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Simplifying Trigonometric Expressions

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

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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 following identities.

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

Let's simplify the following expressions.

  1. \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*}secxsecx11cosx1cosx1

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

\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*}1cosx1cosx11cosx1cosxcosxcosx1cosx1cosxcosx

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*}1cosx1cosxcosx1cosx÷1cosxcosx1cosxcosx1cosx11cosx

  1. \begin{align*}\frac{\sin^4x- \cos^4x}{\sin^2x- \cos^2x}\end{align*}sin4xcos4xsin2xcos2x

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*}sin4xcos4xsin2xcos2x(sin2xcos2x)(sin2x+cos2x)(sin2xcos2x)sin2x+cos2x1

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

  1. \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*}


Example 1

Earlier, you were asked to simplify the trigonometric function \begin{align*}\cos\theta + \cos\theta(\tan^2\theta)\end{align*}.

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 4

\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*}


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. 

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