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: 11



Work Step by Step

$$A=\sin67.5^\circ$$ We can see that $67.5^\circ=\frac{1}{2}\times 135^\circ$. Therefore, we can rewrite A as follows $$A=\sin(\frac{1}{2}\times135^\circ)$$ $$A=\pm\sqrt{\frac{1-\cos135^\circ}{2}}$$ However, since $67.5^\circ$ is in quadrant I, in which $\sin X\gt0$, so $$A=\sqrt{\frac{1-\cos135^\circ}{2}}$$ We also see that $$\cos135^\circ=\cos(90^\circ+45^\circ)$$ $$=\cos90^\circ\cos45^\circ-\sin90^\circ\sin45^\circ$$ $$=0\times\cos45^\circ-1\times\frac{\sqrt 2}{2}$$ $$=-\frac{\sqrt 2}{2}$$ Therefore, apply to A, we have $$A=\sqrt{\frac{1-\frac{-\sqrt 2}{2}}{2}}$$ $$A=\sqrt{\frac{1+\frac{\sqrt 2}{2}}{2}}$$ $$A=\sqrt{\frac{\frac{2+\sqrt 2}{2}}{2}}$$ $$A=\sqrt{\frac{2+\sqrt 2}{4}}$$ $$A=\frac{\sqrt{2+\sqrt2}}{2}$$
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