<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation
Our Terms of Use (click here to view) and Privacy Policy (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use and Privacy Policy.

Simplifying Trigonometric Expressions

Convert to sine/cosine and use basic trig identities to simplify.

Atoms Practice
Estimated8 minsto complete
%
Progress
Practice Simplifying Trigonometric Expressions
Practice
Progress
Estimated8 minsto complete
%
Practice Now
Simplifying Trigonometric Expressions

How could you write the trigonometric function \begin{align*}\cos\theta + \cos\theta(\tan^2\theta)\end{align*}cosθ+cosθ(tan2θ) more simply?

Simplifying Trigonometric Expressions

Now that you are more familiar with trig identities, we can use them to simplify expressions. Remember, that you can use any of the identities in the Introduction to Trig Identities concept. Here is a list of the identities again:

Reciprocal Identities: \begin{align*}\csc \theta=\frac{1}{\sin \theta}, \sec \theta=\frac{1}{\cos \theta},\end{align*}cscθ=1sinθ,secθ=1cosθ, and \begin{align*}\cot \theta=\frac{1}{\tan \theta}\end{align*}cotθ=1tanθ

Tangent and Cotangent Identities: \begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*}tanθ=sinθcosθ and \begin{align*}\cot \theta=\frac{\cos \theta}{\sin \theta}\end{align*}cotθ=cosθsinθ

Pythagorean Identities: \begin{align*}\sin^2 \theta+ \cos^2 \theta=1, 1+ \tan^2 \theta=\sec^2 \theta,\end{align*}sin2θ+cos2θ=1,1+tan2θ=sec2θ, and \begin{align*}1+ \cot^2 \theta=\csc^2 \theta\end{align*}1+cot2θ=csc2θ

Cofunction Identities: \begin{align*}\sin \left(\frac{\pi}{2} - \theta\right)=\cos \theta, \cos \left(\frac{\pi}{2} - \theta\right)=\sin \theta,\end{align*}sin(π2θ)=cosθ,cos(π2θ)=sinθ, and \begin{align*}\tan \left(\frac{\pi}{2} - \theta\right)=\cot \theta\end{align*}tan(π2θ)=cotθ

Negative Angle Identities: \begin{align*}\sin(- \theta)=- \sin \theta, \cos(- \theta)=\cos \theta,\end{align*}sin(θ)=sinθ,cos(θ)=cosθ, and \begin{align*}\tan(- \theta)=- \tan \theta\end{align*}tan(θ)=tanθ

Simplify the following expressions

Simplify \begin{align*}\frac{\sec x}{\sec x-1}\end{align*}secxsecx1.

When simplifying trigonometric expressions, one approach is to change everything into sine or cosine. First, we can change secant to cosine using the Reciprocal Identity.

\begin{align*}\frac{\sec x}{\sec x - 1} \rightarrow \frac{\frac{1}{\cos x}}{\frac{1}{\cos x}-1}\end{align*}

Now, combine the denominator into one fraction by multiplying 1 by \begin{align*}\frac{\cos x}{\cos x}\end{align*}.

\begin{align*}\frac{\frac{1}{\cos x}}{\frac{1}{\cos x}-1} \rightarrow \frac{\frac{1}{\cos x}}{\frac{1}{\cos x}- \frac{\cos x}{\cos x}} \rightarrow \frac{\frac{1}{\cos x}}{\frac{1- \cos x}{\cos x}}\end{align*}

Change this problem into a division problem and simplify.

\begin{align*}\frac{\frac{1}{\cos x}}{\frac{1-\cos x}{\cos x}} \rightarrow & \frac{1}{\cos x} \div \frac{1- \cos x}{\cos x} \\ & \frac{1}{\cancel{\cos x}} \cdot \frac{\cancel{\cos x}}{1- \cos x} \\ & \frac{1}{1- \cos x}\end{align*}

Simplify \begin{align*}\frac{\sin^4x- \cos^4x}{\sin^2x- \cos^2x}\end{align*}.

With this problem, we need to factor the numerator and denominator and see if anything cancels. The rules of factoring a quadratic and the special quadratic formulas can be used in this scenario.

\begin{align*}\frac{\sin^4x - \cos^4x}{\sin^2x-\cos^2x} \rightarrow \frac{\cancel{\left(\sin^2x-\cos^2x\right)} \left(\sin^2x+ \cos^2x\right)}{\cancel{\left(\sin^2x-\cos^2x\right)}} \rightarrow \sin^2x+\cos^2x \rightarrow 1\end{align*}

In the last step, we simplified to the left hand side of the Pythagorean Identity. Therefore, this expression simplifies to 1.

Simplify \begin{align*}\sec \theta \tan^2 \theta+\sec \theta\end{align*}.

First, pull out the GCF.

\begin{align*}\sec \theta \tan^2 \theta+ \sec \theta \rightarrow \sec \theta(\tan^2 \theta+1)\end{align*}

Now, \begin{align*}\tan^2 \theta+1=\sec^2 \theta\end{align*} from the Pythagorean Identities, so simplify further.

\begin{align*}\sec \theta(\tan^2 \theta+1) \rightarrow \sec \theta \cdot \sec^2 \theta \rightarrow \sec^3 \theta\end{align*}

Examples

Example 1

Earlier, you were asked how could you write the trigonometric function \begin{align*}\cos\theta + \cos\theta(\tan^2\theta)\end{align*} more simple. 

Notice that the terms in the expression \begin{align*}\cos\theta + \cos\theta(\tan^2\theta)\end{align*} have a common factor of \begin{align*}\cos\theta\end{align*}, so start by factoring this common term out.

\begin{align*}\cos\theta + \cos\theta(\tan^2\theta)\\ \cos\theta(1 + tan^2\theta)\end{align*}

Now, use the trigonometric identity \begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*}, substitute, and simplify.

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

Simplify the following trigonometric expressions.

Example 2

\begin{align*}\cos \left(\frac{\pi}{2} - x\right) \cot x\end{align*}

Use the Cotangent Identity and the Cofunction Identity \begin{align*}\cos \left(\frac{\pi}{2}- \theta \right)=\sin \theta\end{align*}.

\begin{align*}\cos \left(\frac{\pi}{2}-x\right) \cot x \rightarrow \cancel{\sin x} \cdot \frac{\cos x}{\cancel{\sin x}} \rightarrow \cos x\end{align*}

Example 3

\begin{align*}\frac{\sin \left(-x\right) \cos x}{\tan x}\end{align*}

Use the Negative Angle Identity and the Tangent Identity.

\begin{align*}\frac{\sin \left(-x\right) \cos x}{\tan x} \rightarrow \frac{- \sin x \cos x}{\frac{\sin x}{\cos x}} \rightarrow - \cancel{\sin x} \cos x \cdot \frac{\cos x}{\sin x} \rightarrow - \cos^2x\end{align*}

Example 3

\begin{align*}\frac{\cot x \cos x}{\tan \left(-x\right) \sin \left(\frac{\pi}{2}-x \right)}\end{align*}

In this problem, you will use several identities.

\begin{align*}\frac{\cot x \cos x}{\tan \left(-x\right) \sin \left(\frac{\pi}{2}-x\right)} \rightarrow \frac{\frac{\cos x}{\sin x} \cdot \cos x}{- \frac{\sin x}{\cancel{\cos x}} \cdot \cancel{\cos x}} \rightarrow \frac{\frac{\cos^2 x}{\sin x}}{- \sin x} \rightarrow \frac{\cos^2x}{\sin x} \cdot - \frac{1}{\sin x} \rightarrow - \frac{\cos^2x}{\sin^2 x} \rightarrow - \cot^2x\end{align*}

Review

Simplify the following expressions.

  1. \begin{align*}\cot x \sin x\end{align*}
  2. \begin{align*}\cos^2x \tan(-x)\end{align*}
  3. \begin{align*}\frac{\cos \left(-x\right)}{\sin \left(-x\right)}\end{align*}
  4. \begin{align*}\sec x \cos(-x)- \sin^2x\end{align*}
  5. \begin{align*}\sin x(1+ \cot^2x)\end{align*}
  6. \begin{align*}1- \sin^2 \left(\frac{\pi}{2} - x\right)\end{align*}
  7. \begin{align*}1- \cos^2 \left(\frac{\pi}{2}-x\right)\end{align*}
  8. \begin{align*}\frac{\tan \left(\frac{\pi}{2}-x\right) \sec x}{1- \csc^2 x}\end{align*}
  9. \begin{align*}\frac{\cos^2x \tan^2x-1}{\cos^2x}\end{align*}
  10. \begin{align*}\cot^2x+ \sin^2x+ \cos^2(-x)\end{align*}
  11. \begin{align*}\frac{\sec x \sin x+ \cos \left(\frac{\pi}{2}-x\right)}{1+ \cos x}\end{align*}
  12. \begin{align*}\frac{\cos \left(-x\right)}{1+ \sin \left(-x\right)}\end{align*}
  13. \begin{align*}\frac{\sin^2 \left(-x\right)}{\tan^2x}\end{align*}
  14. \begin{align*}\tan \left(\frac{\pi}{2}-x\right) \cot x- \csc^2 x\end{align*}
  15. \begin{align*}\frac{\csc x \left(1- \cos^2x \right)}{\sin x \cos x}\end{align*}

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.8. 

My Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / notes
Show More

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Simplifying Trigonometric Expressions.
Please wait...
Please wait...