Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 66

Answer

See the explanation below.

Work Step by Step

Let us consider the right side of the given expression: $\frac{1+\cos x}{\sin x}$ By using the trigonometric identity $\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}$, the above expression can be further simplified by multiplying the numerator and denominator by $\frac{1}{1+\cos x}$: $\begin{align} & \frac{1+\cos x}{\sin x}=\frac{\left( 1+\cos x \right)\times \frac{1}{1+\cos x}}{\sin x\times \frac{1}{1+\cos x}} \\ & =\frac{\frac{1+\cos x}{1+\cos x}}{\frac{\sin x}{1+\cos x}} \\ & =\frac{1}{\tan \frac{x}{2}} \\ & =\cot \frac{x}{2} \end{align}$ Hence, the left side of the given expression is equal to the right side, which is $\cot \frac{x}{2}=\frac{1+\cos x}{\sin x}$.
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