University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 58


(a) $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ - Replace $\theta$ with $(A+B)$ in the identity: $\sin\theta=\cos\Big(\frac{\pi}{2}-\theta\Big)$ - Apply the $\cos(A-B)$ formula. - Simplify. (b)$$\cos (A+B)=\cos A\cos B-\sin A\sin B$$ - Apply the $\cos(A-B)$ formula to $\cos(A+B)$ with $B=-(-B)$ - Simplify

Work Step by Step

(a) $$\sin\theta=\cos\Big(\frac{\pi}{2}-\theta\Big)$$ We can replace $\theta$ with $(A+B)$ here. $$\sin(A+B)=\cos\Big(\frac{\pi}{2}-(A+B)\Big)$$ $$\sin(A+B)=\cos\Big[\Big(\frac{\pi}{2}-A\Big)-B\Big]$$ - Here we apply the $\cos(A-B)$ formula for $\cos\Big[\Big(\frac{\pi}{2}-A\Big)-B\Big]$ $$\sin(A+B)=\cos\Big(\frac{\pi}{2}-A\Big)\cos B+\sin\Big(\frac{\pi}{2}-A\Big)\sin B$$ - We already know that $\cos\Big(\frac{\pi}{2}-A\Big)=\sin A$, according to the given identity by the exercise. - $\sin\Big(\frac{\pi}{2}-A\Big)=\sin\frac{\pi}{2}\cos A-\cos\frac{\pi}{2}\sin A=1\times\cos A-0\times\sin A=\cos A$ Therefore, $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ Thus we have obtained the addition formula for $\sin(A+B)$ (b) From Exercise 35: $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ Apply it to $\cos (A+B)$, we have: $$\cos (A+B)=\cos[A-(-B)]$$ $$\cos (A+B)=\cos A\cos(-B)+\sin A\sin(-B)$$ - Recall the identities: $\cos(-B)=\cos B$ and $\sin(-B)=-\sin B$ $$\cos (A+B)=\cos A\cos B-\sin A\sin B$$ We have finished deriving the formula for $\cos (A+B)$.
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