Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Summary Exercises on Verifying Trigonometric Identities - Page 245: 30


$$\sin(60^\circ-x)-\sin(60^\circ+x)=-\sin x$$ The detailed explanation is written in the Work step by step.

Work Step by Step

$$\sin(60^\circ-x)-\sin(60^\circ+x)=-\sin x$$ As the left side is more complicated, it needs to be dealt with first. $$X=\sin(60^\circ-x)-\sin(60^\circ+x)$$ - According to the sum and difference identities for sine, which states $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ as we apply $A=60^\circ$ and $B=x$ to $X$, we would have $$\sin(60^\circ+x)=\sin60^\circ\cos x+\cos60^\circ\sin x$$ $$\sin(60^\circ+x)=\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x$$ and $$\sin(60^\circ-x)=\sin60^\circ\cos x-\cos60^\circ\sin x$$ $$\sin(60^\circ-x)=\frac{\sqrt3}{2}\cos x-\frac{1}{2}\sin x$$ Therefore, $$X=\frac{\sqrt3}{2}\cos x-\frac{1}{2}\sin x-\Big(\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x\Big)$$ $$X=\Big(\frac{\sqrt3}{2}\cos x-\frac{\sqrt3}{2}\cos x\Big)-\frac{1}{2}\sin x-\frac{1}{2}\sin x$$ $$X=0-2\times\frac{1}{2}\sin x$$ $$X=-\sin x$$ Therefore, $$\sin(60^\circ-x)-\sin(60^\circ+x)=-\sin x$$ We have verified that the equation is an identity.
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