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: 1

Answer

$$\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}$$ 1 should be matched with C.

Work Step by Step

$$\sin 15^\circ$$ $15^\circ$ can be rewritten as the difference of $45^\circ$ and $30^\circ$. $$15^\circ=45^\circ-30^\circ$$ Therefore, $$\sin15^\circ=\sin(45^\circ-30^\circ)$$ Now we apply the sine difference identity, which states $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ That makes $$\sin15^\circ=\sin 45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ$$ $$\sin15^\circ=\frac{\sqrt2}{2}\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\frac{1}{2}$$ $$\sin15^\circ=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$$ $$\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}$$ That means we match 1 with C.
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