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 - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 242: 12


$$\sin195^\circ=-\frac{\sqrt{2-\sqrt 3}}{2}$$

Work Step by Step

$$A=\sin195^\circ$$ We can separate $195^\circ$ into $180^\circ$ and $15^\circ$. Therefore, we can rewrite A as follows $$A=\sin(180^\circ+15^\circ)$$ $$A=\sin180^\circ\cos15^\circ+\cos180^\circ\sin15^\circ$$ $$A=0\times\cos15^\circ+(-1)\sin15^\circ$$ $$A=-\sin15^\circ$$ Since $15^\circ$ is half of $30^\circ$, we can rewrite $$A=-\sin\Bigg(\frac{1}{2}30^\circ\Bigg)$$ $$A=-\Bigg[\pm\sqrt{\frac{1-\cos30^\circ}{2}}\Bigg]$$ However, since $15^\circ$ is in quadrant I, in which $\sin X\gt0$, so $$A=-\Bigg[\sqrt{\frac{1-\cos30^\circ}{2}}\Bigg]$$ $$A=-\sqrt{\frac{1-\frac{\sqrt 3}{2}}{2}}$$ $$A=-\sqrt\frac{\frac{2-\sqrt 3}{2}}{2}$$ $$A=-\sqrt{\frac{2-\sqrt 3}{4}}$$ $$A=-\frac{\sqrt{2-\sqrt 3}}{2}$$
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