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: 29


$$\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}=\cot x$$ After proving 2 sides are equal to each other, we can state that the equation is an identity.

Work Step by Step

$$\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}=\cot x$$ From the left first: $$X=\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}$$ - A recount of all the identities here: $$\cos(x+y)=\cos x\cos y-\sin x\sin y$$ $$\cos(y-x)=\cos y\cos x+\sin y\sin x=\cos x\cos y+\sin x\sin y$$ $$\sin(x+y)=\sin x\cos y+\cos x\sin y$$ $$\sin(y-x)=\sin y\cos x-\cos y\sin x=\cos x\sin y-\sin x\cos y$$ Therefore: - Numerator: $$\cos(x+y)+\cos(y-x)=\cos x\cos y-\sin x\sin y+\cos x\cos y+\sin x\sin y$$ $$\cos(x+y)+\cos(y-x)=2\cos x\cos y$$ - Denominator: $$\sin(x+y)-\sin(y-x)=\sin x\cos y+\cos x\sin y-(\cos x\sin y-\sin x\cos y)$$ $$\sin(x+y)-\sin(y-x)=\sin x\cos y+\cos x\sin y-\cos x\sin y+\sin x\cos y$$ $$\sin(x+y)-\sin(y-x)=2\sin x\cos y$$ Apply back to $X$: $$X=\frac{2\cos x\cos y}{2\sin x\cos y}$$ $$X=\frac{\cos x}{\sin x}$$ $$X=\cot x\hspace{1cm}\text{Quotient Identities}$$ Therefore, $$\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}=\cot x$$ That makes the equation eligible to be an identity.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.