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Chapter 5: Problem 22

Find the exact values of the sine, cosine, and tangent of the angle. $$-\frac{7 \pi}{12}$$

Short Answer

Expert verified
\[sin(-\frac{7 \pi}{12}) = \sqrt{6}/4 - \sqrt{2}/4; cos(-\frac{7 \pi}{12}) = \sqrt{6}/4 + \sqrt{2}/4; tan(-\frac{7 \pi}{12}) = \sqrt{2}-1\]

Step by step solution

01

Express as sum/difference of two well-known angles

Express \( -\frac{7 \pi}{12} \) as a difference of two angles, which are multiples of \( \frac{\pi}{6} \) and \( \frac{\pi}{4} \). That would be \( -\frac{5 \pi}{12}-\frac{2 \pi}{12} \) which simplifies into \( -\frac{\pi}{4}-\frac{\pi}{6} \).
02

Find exact values of sine

Use the difference identity for sine, \(sin(a-b) = sin(a)cos(b) - cos(a)sin(b)\), to find sin(\( -\frac{7 \pi}{12} \)). Substituting \(a=-\frac{\pi}{4}\) and \(b=\frac{\pi}{6}\), the result is $\sqrt{2}/2 * \sqrt{3}/2 - \sqrt{2}/2 * 1/2 = \sqrt{6}/4 - \sqrt{2}/4$.
03

Find exact values of cosine

Use the difference identity for cosine, cos(a-b)=cos(a)cos(b)+sin(a)sin(b), to find cos(\( -\frac{7 \pi}{12} \)). Substituting \(a=-\frac{\pi}{4}\) and \(b=\frac{\pi}{6}\), the result is $\sqrt{2}/2 * \sqrt{3}/2 + \sqrt{2}/2 * 1/2 = \sqrt{6}/4 + \sqrt{2}/4$.
04

Find exact values of tangent

Remember that \(tan(\theta) = \frac{sin(\theta)}{cos(\theta)}\). So, \(tan(-\frac{7 \pi}{12}) = \frac{\sqrt{6}/4 - \sqrt{2}/4}{\sqrt{6}/4 + \sqrt{2}/4} = \sqrt{2}-1.\)

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