Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 220: 12

Answer

$$\tan\frac{5\pi}{12}=2+\sqrt3$$

Work Step by Step

*STRATEGY: In this type of exercise, we would try to rewrite the angle in terms of already familiar angles, including $\frac{\pi}{6}$, $\frac{\pi}{4}$ and $\frac{\pi}{3}$. $$\tan\frac{5\pi}{12}$$ Now, we try writing $5\pi$ as the sum of $2\pi$ and $3\pi$. $$\tan\frac{5\pi}{12}=\tan \Big(\frac{2\pi+3\pi}{12}\Big)=\tan\Big(\frac{2\pi}{12}+\frac{3\pi}{12}\Big)=\tan\Big(\frac{\pi}{6}+\frac{\pi}{4}\Big)$$ We then use the tangent sum identity: $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ Therefore, $$\tan\frac{5\pi}{12}=\frac{\tan \frac{\pi}{6}+\tan\frac{\pi}{4}}{1-\tan\frac{\pi}{6}\tan\frac{\pi}{4}}$$ $$\tan\frac{5\pi}{12}=\frac{\frac{1}{\sqrt3}+1}{1-\frac{1}{\sqrt3}\times1}$$ $$\tan\frac{5\pi}{12}=\frac{\frac{1+\sqrt3}{\sqrt3}}{1-\frac{1}{\sqrt3}}$$ $$\tan\frac{5\pi}{12}=\frac{\frac{\sqrt3+1}{\sqrt3}}{\frac{\sqrt3-1}{\sqrt3}}$$ $$\tan\frac{5\pi}{12}=\frac{\sqrt3+1}{\sqrt3-1}$$ Multiply both numerator and denominator with $\sqrt3+1$ $$\tan\frac{5\pi}{12}=\frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)}$$ $$\tan\frac{5\pi}{12}=\frac{3+1+2\sqrt3}{3-1}$$ $$\tan\frac{5\pi}{12}=\frac{4+2\sqrt3}{2}$$ $$\tan\frac{5\pi}{12}=2+\sqrt3$$
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