Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 239: 81

Answer

$cot~72^{\circ} = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}} = 0.325$

Work Step by Step

$cos~x = sin(90^{\circ}-x)$ We can find the value of $cos~72^{\circ}$: $cos~72^{\circ} = sin(90^{\circ}-72^{\circ})$ $cos~72^{\circ} = sin(18^{\circ})$ $cos~72^{\circ} = \frac{\sqrt{5}-1}{4}$ $sin~72^{\circ} = sin~(90^{\circ}-18^{\circ}) = cos~18^{\circ}$ $cos^2~x+sin^2~x = 1$ $cos~x = \sqrt{1-sin^2~x}$ We can find the value of $cos~18^{\circ}$: $cos~18^{\circ} = \sqrt{1-(sin~18^{\circ})^2}$ $cos~18^{\circ} = \sqrt{1-(\frac{\sqrt{5}-1}{4})^2}$ $cos~18^{\circ} = \sqrt{1-(\frac{6-2\sqrt{5}}{16})}$ $cos~18^{\circ} = \sqrt{\frac{10+2\sqrt{5}}{16}}$ $cos~18^{\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4}$ Since $sin~72^{\circ} = cos~18^{\circ}$, then $sin~72^{\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4}$ We can find the value of $cot~72^{\circ}$: $cot~72^{\circ} = \frac{cos~72^{\circ}}{sin~72^{\circ}}$ $cot~72^{\circ} = \frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt{5}}}{4}}$ $cot~72^{\circ} = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$ $cot~72^{\circ} = 0.325$
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