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Trig Identities to Find Exact Trigonometric Values

Pythagorean, Tangent, and Reciprocal Identities used to find values of functions.

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Using Trig Identities to Find Exact Trig Values

You are given the following information about \begin{align*}\theta\end{align*}

\begin{align*}\sin \theta= \frac{2}{3}, \frac{\pi}{2} < \theta < \pi\end{align*}

What are \begin{align*}\cos \theta\end{align*} and \begin{align*}\tan \theta\end{align*}?

Trigonometric Identities

You can use the Pythagorean, Tangent and Reciprocal Identities to find all six trigonometric values for certain angles. Let’s walk through a few problems so that you understand how to do this.

Let's solve the following problems using trigonometric identities.

  1. Given that \begin{align*}\cos \theta=\frac{3}{5}\end{align*} and \begin{align*}0 < \theta < \frac{\pi}{2}\end{align*}, find \begin{align*}\sin \theta\end{align*}.

Use the Pythagorean Identity to find \begin{align*}\sin \theta\end{align*}.

\begin{align*}\sin^2 \theta+\cos^2 \theta&=1 \\ \sin^2 \theta+ \left(\frac{3}{5}\right)^2&=1 \\ \sin^2 \theta&=1- \frac{9}{25} \\ \sin^2 \theta&=\frac{16}{25} \\ \sin \theta&= \pm \frac{4}{5}\end{align*}

Because \begin{align*}\theta\end{align*} is in the first quadrant, we know that sine will be positive. \begin{align*}\sin \theta=\frac{4}{5}\end{align*}

  1. Find \begin{align*}\tan \theta\end{align*} from #1 above.

Use the Tangent Identity to find \begin{align*}\tan \theta\end{align*}.

\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\end{align*}

  1. Find the other three trigonometric functions of \begin{align*}\theta\end{align*} from #1.

To find secant, cosecant, and cotangent use the Reciprocal Identities.

\begin{align*}\csc \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{4}{5}}= \frac{5}{4} \quad \sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{3}{5}}=\frac{5}{3} \quad \cot \theta =\frac{1}{\tan \theta}=\frac{1}{\frac{4}{3}}=\frac{3}{4}\end{align*}

Examples

Example 1

Earlier, you were asked to find \begin{align*}\cos \theta\end{align*} and \begin{align*}\tan \theta\end{align*} of  \begin{align*}\sin \theta= \frac{2}{3}, \frac{\pi}{2} < \theta < \pi\end{align*}

First, use the Pythagorean Identity to find \begin{align*}\cos \theta\end{align*}.

\begin{align*}\sin^2 \theta+\cos^2 \theta&=1 \\ (\frac{2}{3})^2 + \cos^2 \theta =1 \\ \cos^2 \theta&=1- \frac{4}{9} \\ \cos^2 \theta&=\frac{5}{9} \\ \cos \theta&= \pm \frac{\sqrt{5}}{3}\end{align*}

However, because \begin{align*}\theta\end{align*} is restricted to the second quadrant, the cosine must be negative. Therefore, \begin{align*}\cos \theta= -\frac{\sqrt{5}}{3}\end{align*}.

Now use the Tangent Identity to find \begin{align*}\tan \theta\end{align*}.

\begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}=\frac{-2}{\sqrt{5}} = \frac{-2\sqrt{5}}{5}\end{align*}

Find the values of the other five trigonometric functions.

Example 2

\begin{align*}\tan \theta=- \frac{5}{12}, \frac{\pi}{2} < \theta < \pi\end{align*}

First, we know that \begin{align*}\theta\end{align*} is in the second quadrant, making sine positive and cosine negative. For this problem, we will use the Pythagorean Identity \begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*} to find secant.

\begin{align*}1+ \left(- \frac{5}{12}\right)^2&=\sec^2 \theta \\ 1+ \frac{25}{144}&=\sec^2 \theta \\ \frac{169}{144}&=\sec^2 \theta \\ \pm \frac{13}{12}&=\sec \theta \\ - \frac{13}{12}&=\sec \theta\end{align*}

If \begin{align*}\sec \theta=- \frac{13}{12}\end{align*}, then \begin{align*}\cos \theta= - \frac{12}{13}\end{align*}. \begin{align*}\sin \theta=\frac{5}{13}\end{align*} because the numerator value of tangent is the sine and it has the same denominator value as cosine. \begin{align*}\csc \theta=\frac{13}{5}\end{align*} and \begin{align*}\cot \theta=- \frac{12}{5}\end{align*} from the Reciprocal Identities.

Example 3

\begin{align*}\csc \theta=-8, \pi < \theta < \frac{3 \pi}{2}\end{align*}

\begin{align*}\theta\end{align*} is in the third quadrant, so both sine and cosine are negative. The reciprocal of \begin{align*}\csc \theta=-8\end{align*}, will give us \begin{align*}\sin \theta=- \frac{1}{8}\end{align*}. Now, use the Pythagorean Identity \begin{align*}\sin^2 \theta + \cos^2 \theta=1\end{align*} to find cosine.

\begin{align*}\left(- \frac{1}{8}\right)^2+ \cos^2 \theta&=1 \\ \cos^2 \theta&=1- \frac{1}{64} \\ \cos^2 \theta&=\frac{63}{64} \\ \cos \theta&=\pm \frac{3 \sqrt{7}}{8} \\ \cos \theta&=- \frac{3 \sqrt{7}}{8} \\ \end{align*}

\begin{align*}\sec \theta=- \frac{8}{3 \sqrt{7}}=- \frac{8 \sqrt{7}}{21}, \tan \theta= \frac{1}{3 \sqrt{7}}= \frac{\sqrt{7}}{21},\end{align*} and \begin{align*}\cot \theta=3 \sqrt{7}\end{align*}

Review

  1. In which quadrants is the sine value positive? Negative?
  2. In which quadrants is the cosine value positive? Negative?
  3. In which quadrants is the tangent value positive? Negative?

Find the values of the other five trigonometric functions of \begin{align*}\theta\end{align*}.

  1. \begin{align*}\sin \theta=\frac{8}{17},0 < \theta < \frac{\pi}{2}\end{align*}
  2. \begin{align*}\cos \theta=- \frac{5}{6}, \frac{\pi}{2} < \theta < \pi\end{align*}
  3. \begin{align*}\tan \theta= \frac{\sqrt{3}}{4},0 < \theta < \frac{\pi}{2}\end{align*}
  4. \begin{align*}\sec \theta=- \frac{41}{9}, \pi < \theta < \frac{3 \pi}{2}\end{align*}
  5. \begin{align*}\sin \theta=- \frac{11}{14}, \frac{3 \pi}{2} < \theta < 2 \pi\end{align*}
  6. \begin{align*}\cos \theta=\frac{\sqrt{2}}{2},0 < \theta < \frac{\pi}{2}\end{align*}
  7. \begin{align*}\cot \theta= \sqrt{5}, \pi < \theta < \frac{3 \pi}{2}\end{align*}
  8. \begin{align*}\csc \theta=4, \frac{\pi}{2} < \theta < \pi\end{align*}
  9. \begin{align*}\tan \theta=- \frac{7}{10}, \frac{3 \pi}{2} < \theta < 2 \pi\end{align*}
  10. Aside from using the identities, how else can you find the values of the other five trigonometric functions?
  11. Given that \begin{align*}\cos \theta=\frac{6}{11}\end{align*} and \begin{align*}\theta\end{align*} is in the \begin{align*}2^{nd}\end{align*} quadrant, what is \begin{align*}\sin(- \theta)\end{align*}?
  12. Given that \begin{align*}\tan \theta=- \frac{5}{8}\end{align*} and \begin{align*}\theta\end{align*} is in the \begin{align*}4^{th}\end{align*} quadrant, what is \begin{align*}\sec(- \theta)\end{align*}?

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.7. 

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