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: 60


See the explanation below.

Work Step by Step

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