Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.4 - Product-to-Sum and Sum-to-Product Formulas - Exercise Set - Page 689: 20


The required solution is $2\sin \left( {{45}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right)$, $-\frac{\sqrt{2}}{2}$.

Work Step by Step

One of the sum-to-product formula is $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$. So, in this question, according to the above-mentioned formula, the value of $\alpha $ is ${{75}^{\circ }}$ and the value of $\beta $ is ${{15}^{\circ }}$. Now, the expression can be evaluated as provided below: $\begin{align} & \cos {{75}^{\circ }}-\cos {{15}^{\circ }}=-2\sin \left( \frac{{{75}^{\circ }}+{{15}^{\circ }}}{2} \right)\sin \left( \frac{{{75}^{\circ }}-{{15}^{\circ }}}{2} \right) \\ & =-2\sin \left( \frac{{{90}^{\circ }}}{2} \right)\sin \left( \frac{{{60}^{\circ }}}{2} \right) \\ & =-2\sin \left( {{45}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right) \end{align}$ Thus, the values as per the trigonometry table are put to solve the expression further and to find the exact value: $\begin{align} & -2\sin \left( {{45}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right)=-2\left( \frac{\sqrt{2}}{2} \right)\left( \frac{1}{2} \right) \\ & =-\frac{\sqrt{2}}{2} \end{align}$ Hence, the provided expression can be written as $2\sin \left( {{45}^{\circ }} \right)\sin \left( {{30}^{\circ }} \right)$. And the product’s exact value is $-\frac{\sqrt{2}}{2}$.
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