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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*}

Read on, and in this Concept you'll learn four different methods to help you prove identities. You'll be able to apply them to prove the above identity when you are finished.

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Educator.com 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*} can be factored to \begin{align*}\frac{2 (1+ \cos \theta)}{\sin \theta (1 + \cos \theta)} = 2 \csc \theta\end{align*} and in this situation, the factors cancel each other.

Example A

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

Solution: 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*}

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.

Example B

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

Solution: Use the Pythagorean Identity and its alternate form. Manipulate \begin{align*}\sin^2 \theta + \cos^2 \theta = 1\end{align*} to be \begin{align*}\sin^2 \theta = 1 - \cos^2 \theta\end{align*}. Also substitute \begin{align*}\csc^2 x\end{align*} for \begin{align*}1+ \cot^2 x\end{align*}, 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*}

Example C

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

Solution: 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*}, 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*}

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*}

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

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*} can be factored to \begin{align*}\frac{2 (1+ \cos \theta)}{\sin \theta (1 + \cos \theta)} = 2 \csc \theta\end{align*} and in this situation, the factors cancel each other.

Guided Practice

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

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

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


1. 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

2. 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!

3. 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*}

Concept Problem Solution

The original question was to prove that:

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

First remember the Pythagorean Identity:

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


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

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*}

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*}

Explore More

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*}

Answers for Explore More Problems

To view the Explore More 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:

Nov 11, 2015
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