • This one’s tricky since it’s hard to know where to start. Since we have a “\(1-\)“ and a “\(1+\)” on the left, find a conjugate of the numerator and multiply (we could have found the conjugate of the denominator instead)
  • We then get a difference of squares in the numerator
  • For the denominator, note that \(\displaystyle \begin{array}{c}\left( {1+\sin x+\cos x} \right)\left( {1+\sin x-\cos x} \right)\\=1+2\sin x+{{\sin }^{2}}x-{{\cos }^{2}}x\end{array}\)
  • Keep plugging until you can cancel out factors on top and bottom. Remember that \({{\sin }^{2}}x+{{\cos }^{2}}x=1\)
  • Plug in \(\left( {1-{{{\sin }}^{2}}x} \right)\) for \({{\cos }^{2}}x\) to help simplify, since there are more sin‘s on the bottom
  • We did it! Don’t worry if you have to erase a lot before solving these!

Solving with Reciprocal, Quotient and Pythagorean Identities

Here are some problems where we have use reciprocal and/or Pythagorean identities to solve trig equations in the interval \(\left[ {0,2\pi } \right)\):

Solving with Trig Identities Hints
\(\displaystyle 3\cos x+2{{\sin }^{2}}x=3\)

 

\(\displaystyle \begin{align}2{{\sin }^{2}}+3\cos x-3&=0\\2\left( {1-{{{\cos }}^{2}}} \right)+3\cos x-3&=0\\2-2{{\cos }^{2}}+3\cos x-3&=0\\-2{{\cos }^{2}}+3\cos x-1&=0\\2{{\cos }^{2}}-3\cos x+1&=0\\\left( {2\cos x-1} \right)\left( {\cos x-1} \right)&=0\end{align}\)

 

\(\displaystyle \cos x=\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\cos x=1\,\,\,\,\,\,\,\,\,\,\,\,\,x=\left\{ {0,\,\,\frac{\pi }{3},\frac{{5\pi }}{3}} \right\}\)

  • Since we have a quadratic, set everything to 0 and factor
  • Use Pythagorean Identity to convert \({{\sin }^{2}}x\) into \(\left( {1-{{{\cos }}^{2}}x} \right)\), since we have a \(3\cos x\) in the problem (try to match trig functions)
  • Now we can factor and solve!
  • We can check answers in the graphing calculator!

\(\displaystyle \sec x-\tan x=\cot x\)

 

\(\displaystyle \begin{align}\frac{1}{{\cos x}}-\frac{{\sin x}}{{\cos x}}&=\frac{{\cos x}}{{\sin x}}\\\frac{{1-\sin x}}{{\cos x}}-\frac{{\cos x}}{{\sin x}}&=0\\\frac{{\sin x\left( {1-\sin x} \right)}}{{\sin x\cos x}}-\frac{{{{{\cos }}^{2}}x}}{{\sin x\cos x}}&=0\\\frac{{\left( {\sin x-{{{\sin }}^{2}}x-{{{\cos }}^{2}}x} \right)}}{{\sin x\cos x}}&=0\\\sin x-{{\sin }^{2}}x-\left( {1-{{{\sin }}^{2}}x} \right)&=0\\\sin x-1&=0\,\\x&=\left\{ {\frac{\pi }{2}} \right\}\end{align}\)

 

 

This is an extraneous solution, since \(\displaystyle \cos \left( {\frac{\pi }{2}} \right)=0\), and we can’t divide by 0. The answer is no solution or \(\emptyset\).

  • Convert everything to sin and cos, using Reciprocal and Quotient Identities
  • Find common denominator to add terms
  • Turn \({{\cos }^{2}}x\) into \(\left( {1-{{{\sin }}^{2}}x} \right)\) since we have mainly sin; and combine terms
  • Note that we multiplied the whole equation by the denominator \(\displaystyle (\sin x\cos x)\). Since we did this, we’ll have to check for extraneous or extra solutions since \(\sin x\) and \(\cos x\) can’t be 0
  • You can also square each side, but be sure to check for extraneous solutions when you do this:

\(\displaystyle \sec x=\cot x+\tan x\)                   

\(\require{cancel} \displaystyle \begin{align}{{\sec }^{2}}x&={{\left( {\cot x+\tan x} \right)}^{2}}\\1+{{\tan }^{2}}x&={{\cot }^{2}}x+2\cancel{{\cot x\tan x}}+{{\tan }^{2}}x\\1+{{\tan }^{2}}x&={{\cot }^{2}}x+2+{{\tan }^{2}}x\\1+\cancel{{{{{\tan }}^{2}}x}}&={{\cot }^{2}}x+2+\cancel{{{{{\tan }}^{2}}x}}\\{{\cot }^{2}}x&=-1\end{align}\)

 

No solution, since we can’t square something and have it be a negative number.

\(\displaystyle \frac{{{{{\cos }}^{2}}\left( x \right)}}{{{{{\sin }}^{2}}\left( x \right)}}=3\)

 

\(\displaystyle \begin{array}{c}{{\cot }^{2}}\left( x \right)=3\\\cot \left( x \right)=\pm \sqrt{3}\end{array}\)

\(\displaystyle x=\left\{ {\frac{\pi }{3},\,\,\frac{{2\pi }}{3},\,\,\frac{{4\pi }}{3},\,\frac{{5\pi }}{3}} \right\}\)

  • Use Quotient Identity to convert \(\displaystyle \frac{{\cos }}{{\sin }}\) into cot
  • Take square roots of both sides, and don’t forget the \(\pm \)

Here’s another one where we have to check for extraneous solutions. Solve over the reals:

Solving with Trig Identities Hints

\(\displaystyle \frac{{1-\cos x}}{{\sin x}}=\frac{{\sin x}}{{1+\cos x}}\)

 

\(\displaystyle \begin{align}\frac{{1-\cos x}}{{\sin x}}-\frac{{\sin x}}{{1+\cos x}}&=0\\\frac{{\left( {1-\cos x} \right)\left( {1+\cos x} \right)}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}-\frac{{\left( {\sin x} \right)\left( {\sin x} \right)}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}&=0\\\frac{{1-{{{\cos }}^{2}}x}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}-\frac{{{{{\sin }}^{2}}x}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}&=0\\\frac{{{{{\sin }}^{2}}x}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}-\frac{{{{{\sin }}^{2}}x}}{{\left( {\sin x} \right)\left( {1+\cos x} \right)}}&=0\\0&=0\end{align}\)

 

This appears to be all real numbers, but since we have denominators that can’t be 0, we have to check for those cases.

  • Get all terms on one side
  • Find common denominator to add terms
  • Turn \(\left( {1-{{{\cos }}^{2}}x} \right)\) into \({{\sin }^{2}}x\)
  • We get \(0=0\), which means it looks like it works for all real numbers, or \(\mathbb{R}\) (The two terms are an identity, since they equal each other).
  • However, since we have denominators of \(\sin x\) and \(1+\cos x\), we have to check for extraneous solutions, which would be when the denominators are 0:

\(\displaystyle \begin{array}{l}\sin x=0\,\,\,\,\,\,\,\,\cos x=-1\\\,\,\,\,\,0,\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\pi \end{array}\)

 

The answer, over the reals, is all real numbers except 0 and \(\boldsymbol{\pi}\), or \(\displaystyle \left\{ {x|\,\,\,x\ne 0+2\pi k,\,\,\,\pi +2\pi k} \right\}\), which is the same as \(\displaystyle \left\{ {x|\,\,\,x\ne \pi k} \right\}\).

 

Sum and Difference Identities

We use sum and difference identities when we need to split up the angle to make it easier to find the values (for example, to find values on the unit circle).

We also use the identities in conjunction with other identities to prove and solve trig problems.

Here are the sum and difference identities, and tricks to help you memorize them.

Sum and Difference Identities

Hints

\(\begin{array}{l}\cos \left( {A+B} \right)=\cos A\cos B-\sin A\sin B\\\cos \left( {A-B} \right)=\cos A\cos B+\sin A\sin B\end{array}\)

To help memorize this, I remember that since cos is even, we have the cos’s together and the sin’s together on the right side.

Note that the signs (plus or minus) do not match. (We can’t have everything!)

\(\begin{array}{l}\sin \left( {A+B} \right)=\sin A\cos B+\cos A\sin B\\\sin \left( {A-B} \right)=\sin A\cos B-\cos A\sin B\end{array}\)

I remember with sin, the sign’s match! But we don’t have the sin’s and cos’s together on the right side.
\(\displaystyle \tan \left( {A+B} \right)=\frac{{\tan A+\tan B}}{{1-\tan A\tan B}}\)

\(\displaystyle \tan \left( {A-B} \right)=\frac{{\tan A-\tan B}}{{1+\tan A\tan B}}\)

So this one is tricky to memorize. I remember that the numerator is just adding the two tan’s and the signs do match. The denominator always starts with 1 and we subtract products of tan’s, but the signs do not match.
Note: From these identities, you may be asked to be familiar with the Odd/Even Identities:

\(\displaystyle \begin{array}{l}\text{EVEN: }\,\,\,\cos \left( {-x} \right)=\cos \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ODD:}\,\,\,\,\,\,\sin \left( {-x} \right)=-\sin \left( x \right)\,\,\,\,\,\,\,\tan \left( {-x} \right)=-\tan \left( x \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( {-x} \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( {-x} \right)=-\csc \left( x \right)\,\,\,\,\,\,\,\cot \left( {-x} \right)=-\cot \left( x \right)\end{array}\)

 

…..and the Cofunction Identities in radians (trig functions of an angle is equal to the value of the cofunction of the complement). Try to prove these using the identities above!

\(\displaystyle \begin{array}{l}\sin \left( {\frac{\pi }{2}-x} \right)=\cos \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( {\frac{\pi }{2}-x} \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {\frac{\pi }{2}-x} \right)=\cot \left( x \right)\\\cos \left( {\frac{\pi }{2}-x} \right)=\sin \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( {\frac{\pi }{2}-x} \right)=\csc \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cot \left( {\frac{\pi }{2}-x} \right)=\tan \left( x \right)\end{array}\)

First, let’s solve some trig problems using the sum and difference identities:

Sum and Difference Identity Sum and Difference Identity

Simplify \(\displaystyle \sin \left( {\frac{{4\pi }}{5}} \right)\cos \left( {\frac{{2\pi }}{7}} \right)+\cos \left( {\frac{{4\pi }}{5}} \right)\sin \left( {\frac{{2\pi }}{7}} \right)\)

 

Since \(\sin A\cos B+\cos A\sin B=\sin \left( {A+B} \right)\), we have:

\(\displaystyle \begin{align}\sin \left( {\frac{{4\pi }}{5}} \right)\cos \left( {\frac{{2\pi }}{7}} \right)+\cos \left( {\frac{{4\pi }}{5}} \right)\sin \left( {\frac{{2\pi }}{7}} \right)\\=\sin \left( {\frac{{4\pi }}{5}+\frac{{2\pi }}{7}} \right)=\sin \left( {\frac{{38\pi }}{{35}}} \right)\end{align}\)

Find exact value of \(\tan \left( {165{}^\circ } \right)\)

 

Since \(\displaystyle \tan \left( {A+B} \right)=\frac{{\tan A+\tan B}}{{1-\tan A\tan B}}\), we can find two values on the Unit Circle whose sum is 165°120° and 45°:

\(\begin{align}\tan \left( {165} \right)=\tan \left( {120+45} \right)&=\frac{{\tan 120+\tan 45}}{{1-\left( {\tan 120} \right)\left( {\tan 45} \right)}}\\&=\frac{{-\sqrt{3}+1}}{{1-\left( {-\sqrt{3}} \right)\left( 1 \right)}}=\frac{{1-\sqrt{3}}}{{1+\sqrt{3}}}\end{align}\)

Find the exact value of \(\displaystyle \cos \left( {\frac{\pi }{{12}}} \right)\)

 

Divide up \(\displaystyle \frac{\pi }{{12}}\) into two fractions that can be found on the Unit Circle:

\(\displaystyle \begin{align}\cos \left( {\frac{\pi }{{12}}} \right)&=\cos \left( {\frac{{4\pi }}{{12}}-\frac{{3\pi }}{{12}}} \right)=\cos \left( {\frac{\pi }{3}-\frac{\pi }{4}} \right)\\&=\cos \left( {\frac{\pi }{3}} \right)\cos \left( {\frac{\pi }{4}} \right)+\sin \left( {\frac{\pi }{3}} \right)\sin \left( {\frac{\pi }{4}} \right)\\&=\left( {\frac{1}{2}} \right)\left( {\frac{{\sqrt{2}}}{2}} \right)+\left( {\frac{{\sqrt{3}}}{2}} \right)\left( {\frac{{\sqrt{2}}}{2}} \right)\\&=\frac{{\sqrt{2}+\sqrt{6}}}{4}\end{align}\)

Simplify \(\displaystyle \frac{{\tan \frac{\pi }{{12}}-\tan \frac{{5\pi }}{4}}}{{1+\left( {\tan \frac{{5\pi }}{4}} \right)\left( {\tan \frac{\pi }{{12}}} \right)}}\)

 

Since \(\displaystyle \frac{{\tan A-\tan B}}{{1+\tan A\tan B}}=\tan \left( {A-B} \right)\), we have:

\(\displaystyle \frac{{\tan \frac{\pi }{{12}}-\tan \frac{{5\pi }}{4}}}{{1+\left( {\tan \frac{{5\pi }}{4}} \right)\left( {\tan \frac{\pi }{{12}}} \right)}}=\tan \left( {\frac{\pi }{{12}}-\frac{{5\pi }}{4}} \right)=\,\,\tan \left( {\frac{{-7\pi }}{6}} \right)\)

Find the exact value of \(\displaystyle \sec \left( {-\frac{\pi }{{12}}} \right)\)

 

Divide up \(\displaystyle -\frac{\pi }{{12}}\) into two fractions that can be found on the Unit Circle; use cos first, and then take reciprocal:

\(\displaystyle \begin{align}\cos \left( {-\frac{\pi }{{12}}} \right)&=\cos \left( {\frac{{3\pi }}{{12}}-\frac{{4\pi }}{{12}}} \right)=\cos \left( {\frac{\pi }{4}-\frac{\pi }{3}} \right)\\&=\cos \left( {\frac{\pi }{4}} \right)\cos \left( {\frac{\pi }{3}} \right)+\sin \left( {\frac{\pi }{4}} \right)\sin \left( {\frac{\pi }{3}} \right)\\&=\left( {\frac{{\sqrt{2}}}{2}} \right)\left( {\frac{1}{2}} \right)+\left( {\frac{{\sqrt{2}}}{2}} \right)\left( {\frac{{\sqrt{3}}}{2}} \right)\\&=\frac{{\sqrt{2}+\sqrt{6}}}{4}\end{align}\)

\(\displaystyle \begin{align}\sec \left( {-\frac{\pi }{{12}}} \right)&=\frac{1}{{\cos \left( {-\frac{\pi }{{12}}} \right)}}=\frac{4}{{\sqrt{2}+\sqrt{6}}}\left( {\frac{{\sqrt{2}-\sqrt{6}}}{{\sqrt{2}-\sqrt{6}}}} \right)\\&=\sqrt{6}-\sqrt{2}\end{align}\)

 

Find the exact value of \(\displaystyle \cot \left( {-\frac{{5\pi }}{{12}}} \right)\)

 

Divide up \(\displaystyle -\frac{{5\pi }}{{12}}\) into two fractions that can be found on the Unit Circle; use tan first, and then take the reciprocal:

\(\displaystyle \begin{align}\tan \left( {-\frac{{5\pi }}{{12}}} \right)&=\tan \left( {-\frac{{2\pi }}{{12}}+-\frac{{3\pi }}{{12}}} \right)=\tan \left( {-\frac{\pi }{6}+-\frac{\pi }{4}} \right)\\&=\frac{{\tan \left( {-\frac{\pi }{6}} \right)+\tan \left( {-\frac{\pi }{4}} \right)}}{{1-\tan \left( {-\frac{\pi }{6}} \right)\tan \left( {-\frac{\pi }{4}} \right)}}=\frac{{\left( {\frac{{-1}}{{\sqrt{3}}}} \right)+\left( {-1} \right)}}{{1-\left( {\frac{{-1}}{{\sqrt{3}}}} \right)\left( {-1} \right)}}\\&=\frac{{-\frac{1}{{\sqrt{3}}}+-\frac{{\sqrt{3}}}{{\sqrt{3}}}}}{{\frac{{\sqrt{3}}}{{\sqrt{3}}}-\frac{1}{{\sqrt{3}}}}}=\frac{{-\sqrt{3}-1}}{{\sqrt{3}-1}}\end{align}\)

 

\(\displaystyle \cot \left( {-\frac{{5\pi }}{{12}}} \right)=\frac{1}{{\tan \left( {-\frac{{5\pi }}{{12}}} \right)}}=\frac{{1-\sqrt{3}}}{{1+\sqrt{3}}}\)

And let’s do some sum and difference identity proofs. The last two are quite tricky!

Sum and Difference Identity Proofs
                  \(\displaystyle \cos \left( {x+\pi } \right)=-\cos x\)

 

\(\displaystyle \begin{align}\cos x\cos \pi -\sin x\sin \pi &=\\\left( {\cos x} \right)\left( {-1} \right)-\left( {\sin x} \right)\left( 0 \right)&=\\\left( {\cos x} \right)\left( {-1} \right)-0&=-\cos x\,\,\,\,\,\,\surd \end{align}\)

 

Note:  Evaluate any expressions that turn into constants (like \(\sin \pi \)).

                                                     \(\displaystyle \sin \left( {\alpha +\beta } \right)\sin \left( {\alpha -\beta } \right)={{\sin }^{2}}\alpha -{{\sin }^{2}}\beta \)       

 

\(\require{cancel} \displaystyle \begin{align}\left( {\sin \alpha \cos \beta +\cos \alpha \sin \beta } \right)\left( {\sin \alpha \cos \beta -\cos \alpha \sin \beta } \right)&=\\{{\left( {\sin \alpha \cos \beta } \right)}^{2}}-{{\left( {\cos \alpha \sin \beta } \right)}^{2}}&=\\\left( {{{{\sin }}^{2}}\alpha } \right)\left( {{{{\cos }}^{2}}\beta } \right)-\left( {{{{\cos }}^{2}}\alpha } \right)\left( {{{{\sin }}^{2}}\beta } \right)&=\\\left( {{{{\sin }}^{2}}\alpha } \right)\left( {1-{{{\sin }}^{2}}\beta } \right)-\left( {1-{{{\sin }}^{2}}\alpha } \right)\left( {{{{\sin }}^{2}}\beta } \right)&=\\{{\sin }^{2}}\alpha -\cancel{{\left( {{{{\sin }}^{2}}\alpha } \right)\left( {{{{\sin }}^{2}}\beta } \right)}}-\\{{\sin }^{2}}\beta +\cancel{{\left( {{{{\sin }}^{2}}\alpha } \right)\left( {{{{\sin }}^{2}}\beta } \right)}}&={{\sin }^{2}}\alpha -{{\sin }^{2}}\beta \,\,\,\,\surd \end{align}\)

 

Note:  We knew to eventually turn everything into sin, since the right-hand side of the identity only had sin in it.

                \(\displaystyle \frac{{\sin \left( {x+y} \right)}}{{\sin x\cos y}}=\cot x\tan y+1\)

 

\(\displaystyle \begin{align}\frac{{\sin x\cos y+\cos x\sin y}}{{\sin x\cos y}}&=\\\frac{{\sin x\cos y}}{{\sin x\cos y}}+\frac{{\cos x\sin y}}{{\sin x\cos y}}&=\\1+\frac{{\cos x}}{{\sin x}}\cdot \frac{{\sin y}}{{\cos y}}&=\cot x\tan y+1\,\,\,\,\,\surd \end{align}\)

 

Note:  We had to split the fraction on the left to get to the two terms on the right.

                                                                           \(\displaystyle \sec \left( {x-y} \right)=\frac{{\left( {\sec x} \right)\left( {\sec y} \right)}}{{\tan x\tan y+1}}\)

 

\(\displaystyle \begin{align}\frac{1}{{\cos \left( {x-y} \right)}}&=\\\frac{1}{{\cos x\cos y+\sin x\sin y}}&=\\\frac{1}{{\cos x\cos y+\sin x\sin y}}\cdot \frac{{\sec x\sec y}}{{\sec x\sec y}}&=\\\frac{{\sec x\sec y}}{{\cos x\cos y\cdot \sec x\sec y+\sin x\sin y\cdot \sec x\sec y}}&=\\\frac{{\sec x\sec y}}{{\cancel{{\cos x}}\cancel{{\cos y}}\cdot \frac{1}{{\cancel{{\cos x}}}}\cdot \frac{1}{{\cancel{{\cos y}}}}+\sin x\sin y\cdot \frac{1}{{\cos x}}\cdot \frac{1}{{\cos y}}}}&=\\\frac{{\sec x\sec y}}{{1+\tan x\tan y}}&=\frac{{\sec x\sec y}}{{\tan x\tan y+1}}\end{align}\)

 

Note:  We thought we’d try to multiply the top and bottom by \(\sec x\sec y\) to get the right-hand side numerator; it worked!

 

Solving with Sum and Difference Identities

Here are some problems where we use Sum and Difference identities to solve trig equations in the indicated interval:

Solving Sum and Difference Identity Trig Equations

\(\displaystyle \tan 3x-\tan 63{}^\circ =\sqrt{3}+\sqrt{3}\left( {\tan 3x} \right)\left( {\tan 63{}^\circ } \right)\)

(over the reals)

 

\(\displaystyle \begin{align}\tan 3x-\tan 63{}^\circ &=\sqrt{3}\left( {1+\left( {\tan 3x} \right)\left( {\tan 63{}^\circ } \right)} \right)\\\frac{{\tan 3x-\tan 63{}^\circ }}{{1+\left( {\tan 3x} \right)\left( {\tan 63{}^\circ } \right)}}&=\sqrt{3}\\\tan \left( {3x-63{}^\circ } \right)&=\sqrt{3}\\\\3x-63{}^\circ &=60{}^\circ +180{}^\circ k\\3x&=123{}^\circ +180{}^\circ k\\x&=\frac{{123{}^\circ }}{3}+\frac{{180{}^\circ k}}{3}\end{align}\)

 

\(\displaystyle \left\{ {x|x=41{}^\circ +60{}^\circ k} \right\}\)

 

    \(\displaystyle 2\cos \left( {x+\frac{\pi }{6}} \right)\cos \left( {x-\frac{\pi }{6}} \right)=1\)

(interval \(\left[ {0,2\pi } \right)\))

 

\(\displaystyle \begin{align}\cos \left( {x+\frac{\pi }{6}} \right)\cos \left( {x-\frac{\pi }{6}} \right)&=\frac{1}{2}\\\left( {\cos x\cos \frac{\pi }{6}-\sin x\sin \frac{\pi }{6}} \right)\left( {\cos x\cos \frac{\pi }{6}+\sin x\sin \frac{\pi }{6}} \right)&=\frac{1}{2}\\\left( {\frac{{\sqrt{3}}}{2}\cos x-\frac{1}{2}\sin x} \right)\left( {\frac{{\sqrt{3}}}{2}\cos x+\frac{1}{2}\sin x} \right)&=\frac{1}{2}\\\frac{3}{4}{{\cos }^{2}}x-\frac{1}{4}{{\sin }^{2}}x&=\frac{1}{2}\\\frac{3}{4}\left( {1-{{{\sin }}^{2}}x} \right)-\frac{1}{4}{{\sin }^{2}}x&=\frac{1}{2}\\3-3{{\sin }^{2}}x-{{\sin }^{2}}x&=2\\-4{{\sin }^{2}}x&=-1\\{{\sin }^{2}}x&=\frac{1}{4}\\\sin x&=\pm \frac{1}{2}\end{align}\)

 

\(\displaystyle x=\left\{ {\frac{\pi }{6},\,\,\frac{{5\pi }}{6},\,\,\frac{{7\pi }}{6},\,\,\frac{{11\pi }}{6}} \right\}\)

Double Angle and Half Angle Identities

We use double angle and half angle identities the same way we used sum and difference identities when we need to split up the angle to make it easier to find the values (for example, to find values on the unit circle).

We also use the identities in conjunction with other identities to prove and solve trig problems.

Here are the double angle and half angle identities, and tricks to help you memorize them:

Double and Half Angle Identities Hints
\(\begin{align}\sin \left( {2A} \right)&=2\sin A\cos A\\\\\cos \left( {2A} \right)&={{\cos }^{2}}A-{{\sin }^{2}}A\\&=1-2{{\sin }^{2}}A\\&=2{{\cos }^{2}}A-1\\\\\tan \left( {2A} \right)&=\frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}\end{align}\) \(\sin \left( {2A} \right)\) is pretty easy to remember; just take the “2” out and put it in front of a sin and cos.

 

\(\cos \left( {2A} \right)\) is a little more complicated, especially since it can be written in three ways. I memorize the \({{\cos }^{2}}A-{{\sin }^{2}}A\) part  and then remember that I can use the Pythagorean Identity \({{\sin }^{2}}A+{{\cos }^{2}}A=1\) to substitute and do the algebra to arrive at the other expressions.

 

For \(\tan \left( {2A} \right)\), the identity only has tan’s in it, with the “2” in the numerator, and the \(1-{{\tan }^{2}}A\) in the denominator.

\(\begin{align}\sin \left( {\frac{A}{2}} \right)&=\pm \sqrt{{\frac{{1-\cos A}}{2}}}\\\\\\\cos \left( {\frac{A}{2}} \right)&=\pm \sqrt{{\frac{{1+\cos A}}{2}}}\\\\\\\tan \left( {\frac{A}{2}} \right)&=\frac{{\sin A}}{{1+\cos A}}=\frac{{1-\cos A}}{{\sin A}}\end{align}\) These are a little more complicated. For the \(\displaystyle \sin \left( {\frac{A}{2}} \right)\) and \(\displaystyle \cos \left( {\frac{A}{2}} \right)\), they both contain a cos, and the cos half angle has the plus sign inside the radical.

 

The tricky thing is that the sign of the whole identity depends on the quadrant where the angle \(\displaystyle \boldsymbol{\frac{A}{2}}\) is (not angle \(A\), but \(\displaystyle \frac{A}{2}\)). For example, for \(\displaystyle \sin \left( {\frac{A}{2}} \right)\), if \(\displaystyle \frac{A}{2}\) is in the first or second quadrants, use \(\displaystyle +\sqrt{{\frac{{1-\cos A}}{2}}}\), and if \(\displaystyle \frac{A}{2}\) is in the third or fourth quadrants, use \(\displaystyle -\sqrt{{\frac{{1-\cos A}}{2}}}\). Similarly for \(\displaystyle \cos \left( {\frac{A}{2}} \right)\), if \(\displaystyle \frac{A}{2}\) is in the first or fourth quadrants, use \(\displaystyle +\sqrt{{\frac{{1+\cos A}}{2}}}\), and if \(\displaystyle \frac{A}{2}\) is in the second or third quadrants, use \(\displaystyle -\sqrt{{\frac{{1+\cos A}}{2}}}\).

 

For \(\displaystyle \tan \left( {\frac{A}{2}} \right)\), you’ll want to memorize both parts of the identity, but you can derive the second part by multiplying the first by \(\displaystyle \frac{{1-\cos A}}{{1-\cos A}}\) (conjugate of denominator).

Let’s do some double and half angle identity proofs. Notice how we always try to start on the more complicated side.

Double and Half Angle Identity Proofs

\(\displaystyle \frac{{\sin \left( {2x} \right)}}{2}=\frac{{\tan x}}{{{{{\tan }}^{2}}x+1}}\)                

      

\(\require{cancel} \displaystyle \begin{align}&=\frac{{\tan x}}{{{{{\sec }}^{2}}x}}\\&=\frac{{\sin x}}{{\cancel{{\cos x}}}}\cdot \frac{{{{{\cancel{{{{{\cos }}^{2}}x}}}}^{{\cos x}}}}}{1}\\&=\sin x\cos x\\&=\sin x\cos x\cdot \frac{2}{2}\\\frac{{\sin \left( {2x} \right)}}{2}&=\frac{{2\sin x\cos x}}{2}\,\,\,\,\,\surd \end{align}\)

 

Note:  To get \(\displaystyle \sin x\cos x\) into the right form for the identity \(\displaystyle (2\sin x\cos x)\), we had to multiply by \(\displaystyle \frac{2}{2}\).

 \(\displaystyle \frac{{1-\cos \left( {2A} \right)}}{{\sin \left( {2A} \right)}}=\tan A\)

 

\(\displaystyle \begin{align}\frac{{1-\left( {1-2{{{\sin }}^{2}}A} \right)}}{{\sin \left( {2A} \right)}}&=\\\frac{{2{{{\sin }}^{2}}A}}{{\sin \left( {2A} \right)}}&=\\\frac{{{}^{{\sin A}}\cancel{{2{{{\sin }}^{2}}A}}}}{{\cancel{{2\sin A}}\cos A}}&=\tan A\,\,\,\,\,\surd \end{align}\)

 

Note:  We knew to use the \(\displaystyle 1-2{{\sin }^{2}}A\) (instead of \(\displaystyle 2{{\cos }^{2}}A-1\)) for \(\cos 2A\) to get rid of the 1’s, and also because we have sin in the denominator instead of cos.

  \(\displaystyle \tan \left( {\frac{1}{2}\theta } \right)=\csc \theta -\cot \theta \)      

 

\(\displaystyle \begin{align}&=\frac{1}{{\sin \theta }}-\frac{{\cos \theta }}{{\sin \theta }}\\\tan \left( {\frac{1}{2}\theta } \right)&=\frac{{1-\cos \theta }}{{\sin \theta }}\,\,\,\,\,\surd \end{align}\)

 

Note:  Start from the right side, and turn everything into sin and cos since the Half Angle tan identity is written in terms of sin and cos.

                       \(\displaystyle \frac{{\cos \left( {2\theta } \right)}}{{\cos \theta -\sin \theta }}=\cos \theta +\sin \theta \)

 

\(\displaystyle \begin{align}\frac{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}{{\cos \theta -\sin \theta }}&=\\\frac{{\cancel{{\left( {\cos \theta -\sin \theta } \right)}}\left( {\cos \theta +\sin \theta } \right)}}{{\cancel{{\cos \theta -\sin \theta }}}}&=\cos \theta +\sin \theta \,\,\,\,\,\surd \end{align}\)

 

Note:  We knew to use \(\displaystyle {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \) and difference of squares  for \(\cos 2\theta \) since the denominator contains both cos and sin.

 \(\displaystyle \frac{{1+\sin \left( {2\theta } \right)}}{{\cos \left( {2\theta } \right)}}=\frac{{\cos \theta +\sin \theta }}{{\cos \theta -\sin \theta }}\)                            

 

\(\displaystyle \begin{align}&=\frac{{\cos \theta +\sin \theta }}{{\cos \theta -\sin \theta }}\cdot \frac{{\cos \theta +\sin \theta }}{{\cos \theta +\sin \theta }}\\&=\frac{{{{{\cos }}^{2}}\theta +2\sin \theta \cos \theta +{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}\\&=\frac{{1+2\sin \theta \cos \theta }}{{\cos \left( {2\theta } \right)}}\\\frac{{1+\sin \left( {2\theta } \right)}}{{\cos \left( {2\theta } \right)}}&=\frac{{1+\sin \left( {2\theta } \right)}}{{\cos \left( {2\theta } \right)}}\,\,\,\,\,\,\surd \end{align}\)

 

Note:  The right-hand side looked a little more complicated so we started there. We multiplied by 1, using the conjugate of the bottom; this got us the double angle identity for \(\displaystyle \cos \left( {2\theta } \right)\) in the denominator. We also used the Pythagorean Identity \({{\sin }^{2}}x+{{\cos }^{2}}x=1\).

         \(\displaystyle \frac{{\cos \left( {2x} \right)+\cos x}}{{\sin \left( {2x} \right)-\sin x}}=\frac{{\sin x}}{{1-\cos x}}\)

 

\(\displaystyle \begin{align}\frac{{2{{{\cos }}^{2}}x-1+\cos x}}{{2\sin x\cos x-\sin x}}&=\\\frac{{\cancel{{\left( {2\cos x-1} \right)}}\left( {\cos x+1} \right)}}{{\sin x\cancel{{\left( {2\cos x-1} \right)}}}}&=\\\frac{{\cos x+1}}{{\sin x}}\cdot \frac{{\cos x-1}}{{\cos x-1}}&=\\\frac{{{{{\cos }}^{2}}x-1}}{{\sin x\left( {\cos x-1} \right)}}&=\\\frac{{{{{\cancel{{-{{{\sin }}^{2}}x}}}}^{{-\sin x}}}}}{{\cancel{{\sin x}}\left( {\cos x-1} \right)}}&=\frac{{\sin x}}{{1-\cos x}}\,\,\,\,\,\,\surd \end{align}\)

 

Note:  We knew to use the \(\displaystyle 2{{\cos }^{2}}A-1\) for \(\cos 2A\) because of the cos in the numerator; then we could factor. We multiplied by 1, using the conjugate of the numerator, and had to use the Pythagorean identity to cancel out the sin’s. Tricky!

 

Now let’s use these identities to find exact values for the following expressions using triangles, similar to what we did here in the in The Inverse Trigonometric Functions section:

Double and Half Angle “Triangle” Identity Problems
Find an expression for \(\displaystyle \cos \left( {\frac{A}{2}} \right)\), given \(\displaystyle \sin A=-\frac{3}{5}\) and \(\displaystyle\frac{{3\pi }}{2}<A<2\pi \):   

 

Since \(\displaystyle \cos \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1+\cos A}}{2}}}\), we need to get \(\cos A\), and then find the quadrant of \(\displaystyle \cos \left( {\frac{A}{2}} \right)\) to get the correct sign.

From the triangle, \(\displaystyle \pm \sqrt{{\frac{{1+\cos A}}{2}}}=\pm \sqrt{{\frac{{1+\frac{4}{5}}}{2}}}=\pm \sqrt{{\frac{9}{{10}}}}\).

But since angle A is between \(\displaystyle \frac{{3\pi }}{2}\) and \(\displaystyle 2\pi \), angle \(\displaystyle \frac{A}{2}\) will be between \(\displaystyle \frac{{3\pi }}{4}\) and \(\displaystyle \pi \), which is in the 2nd quadrant (you have to think of these positive angles going counter-clockwise from 0, or the positive \(x\)-axis). In the 2nd quadrant, cos is negative. So, \(\displaystyle \cos \left( {\frac{A}{2}} \right)=-\sqrt{{\frac{9}{{10}}}}\).

Find an expression for \(\cot \left( {2A} \right)\), given \(\displaystyle \tan A=\frac{2}{7}\) and \(\displaystyle \pi <A<\frac{{3\pi }}{2}\):

 

 

Since \(\displaystyle \tan \left( {2A} \right)=\frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}\), \(\displaystyle \cot \left( {2A} \right)=\frac{1}{{\frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}}}=\frac{{1-{{{\tan }}^{2}}A}}{{2\tan A}}\) (the reciprocal). We can just use \(\tan A\) and plug in.

Note that tan is positive in the 3rd quadrant:   \(\displaystyle \cot \left( {2A} \right)=\frac{{1-{{{\left( {\frac{2}{7}} \right)}}^{2}}}}{{2\left( {\frac{2}{7}} \right)}}=\frac{{45}}{{28}}\).

Find an expression for \(\displaystyle \sin \left( {\frac{A}{2}} \right)\), given \(\displaystyle \tan A=-\frac{1}{5}\) and \(\displaystyle \frac{\pi }{2}<A<\pi \):

 

 

Since \(\displaystyle \sin \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1-\cos A}}{2}}}\), we need to get \(\cos A\), and then find the quadrant of \(\displaystyle \sin \left( {\frac{A}{2}} \right)\) to get the correct sign.

From the triangle, \(\displaystyle \pm \sqrt{{\frac{{1-\cos A}}{2}}}=\pm \sqrt{{\frac{{1-\frac{{-5}}{{\sqrt{{26}}}}}}{2}}}=\pm \sqrt{{\frac{{\frac{{\sqrt{{26}}}}{{\sqrt{{26}}}}-\frac{{-5}}{{\sqrt{{26}}}}}}{2}}}=\pm \sqrt{{\frac{{\sqrt{{26}}+5}}{{2\sqrt{{26}}}}}}\). Since angle A is between \(\displaystyle \frac{\pi }{2}\) and \(\displaystyle \pi \), angle \(\displaystyle \frac{A}{2}\) will be between 0 and \(\displaystyle \frac{\pi }{2}\), which is in the 1st quadrant (you have to think of these positive angles going counter-clockwise from 0, or the positive \(x\)-axis). So, \(\displaystyle \sin \left( {\frac{A}{2}} \right)=\,\,\sqrt{{\frac{{\sqrt{{26}}+5}}{{2\sqrt{{26}}}}}}\). We can simplify this to \(\displaystyle \sqrt{{\frac{{\sqrt{{26}}+5}}{{2\sqrt{{26}}}}}}\cdot \frac{{\sqrt{{26}}}}{{\sqrt{{26}}}}=\sqrt{{\frac{{26+5\sqrt{{26}}}}{{52}}}}\).

 

Solving with Double and Half Angle Identities

Here are some problems where we have use Double and Half Angle identities to solve trig equations in the indicated interval:

Solving Trig Equations with Double and Half Angle Identities

   \(\displaystyle {{\cos }^{2}}\left( {\frac{x}{2}} \right)=\cos x+1\) over the reals

 

\(\displaystyle \begin{align}{{\cos }^{2}}\left( {\frac{x}{2}} \right)-\cos x-1&=0\\{{\left( {\pm \sqrt{{\frac{{1+\cos x}}{2}}}} \right)}^{2}}-\cos x-1&=0\\\frac{{1+\cos x}}{2}-\cos x-1&=0\\1+\cos x-2\cos x-2&=0\\\cos x&=-1\end{align}\)

 

\(\displaystyle \left\{ {x|x=\pi +2\pi k} \right\}\)

 \(\displaystyle \frac{{1+\cos \theta }}{{\sin x}}+1=0\)   interval \(\left[ {0,2\pi } \right)\)

 

$\(\displaystyle \begin{align}\frac{1}{{\left( {\frac{{\sin \theta }}{{1+\cos \theta }}} \right)}}&=-1\\\frac{1}{{\tan \left( {\frac{\theta }{2}} \right)}}&=-1\\\cot \left( {\frac{\theta }{2}} \right)&=-1\end{align}\)

\(\displaystyle \frac{\theta }{2}=\frac{{3\pi }}{4}+\pi k\,\,\,\,\,\,\,\,\theta =\frac{{3\pi }}{2}+2\pi k\)

\(\displaystyle \theta =\left\{ {\frac{{3\pi }}{2}} \right\}\)

(Remember from Trig Solving of Multiple Angles, if the coefficient in the trig argument isn’t 1 – we have \(\displaystyle \frac{1}{2}\) – we have to check for reals first, and then narrow down to \(\left[ {0,2\pi } \right)\) interval.)

   \(\displaystyle \cos \left( {2A} \right)+\cos A=0\) over the reals

 

\(\displaystyle \begin{align}2{{\cos }^{2}}A-1+\cos A&=0\\2{{\cos }^{2}}A+\cos A-1&=0\\\left( {2\cos A-1} \right)\left( {\cos A+1} \right)&=0\end{align}\)

\(\displaystyle \cos A=\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\cos A=-1\)

\(\displaystyle \left\{ {A|A=\frac{\pi }{3}+2\pi k,\,\,\,\,\frac{{5\pi }}{3}+2\pi k,\,\,\pi +2\pi k} \right\}\)

   \(\displaystyle \sin \left( {2x} \right)-\cos x=0\)   interval \(\left[ {0,2\pi } \right)\)

 

\(\displaystyle \begin{align}2\sin x\cos x-\cos x&=0\\\cos x\left( {2\sin x-1} \right)&=0\end{align}\)

\(\displaystyle \cos x=0\,\,\,\,\,\,\,\,\,\,\,\,\,\sin x=\frac{1}{2}\)

\(\displaystyle x=\left\{ {\frac{\pi }{2},\,\,\frac{{3\pi }}{2},\,\,\frac{\pi }{6},\,\,\frac{{5\pi }}{6}} \right\}\)

    \(\displaystyle \tan \left( {\frac{A}{2}} \right)\sin A=1-\cos A\) over the reals

 

\(\require{cancel} \displaystyle \begin{align}\left( {\frac{{1-\cos A}}{{\cancel{{\sin A}}}}} \right)\cancel{{\sin A}}&=1-\cos A\\1-\cos A&=1-\cos A\end{align}\)

 

This would appear to be all real numbers, but we have to restrict the answers by the asymptotes of tan, which are \(\displaystyle \frac{\pi }{2}+\pi k\) for the parent function. Since we have \(\displaystyle \tan \left( {\frac{A}{2}} \right)\), the asymptotes are \(\pi +2\pi k\). So, the answer is \(\displaystyle \left\{ {A|A\ne \pi +2\pi k} \right\}\). Tricky!

    \(\displaystyle 4\sin x\cos x=1\)   interval \(\left[ {0,2\pi } \right)\)

 

\(\displaystyle \begin{array}{c}2\left( {2\sin x\cos x} \right)=1\\2\left( {\sin \left( {2x} \right)} \right)=1\end{array}\)

\(\displaystyle \sin \left( {2x} \right)=\frac{1}{2}\)

\(\displaystyle \,2x=\frac{\pi }{6}+2\pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac{{5\pi }}{6}+2\pi k\)

\(\displaystyle x=\frac{\pi }{{12}}+\pi k\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{5\pi }}{{12}}+\pi k\)

In Unit Circle, this will be \(\displaystyle x=\left\{ {\frac{\pi }{{12}},\,\,\frac{{5\pi }}{{12}},\,\,\frac{{13\pi }}{{12}},\,\,\frac{{17\pi }}{{12}}} \right\}\).

Trig identity Summary and Mixed Identity Proofs

Now let’s put it all together.  First, here is a table with all the identities we’ve talked about:

Reciprocal and Quotient Identities

\(\displaystyle \sin \theta =\frac{1}{{\csc \theta }}\,\,\,\,\,\,\,\,\,\,\,\csc \theta =\frac{1}{{\sin \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac{1}{{\sec \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \theta =\frac{1}{{\cos \theta }}\,\)

\(\displaystyle \tan \theta =\frac{1}{{\cot \theta }}=\frac{{\sin \theta }}{{\cos \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cot \theta =\frac{1}{{\tan \theta }}=\frac{{\cos \theta }}{{\sin \theta }}\)

Pythagorean Identities

\(\begin{array}{l}\sin \left( {A+B} \right)=\sin A\cos B+\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {A+B} \right)=\cos A\cos B-\sin A\sin B\\\sin \left( {A-B} \right)=\sin A\cos B-\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {A-B} \right)=\cos A\cos B+\sin A\sin B\end{array}\)

 

\(\displaystyle {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\cot }^{2}}\theta +1={{\csc }^{2}}\theta \)

Sum and Difference Identities

\(\displaystyle \tan \left( {A+B} \right)=\frac{{\tan A+\tan B}}{{1-\tan A\tan B}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {A-B} \right)=\frac{{\tan A-\tan B}}{{1+\tan A\tan B}}\)

Double Angle Identities

\(\sin \left( {2A} \right)=2\sin A\cos A\)          \(\begin{align}\cos \left( {2A} \right)&={{\cos }^{2}}A-{{\sin }^{2}}A\\&=1-2{{\sin }^{2}}A\\&=2{{\cos }^{2}}A-1\end{align}\)               \(\displaystyle \tan \left( {2A} \right)=\frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}\)

Half Angle Identities

    \(\displaystyle \sin \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1-\cos A}}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( {\frac{A}{2}} \right)=\pm \sqrt{{\frac{{1+\cos A}}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {\frac{A}{2}} \right)=\frac{{\sin A}}{{1+\cos A}}=\frac{{1-\cos A}}{{\sin A}}\)

 

Here are a set of “hints” that might help you prove and solve trig identity problems:

  • Start with the more complicated side. If you absolutely can’t get to the other side, go down both sides, see where the two sides are identical and then move up one of the sides. Some teachers will let you work down both sides until the two sides match up.
  • Turn everything into sin and cos, for example if you have tan or reciprocal functions that you can simplify.
  • Match trig functions (like tan) to what’s on the other side. For example, if you have a \({{\tan }^{2}}\) on one side, and a \(\displaystyle \frac{{\sin }}{{\cos }}\) on the other, change the \(\displaystyle \frac{{\sin }}{{\cos }}\) to a tan.
  • Look at the other side of the identity to see what direction to go on the more complicated side. For example, if there is one term on the right side, strive for one term on the left.
  • Use Pythagorean Identities when you see you can cancel something out (like a “1”) or you see a trig function that is squared that you can eliminate.
  • Find common denominators if the number of terms don’t match on each side. For example, if you have two terms on the left side and only one term on the right side, find the common denominator and add the two terms on the left side so they become one. If you have two terms on both sides, for example, you may want to leave them alone. You may also need to “break apart” terms when there is more than one term in the numerator (using the same denominator), for example,  \(\displaystyle \frac{{x+2}}{x}=\frac{x}{x}+\frac{2}{x}=1+\frac{2}{x}\). It’s a good idea to simplify fractions (for example, using reciprocal identities) before finding common denominators and adding or subtracting fractions.
  • Divide numerator and denominator by something that makes the term simplified, for example, if you have a difference of squares in the numerator, divide numerator and denominator by one of these factors.
  • Cross out (simplify) anything you can earlier rather than later.
  • For \(\cos \left( {2A} \right)\), to know which version of identity to use, check to see if there’s a “\(-1\)” or a “\(+1\)” on same side; you’ll probably want to cancel these out. For example, use \(1-2{{\sin }^{2}}x\)  if there’s a “\(-1\)” following, and use \(2{{\cos }^{2}}x-1\)  if there’s a  “\(+1\)” following. If there’s a  “\(-\,\,{{\cos }^{2}}x\)”, or a  “\(+\,\,{{\sin }^{2}}x\)”  following, you may want to use  \({{\cos }^{2}}x-{{\sin }^{2}}x\), to be able to simplify.
  • Watch for difference of squares, such as  \(\left( {\cos x-1} \right)\left( {\cos x+1} \right)={{\cos }^{2}}x-1\).
  • With quadratics, get everything to one side and try to factor.
  • Multiply by conjugates, usually in denominators, but sometimes in numerators, to get difference of squares. (Multiply by 1, with the conjugate in both the numerator and denominator).
  • Factor out Greatest Common Factors (GCFs) if can.
  • Again, if there’s a random “\(+1\)” or “\(-1\)” that you need to get rid of, find an identity that, when added to it, eliminates it.
  • When solving, simplify with identities first, if you can.
  • When solving, you can square each side, or multiply both sides by what’s in a denominator, but check for extraneous solutions (denominators can’t be 0).
  • When solving, if you get answers for any trig function that has asymptotes (like tan), check for extraneous solutions (solutions that would be asymptotes). Also check for solutions that make a denominator 0; these should be eliminated.

Here are some Mixed Identity Proof problems:

Trig Identity Proofs Hints
                   \(\displaystyle \frac{{{{{\cos }}^{4}}\theta -{{{\sin }}^{4}}\theta }}{{{{{\left( {\cos \theta +\sin \theta } \right)}}^{2}}}}=\frac{{\cos 2\theta }}{{1+\sin 2\theta }}\)

 

\(\displaystyle \begin{align}\frac{{\left( {{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta } \right)\left( {{{{\cos }}^{2}}\theta +{{{\sin }}^{2}}\theta } \right)}}{{{{{\cos }}^{2}}\theta +2\cos \theta \sin \theta +{{{\sin }}^{2}}\theta }}&=\\\frac{{\left( {{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta } \right)\left( 1 \right)}}{{2\cos \theta \sin \theta +1}}&=\frac{{\cos 2\theta }}{{1+\sin 2\theta }}\,\,\,\,\,\,\,\,\,\,\surd \end{align}\)

  • Factor difference of squares in numerator
  • Multiply binomial in denominator
  • Use Pythagorean Identities to simplify
  • Use Double Angle Identities
 

                       \(\displaystyle \frac{1}{{1-\sin \theta }}=\sec \theta \tan \theta +{{\sec }^{2}}\theta \,\,\)

 

\(\displaystyle \begin{align}\left( {\frac{1}{{1-\sin \theta }}} \right)\left( {\frac{{1+\sin \theta }}{{1+\sin \theta }}} \right)&=\\\frac{{1+\sin \theta }}{{1-{{{\sin }}^{2}}\theta }}&=\\\frac{{1+\sin \theta }}{{{{{\cos }}^{2}}\theta }}&=\\\frac{1}{{{{{\cos }}^{2}}\theta }}+\frac{{\sin \theta }}{{{{{\cos }}^{2}}\theta }}&=\\{{\sec }^{2}}\theta +\tan \theta \left( {\frac{1}{{\cos \theta }}} \right)&=\sec \theta \tan \theta +{{\sec }^{2}}\theta \,\,\,\,\,\,\,\surd \end{align}\)

  • Multiply by 1 using conjugate of the denominator
  • Use Pythagorean Identity to simplify
  • Separate fraction into two terms, since there are two terms on the other side
  • Use Reciprocal and Quotient Identities to simplify
\(\displaystyle \frac{{1+\sin \left( {2y} \right)}}{{\cos \left( {2y} \right)}}=\frac{{\cot y+1}}{{\cot y-1}}\)                                             

 

\(\displaystyle \begin{align}&=\frac{{\left( {\frac{{\cos y}}{{\sin y}}+1} \right)}}{{\left( {\frac{{\cos y}}{{\sin y}}-1} \right)}}\\&=\frac{{\left( {\frac{{\cos y+\sin y}}{{\sin y}}} \right)}}{{\left( {\frac{{\cos y-\sin y}}{{\sin y}}} \right)}}\\&=\frac{{\cos y+\sin y}}{{\cos y-\sin y}}\\&=\frac{{\cos y+\sin y}}{{\cos y-\sin y}}\left( {\frac{{\cos y+\sin y}}{{\cos y+\sin y}}} \right)\\\frac{{1+\sin \left( {2y} \right)}}{{\cos \left( {2y} \right)}}&=\frac{{{{{\cos }}^{2}}y+2\sin y\cos y+{{{\sin }}^{2}}y}}{{{{{\cos }}^{2}}y-{{{\sin }}^{2}}y}}\,\,\,\,\,\,\,\,\,\surd \end{align}\)

  • Even though the left side looks more complicated, when I tried it, it didn’t go anywhere
  • It turns out the right side can be simplified more easily; turn everything into sin and cos
  • Find common denominators for the fractions, and simplify them
  • Since we can use a \({{\cos }^{2}}y-{{\sin }^{2}}y\) on the bottom for the \(\cos \left( {2y} \right)\) on the left, multiply by 1 using conjugate of the denominator
  • Use a Pythagorean Identity, and then Double Angle Identities to simplify

Here are more Mixed Identity Proofs:

Trig Identity Proofs Hints
\(\displaystyle \frac{{\cos 2x+\cos x-20}}{{\sin 2x-6\sin x}}=\cot x+\frac{7}{2}\csc x\)

 

\(\require{cancel} \displaystyle \begin{align}\frac{{2{{{\cos }}^{2}}x-1+\cos x-20}}{{2\cos x\sin x-6\sin x}}&=\\\frac{{2\cos {{x}^{2}}+\cos x-21}}{{2\sin x\left( {\cos x-3} \right)}}&=\\\frac{{\left( {2\cos x+7} \right)\cancel{{\left( {\cos x-3} \right)}}}}{{2\sin x\cancel{{\left( {\cos x-3} \right)}}}}&=\\\frac{{\cancel{2}\cos x}}{{\cancel{2}\sin x}}+\frac{7}{{2\sin x}}&=\cot x+\frac{7}{2}\csc x\,\,\,\,\,\,\,\,\,\surd \end{align}\)

 

  • Use Double Angle identity to get everything in numerator to cos
  • Factor in numerator and denominator and we can simplify!
  • Separate fraction since we have two terms on the right-hand side
  • Use Reciprocal and Quotient Identities to simplify
\(\displaystyle \frac{{\tan A}}{{\sec A+1}}+\frac{{\sec A+1}}{{\tan A}}=\frac{2}{{\sin A}}\)

 

\(\begin{align}\frac{{\left( {\tan A} \right)\left( {\tan A} \right)}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}+\frac{{\left( {\sec A+1} \right)\left( {\sec A+1} \right)}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}&=\\\frac{{{{{\tan }}^{2}}A+{{{\sec }}^{2}}A+2\sec A+1}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}&=\\\frac{{{{{\sec }}^{2}}A+{{{\sec }}^{2}}A+2\sec A}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}&=\\\frac{{2{{{\sec }}^{2}}A+2\sec A}}{{\left( {\sec A+1} \right)\left( {\tan A} \right)}}&=\\\frac{{2\sec A\cancel{{\left( {\sec A+1} \right)}}}}{{\cancel{{\left( {\sec A+1} \right)}}\left( {\tan A} \right)}}&=\\\frac{{2\left( {\frac{1}{{\cos A}}} \right)}}{{\frac{{\sin A}}{{\cos A}}}}=\frac{2}{{\sin A}}\,\,\,\,\,\,\,\,\,\,\surd \end{align}\)

  • Find common denominators for the fractions, and combine them, since we only have one term on the right
  • Turn \({{\tan }^{2}}A+1\) into \({{\sec }^{2}}A\) and combine with other \({{\sec }^{2}}A\)
  • Factor the numerator and simplify
  • Use Reciprocal and Quotient Identities to simplify

And a couple more:

Trig Identity Proofs Hints
                         \(\displaystyle \frac{{1-{{{\tan }}^{2}}x}}{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}={{\sec }^{2}}x\)

 

\(\require{cancel} \displaystyle \begin{align}\frac{{1-\frac{{{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}}}{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}&=\\\frac{{\frac{{{{{\cos }}^{2}}x}}{{{{{\cos }}^{2}}x}}-\frac{{{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}}}{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}&=\\\frac{{\frac{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}{{{{{\cos }}^{2}}x}}}}{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}&=\\\frac{{\cancel{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}}}{{{{{\cos }}^{2}}x}}\cdot \frac{1}{{\cancel{{{{{\cos }}^{2}}x-{{{\sin }}^{2}}x}}}}&=\\\frac{1}{{{{{\cos }}^{2}}x}}&={{\sec }^{2}}x\,\,\,\,\,\,\,\surd \end{align}\)

  • Even though it looks like we should use the \({{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x\) identity, we don’t have a “\(2x\)” angle anywhere else, so it probably wouldn’t help
  • Turn everything into sin and cos and get common denominator to subtract terms in numerator
  • Divide fraction and use Reciprocal Identity
\(\displaystyle \frac{{1+{{{\tan }}^{2}}\theta }}{{1-{{{\tan }}^{2}}\theta }}=\sec \left( {2\theta } \right)\)   

 

\(\displaystyle \begin{align}\frac{{{{{\sec }}^{2}}\theta }}{{1-{{{\tan }}^{2}}\theta }}&=\\\frac{{\frac{1}{{{{{\cos }}^{2}}\theta }}}}{{\frac{{{{{\cos }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}-\frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}}}&=\\\frac{{\frac{1}{{{{{\cos }}^{2}}\theta }}}}{{\frac{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}}}&=\\\frac{1}{{{{{\cos }}^{2}}\theta -{{{\sin }}^{2}}\theta }}&=\\\frac{1}{{\cos \left( {2\theta } \right)}}&=\sec \left( {2\theta } \right)\,\,\,\,\,\,\surd \end{align}\)

  • Use a Pythagorean Identity, and then turn everything into sin and cos
  • Get common denominator to subtract terms in denominator
  • Divide fraction and use Double Angle and Reciprocal Identities

Understand these problems, and practice, practice, practice!


For Practice: Use the Mathway widget below to try a Trig Identity problem. Click on Submit (the blue arrow to the right of the problem) and click on Verify the Identity to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

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On to Law of Sines and Cosines, and Areas of Triangles  – you’re ready!