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: 65


See the explanation below.

Work Step by Step

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