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 223: 77b

Answer

First we use the sine sum identity to change $\sin(\theta+90^\circ)$ into $\cos\theta$. Then we calculate $\frac{0.6}{\sin12^\circ}\approx2.9$ Apply back to $F$ and the statement is proved.

Work Step by Step

$$F=\frac{0.6W\sin(\theta+90^\circ)}{\sin12^\circ}$$ For $\sin(\theta+90^\circ)$, in fact we can apply the sine sum identity: $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ That means $$\sin(\theta+90^\circ)=\sin\theta\cos90^\circ+\cos\theta\sin90^\circ$$ $$\sin(\theta+90^\circ)=\sin\theta\times0+\cos\theta\times1$$ $$\sin(\theta+90^\circ)=\cos\theta$$ Therefore, the formula for $F$ can be written as $$F=\frac{0.6W\cos\theta}{\sin12^\circ}=\frac{0.6}{\sin12^\circ}W\cos\theta$$ $\frac{0.6}{\sin12^\circ}\approx2.9$. Therefore, $$F\approx2.9W\cos\theta$$ So $F$ is approximately equal to $2.9W\cos\theta$.
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