# 3.2.1:Trig Identities to Find Exact Trigonometric Values

- Page ID
- 4228

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

## Using Trig Identities to Find Exact Trig Values

You are given the following information about \(\theta\)

\(\sin\theta =\dfrac{2}{3}\), \(\dfrac{\pi}{2}<\theta <\pi\)

What are \(\cos\theta\) and \(\tan\theta \)?

**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.

- Given that \(\cos\theta =\dfrac{3}{5}\) and \(0<\theta <\dfrac{\pi}{2}\), find \(\sin \theta\).

Use the Pythagorean Identity to find sin\theta .

\(\begin{aligned}

\sin ^{2} \theta+\cos ^{2} \theta &=1 \\

\sin ^{2} \theta+\left(\dfrac{3}{5}\right)^{2} &=1 \\

\sin ^{2} \theta &=1-\dfrac{9}{25} \\

\sin ^{2} \theta &=\dfrac{16}{25} \\

\sin \theta &=\pm \dfrac{4}{5}

\end{aligned}\)

Because \(\theta\) is in the first quadrant, we know that sine will be positive. \(\sin\theta =\dfrac{4}{5}\)

- Find \(\tan\theta \) from #1 above.

Use the Tangent Identity to find \(\tan\theta\).

\(\tan\theta =\dfrac{\sin\theta}{\cos\theta}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}\)

- Find the other three trigonometric functions of \(\theta\) from #1.

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

\(\csc \theta=\dfrac{1}{\sin \theta}=\dfrac{1}{\dfrac{4}{5}}=\dfrac{5}{4} \quad \sec \theta=\dfrac{1}{\cos \theta}=\dfrac{1}{\dfrac{3}{5}}=\dfrac{5}{3} \quad \cot \theta=\dfrac{1}{\tan \theta}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)

Example \(\PageIndex{1}\)

Earlier, you were asked to find \(\cos\theta\) and \(\tan\theta\) of \(\sin\theta =\dfrac{2}{3}\), \(\dfrac{\pi}{2}<\theta <\pi \).

**Solution**

First, use the Pythagorean Identity to find \(\cos\theta\).

\(\begin{aligned}

\sin ^{2} \theta+\cos ^{2} \theta &=1 \\

\left(\dfrac{2}{3}\right)^{2}+\cos ^{2} \theta&=1 \\

\cos ^{2} \theta &=1-\dfrac{4}{9} \\

\cos ^{2} \theta &=\dfrac{5}{9} \\

\cos \theta &=\pm \dfrac{\sqrt{5}}{3}

\end{aligned}\)

However, because \(\theta\) is restricted to the second quadrant, the cosine must be negative. Therefore, \(\cos \theta =−\dfrac{\sqrt{5}}{3}\).

Now use the Tangent Identity to find tan\theta .

\(\tan\theta =\dfrac{\sin\theta}{\cos\theta}=\dfrac{\dfrac{2}{3}}{−\dfrac{\sqrt{5}}{3}}=−2\sqrt{5}=\dfrac{−2\sqrt{5}}{5}\)

**Find the values of the other five trigonometric functions.**

Example \(\PageIndex{2}\)

\(\tan\theta =−\dfrac{5}{12}\), \(\dfrac{\pi}{ 2}<\theta <\pi\)

**Solution**

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

\(\begin{aligned}

1+\left(-\dfrac{5}{12}\right)^{2} &=\sec ^{2} \theta \\

1+\dfrac{25}{144} &=\sec ^{2} \theta \\

\dfrac{169}{144} &=\sec ^{2} \theta \\

\pm \dfrac{13}{12} &=\sec \theta \\

-\dfrac{13}{12} &=\sec \theta

\end{aligned}\)

If \(\sec\theta =−\dfrac{13}{12}\), then \(\cos\theta =−\dfrac{12}{13}\). \(\sin\theta =\dfrac{5}{13}\) because the numerator value of tangent is the sine and it has the same denominator value as cosine. \(\csc\theta =\dfrac{13}{5}\) and \(\cot\theta =−\dfrac{12}{5}\) from the Reciprocal Identities.

Example \(\PageIndex{3}\)

\(\csc \theta =−8\), \(\pi <\theta <\dfrac{3 \pi}{2}\)

**Solution**

\theta is in the third quadrant, so both sine and cosine are negative. The reciprocal of \(\csc\theta =−8\), will give us \(\sin\theta =−\dfrac{1}{8}\). Now, use the Pythagorean Identity \(sin^2 \theta +\cos^2 \theta =1\) to find cosine.

\(\begin{aligned}

\left(-\dfrac{1}{8}\right)^{2}+\cos ^{2} \theta &=1 \\

\cos ^{2} \theta &=1-\dfrac{1}{64} \\

\cos ^{2} \theta &=\dfrac{63}{64} \\

\cos \theta &=\pm \dfrac{3 \sqrt{7}}{8} \\

\cos \theta &=-\dfrac{3 \sqrt{7}}{8}

\end{aligned}\)

\(\sec\theta =−\dfrac{8}{3\sqrt{7}}=−\dfrac{8\sqrt{7}}{21}\), \(\tan\theta =\dfrac{1}{3\sqrt{7}}=\dfrac{\sqrt{7}}{ 21}\), and \(\cot\theta =3\sqrt{7}\)

**Review **

- In which quadrants is the sine value positive? Negative?
- In which quadrants is the cosine value positive? Negative?
- In which quadrants is the tangent value positive? Negative?

Find the values of the other five trigonometric functions of \(\theta \).

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

**Answers for Review Problems**

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