## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.3 Sum and Difference Identities for Cosine - 5.3 Exercises - Page 219: 56

#### Answer

$$\cos(s-t)=\frac{\sqrt{62}+\sqrt{70}}{24}$$ $$\cos(s+t)=\frac{\sqrt{62}-\sqrt{70}}{24}$$

#### Work Step by Step

To find $\cos(s+t)$ and $\cos(s-t)$, cosine sum and difference identities need to be applied. Yet, both cosine sum and difference identities rely on the fact that you already know $\cos s$, $\sin s$, $\cos t$ and $\sin t$. Therefore, the first job in this type of exercise is to find out all 4 values: $\cos s$, $\sin s$, $\cos t$ and $\sin t$. 1) Find $\cos s$, $\sin s$, $\cos t$ and $\sin t$. As $\cos s$ and $\sin t$ are known, we would use Pythagorean Identities for $\sin$ and $\cos$ to find the rest: $$\sin^2 s=1-\cos^2 s=1-\Big(\frac{\sqrt2}{4}\Big)^2=1-\frac{2}{16}=\frac{14}{16}$$ $$\sin s=\pm\frac{\sqrt{14}}{4}$$ $$\cos^2 t=1-\sin^2 t=1-\Big(-\frac{\sqrt5}{6}\Big)^2=1-\frac{5}{36}=\frac{31}{36}$$ $$\cos t=\pm\frac{\sqrt{31}}{6}$$ Now about the signs of $\sin s$ and $\cos t$: As both $s$ and $t$ are in quadrant IV, $\sin s\lt0$ while $\cos t\gt0$, so $$\sin s=-\frac{\sqrt{14}}{4}\hspace{2cm}\cos t=\frac{\sqrt{31}}{6}$$ 2) Now we can apply cosine sum and difference identities: $$\cos(s+t)=\cos s\cos t-\sin s\sin t=\Big(\frac{\sqrt{2}}{4}\Big)\Big(\frac{\sqrt{31}}{6}\Big)-\Big(-\frac{\sqrt{14}}{4}\Big)\Big(-\frac{\sqrt5}{6}\Big)$$ $$\cos(s+t)=\frac{\sqrt{62}}{24}-\frac{\sqrt{70}}{24}=\frac{\sqrt{62}-\sqrt{70}}{24}$$ $$\cos(s-t)=\cos s\cos t+\sin s\sin t=\Big(\frac{\sqrt{2}}{4}\Big)\Big(\frac{\sqrt{31}}{6}\Big)+\Big(-\frac{\sqrt{14}}{4}\Big)\Big(-\frac{\sqrt5}{6}\Big)$$ $$\cos(s-t)=\frac{\sqrt{62}}{24}+\frac{\sqrt{70}}{24}=\frac{\sqrt{62}+\sqrt{70}}{24}$$

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