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3.2: Proofs of Trigonometric Identities

Difficulty Level: At Grade Created by: CK-12
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What if your instructor gave you two trigonometric expressions and asked you to prove that they were true. Could you do this? For example, can you show that

\begin{align*}\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\end{align*}sin2θ=1cos2θ2

Trigonometric Identities

In Trigonometry you will see complex trigonometric expressions. Often, complex trigonometric expressions can be equivalent to less complex expressions. The process for showing two trigonometric expressions to be equivalent (regardless of the value of the angle) is known as validating or proving trigonometric identities.

There are several options a student can use when proving a trigonometric identity.

Option One: Often one of the steps for proving identities is to change each term into their sine and cosine equivalents.

Option Two: Use the Trigonometric Pythagorean Theorem and other Fundamental Identities.

Option Three: When working with identities where there are fractions- combine using algebraic techniques for adding expressions with unlike denominators.

Option Four: If possible, factor trigonometric expressions. For example, \begin{align*}\frac{2 + 2 \cos \theta}{\sin \theta(1+ \cos \theta)} = 2 \csc \theta\end{align*}2+2cosθsinθ(1+cosθ)=2cscθ can be factored to \begin{align*}\frac{2 (1+ \cos \theta)}{\sin \theta (1 + \cos \theta)} = 2 \csc \theta\end{align*}2(1+cosθ)sinθ(1+cosθ)=2cscθ and in this situation, the factors cancel each other.













Prove the identity: \begin{align*}\csc \theta \times \tan \theta = \sec \theta\end{align*}cscθ×tanθ=secθ

Reducing each side separately. It might be helpful to put a line down, through the equals sign. Because we are proving this identity, we don’t know if the two sides are equal, so wait until the end to include the equality.

\begin{align*}\begin{array}{c|c } \csc x \times \tan x & \sec x \\ \frac{1}{\sin x} \times \frac{\sin x}{\cos x} & \frac{1}{\cos x} \\ \frac{1}{\cancel{\sin x}} \times \frac{\cancel{\sin x}}{\cos x}& \frac{1}{\cos x} \\ \frac{1}{\cos x} & \frac{1}{\cos x} \end{array}\end{align*}cscx×tanx1sinx×sinxcosx1sinx×sinxcosx1cosxsecx1cosx1cosx1cosx

At the end we ended up with the same thing, so we know that this is a valid identity.

Notice when working with identities, unlike equations, conversions and mathematical operations are performed only on one side of the identity. In more complex identities sometimes both sides of the identity are simplified or expanded. The thought process for establishing identities is to view each side of the identity separately, and at the end to show that both sides do in fact transform into identical mathematical statements.

Prove the identity: \begin{align*}(1-\cos^2 x)(1+\cot^2 x) = 1\end{align*}(1cos2x)(1+cot2x)=1

Use the Pythagorean Identity and its alternate form. Manipulate \begin{align*}\sin^2 \theta + \cos^2 \theta = 1\end{align*}sin2θ+cos2θ=1 to be \begin{align*}\sin^2 \theta = 1 - \cos^2 \theta\end{align*}sin2θ=1cos2θ. Also substitute \begin{align*}\csc^2 x\end{align*}csc2x for \begin{align*}1+ \cot^2 x\end{align*}1+cot2x, then cross-cancel.

\begin{align*}\begin{array}{c|c } (1-\cos^2 x)(1+\cot^2 x) & 1 \\ \sin^2 x \cdot \csc^2 x & 1 \\ \sin^2 x \cdot \frac{1}{\sin^2 x}& 1 \\ 1 & 1 \end{array}\end{align*}(1cos2x)(1+cot2x)sin2xcsc2xsin2x1sin2x11111

Prove the identity: \begin{align*}\frac{\sin \theta}{1+ \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \csc \theta\end{align*}sinθ1+cosθ+1+cosθsinθ=2cscθ.

Combine the two fractions on the left side of the equation by finding the common denominator: \begin{align*}(1 + \cos \theta) \times \sin \theta\end{align*}(1+cosθ)×sinθ, and the change the right side into terms of sine.

\begin{align*}\begin{array}{c|c } \frac{\sin \theta}{1+ \cos \theta} + \frac{1+ \cos \theta}{\sin \theta} & 2 \csc \theta \\ \frac{\sin \theta}{\sin \theta} \cdot \frac{\sin \theta}{1+ \cos \theta}\ \ + \frac{1+ \cos \theta}{\sin \theta} \cdot \frac{1+ \cos \theta}{1+ \cos \theta} & 2 \csc \theta \\ \quad \ \frac{\sin^2 \theta + (1+ \cos \theta)^2}{\sin \theta (1 + \cos \theta)}& 2 \csc \theta \end{array}\end{align*}sinθ1+cosθ+1+cosθsinθsinθsinθsinθ1+cosθ  +1+cosθsinθ1+cosθ1+cosθ sin2θ+(1+cosθ)2sinθ(1+cosθ)2cscθ2cscθ2cscθ

Now, we need to apply another algebraic technique, FOIL. (FOIL is a memory device that describes the process for multiplying two binomials, meaning multiplying the First two terms, the Outer two terms, the Inner two terms, and then the Last two terms, and then summing the four products.) Always leave the denominator factored, because you might be able to cancel something out at the end.

\begin{align*}\begin{array}{c|c } \frac{\sin^2 \theta + 1 + 2 \cos \theta + \cos^2 \theta}{\sin \theta (1+ \cos \theta)}& 2 \csc \theta\end{array}\end{align*}sin2θ+1+2cosθ+cos2θsinθ(1+cosθ)2cscθ

Using the second option, substitute \begin{align*}\sin^2 \theta + \cos^2 \theta = 1\end{align*}sin2θ+cos2θ=1 and simplify.

\begin{align*}\begin{array}{c|c } \frac{1 + 1 + 2 \cos \theta}{\sin \theta(1+ \cos \theta)} & 2 \csc \theta \\ \frac{2 + 2 \cos \theta}{\sin \theta(1+ \cos \theta)} & 2 \csc \theta \\ \frac{2 (1+ \cos \theta)}{\sin \theta (1 + \cos \theta)}& 2 \csc \theta \\ \frac{2}{\sin \theta}& \frac{2}{\sin \theta} \end{array}\end{align*}1+1+2cosθsinθ(1+cosθ)2+2cosθsinθ(1+cosθ)2(1+cosθ)sinθ(1+cosθ)2sinθ2cscθ2cscθ2cscθ2sinθ

Option Four: If possible, factor trigonometric expressions. Actually procedure four was used in the above example: \begin{align*}\frac{2 + 2 \cos \theta}{\sin \theta(1+ \cos \theta)} = 2 \csc \theta\end{align*}2+2cosθsinθ(1+cosθ)=2cscθ can be factored to \begin{align*}\frac{2 (1+ \cos \theta)}{\sin \theta (1 + \cos \theta)} = 2 \csc \theta\end{align*}2(1+cosθ)sinθ(1+cosθ)=2cscθ and in this situation, the factors cancel each other.


Example 1

Earlier, you were asked to prove 

\begin{align*}\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\end{align*}sin2θ=1cos2θ2

First remember the Pythagorean Identity:

\begin{align*}\sin^2 \theta + \cos^2 \theta = 1\end{align*}sin2θ+cos2θ=1


\begin{align*}\sin^2 \theta = 1 - \cos^2 \theta\end{align*}sin2θ=1cos2θ

From the Double Angle Identities, we know that

\begin{align*} \cos 2\theta = \cos^2 \theta - \sin^2 \theta\\ \cos^2 \theta = \cos 2\theta + \sin^2 \theta\\ \end{align*}cos2θ=cos2θsin2θcos2θ=cos2θ+sin2θ

Substituting this into the above equation for \begin{align*}\sin^2\end{align*},

\begin{align*} \sin^2 \theta = 1 - (\cos 2\theta + \sin^2 \theta)\\ \sin^2 \theta = 1 - \cos 2\theta - \sin^2 \theta\\ 2\sin^2 \theta = 1 - \cos 2\theta\\ \sin^2 \theta = \frac{1 - \cos 2\theta}{2}\\ \end{align*}

Example 2

Prove the identity: \begin{align*}\sin x \tan x + \cos x = \sec x\end{align*}

Step 1: Change everything into sine and cosine

\begin{align*}\sin x \tan x + \cos x & = \sec x \\ \sin x \cdot \frac{\sin x}{\cos x} + \cos x & = \frac{1}{\cos x} \end{align*}

Step 2: Give everything a common denominator, \begin{align*}\cos x\end{align*}.

\begin{align*}\frac{\sin^2 x}{\cos x} + \frac{\cos^2 x}{\cos x} = \frac{1}{\cos x}\end{align*}

Step 3: Because the denominators are all the same, we can eliminate them.

\begin{align*}\sin^2x + \cos^2x = 1\end{align*}

We know this is true because it is the Trig Pythagorean Theorem


Example 3

Prove the identity: \begin{align*}\cos x - \cos x \sin^2 x = \cos^3 x\end{align*}

Step 1: Pull out a \begin{align*}\cos x\end{align*}

\begin{align*}\cos x - \cos x \sin^2x & = \cos^3x \\ \cos x (1 - \sin^2x) & = \cos^3 x\end{align*}

Step 2: We know \begin{align*}\sin^2x + \cos^2x = 1\end{align*}, so \begin{align*}\cos^2x = 1 - \sin^2x\end{align*} is also true, therefore \begin{align*}\cos x (\cos^2x) = \cos^3x\end{align*}. This, of course, is true, we are finished!


Example 4

Prove the identity: \begin{align*}\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2 \csc x\end{align*}

Step 1: Change everything in to sine and cosine and find a common denominator for left hand side.

\begin{align*}& \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2 \csc x \\ & \frac{\sin x}{1+\cos x} + \frac{1 + \cos x}{\sin x} = \frac{2}{\sin x} \leftarrow \text{LCD}:\ \sin x (1 + \cos x) \\ & \frac{\sin^2 x + (1 + \cos x)^2}{\sin x (1 + \cos x)}\end{align*}

Step 2: Working with the left side, FOIL and simplify.

\begin{align*}& \frac{\sin^2 x + 1 + 2 \cos x + \cos^2 x}{\sin x (1 + \cos x)} && \rightarrow \text{FOIL}\ (1 + \cos x)^2 \\ & \frac{\sin^2 x + \cos^2 x + 1 + 2 \cos x }{\sin x (1 + \cos x)} && \rightarrow \text{move}\ \cos^2 x \\ & \frac{1 + 1 + 2 \cos x}{\sin x (1 + \cos x)} && \rightarrow \sin^2 x + \cos^2 x = 1 \\ & \frac{2 + 2 \cos x}{\sin x (1 + \cos x)} && \rightarrow \text{add} \\ & \frac{2(1+ \cos x)}{\sin x (1 + \cos x)} && \rightarrow \text{factor out}\ 2 \\ & \frac{2}{\sin x} && \rightarrow \text{cancel}\ (1 + \cos x)\end{align*}


Use trigonometric identities to simplify each expression as much as possible.

  1. \begin{align*}\tan(x)\cos(x)\end{align*}
  2. \begin{align*}\cos(x)-\cos^3(x)\end{align*}
  3. \begin{align*}\frac{1-\cos^2(x)}{\sin(x)}\end{align*}
  4. \begin{align*}\cot(x)\sin(x)\end{align*}
  5. \begin{align*}\frac{1-\sin^2(x)}{\cos(x)}\end{align*}
  6. \begin{align*}\sin(x)\csc(x)\end{align*}
  7. \begin{align*}\tan(-x)\cot(x)\end{align*}
  8. \begin{align*}\frac{\sec^2(x)-\tan^2(x)}{\cos^2(x)+\sin^2(x)}\end{align*}

Prove each identity.

  1. \begin{align*}\tan(x)+\cot(x)=\sec(x)\csc(x)\end{align*}
  2. \begin{align*}\sin(x)=\frac{\sin^2(x)+\cos^2(x)}{\csc(x)}\end{align*}
  3. \begin{align*}\frac{1}{\sec(x)-1}+\frac{1}{\sec(x)+1}=2\cot(x)\csc(x)\end{align*}
  4. \begin{align*}(\cos(x))(\tan(x)+\sin(x)\cot(x))=\sin(x)+\cos^2(x)\end{align*}
  5. \begin{align*}\sin^4(x)-\cos^4(x)=\sin^2(x)-\cos^2(x)\end{align*}
  6. \begin{align*}\sin^2(x)\cos^3(x)=(\sin^2(x)-\sin^4(x))(\cos(x))\end{align*}
  7. \begin{align*}\frac{\sin(x)}{\csc(x)}=1-\frac{\cos(x)}{\sec(x)}\end{align*}

Review (Answers)

To see the Review answers, open this PDF file and look for section 3.2. 




FOIL is an acronym used to remember a technique for multiplying two binomials. You multiply the FIRST terms, OUTSIDE terms, INSIDE terms, and LAST terms and then combine any like terms.
Trigonometric Identity

Trigonometric Identity

A trigonometric identity is an equation that relates two or more trigonometric functions.

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Difficulty Level:
At Grade
Date Created:
Sep 26, 2012
Last Modified:
Mar 23, 2016
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