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: 20


$$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$ The equation is an identity, as verified below.

Work Step by Step

$$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$ 1) We start from the left side. $$X=\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)$$ - Recall the sum identity for tangent, which states $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ Replace the identity into $X$ with $A=\frac{x}{2}$ and $B=\frac{\pi}{4}$, we have $$X=\frac{\tan\frac{x}{2}+\tan\frac{\pi}{4}}{1-\tan\frac{x}{2}\tan\frac{\pi}{4}}$$ $$X=\frac{\tan\frac{x}{2}+1}{1-\tan\frac{x}{2}\times1}$$ $$X=\frac{\tan\frac{x}{2}+1}{1-\tan\frac{x}{2}}$$ - Now use the half-angle identity for tangent: $$\tan\frac{x}{2}=\frac{1-\cos x}{\sin x}$$ Thus, $$X=\frac{\frac{1-\cos x}{\sin x}+1}{1-\frac{1-\cos x}{\sin x}}$$ $$X=\frac{\frac{1-\cos x+\sin x}{\sin x}}{\frac{\sin x-1+\cos x}{\sin x}}$$ $$X=\frac{1-\cos x+\sin x}{\sin x-1+\cos x}$$ $$X=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$$ We stop here as there seems no potential in simplifying anymore. 2) Then we look at the right side. $$Y=\sec x+\tan x$$ - Apply the following identities: $$\sec x=\frac{1}{\cos x}\hspace{2cm}\tan x=\frac{\sin x}{\cos x}$$ $$Y=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$$ $$Y=\frac{1+\sin x}{\cos x}$$ So the job here is to prove that $\frac{1+\sin x}{\cos x}=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$, which seems so different. So in a seemingly dead-end case like this, we would try multiplying both the numerator and denominator of $Y$ with $\frac{\sin x+\cos x-1}{\sin x+\cos x-1}$. $$Y=\frac{1+\sin x}{\cos x}\times\frac{\sin x+\cos x-1}{\sin x+\cos x-1}$$ $$Y=\frac{\sin x+\cos x-1+\sin^2x+\sin x\cos x-\sin x}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{\sin^2x+\sin x\cos x+\cos x-1}{\cos x(\sin x+\cos x-1)}$$ - Now remember that $\sin^2x=1-\cos^2x$, which can be used to rewrite $\sin^2x$. $$Y=\frac{1-\cos^2x+\sin x\cos x+\cos x-1}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{-\cos^2x+\sin x\cos x+\cos x}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{\cos x(\sin x-\cos x+1)}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$$ So after a rather painstaking process of transformations, we show that $X=Y$. Therefore, $$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$ The equation is therefore an identity.
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