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: 80

Answer

$ \tan 72^{\circ}=\frac{\sqrt{10+2\sqrt{5}} }{\sqrt{5}-1}=3.078 $

Work Step by Step

By using trigonometric ratios of complementary angles. $ \cos \theta = \sin (90 ^{\circ} - \theta ) $ Let $ \theta =72^{\circ} $ . $ \cos 72 ^{\circ} = \sin (90 ^{\circ} - 72 ) $ Add like terms. $ \cos 72 ^{\circ} = \sin (18 ^{\circ}) $ Substitute the value of $ \sin (18 ^{\circ}) $ . $ \cos 72 ^{\circ} = \frac{\sqrt{5}-1}{4} $ By using trigonometric basic identity. $ \sin ^2 \theta + \cos ^2 \theta =1 $ Isolate $ \sin \theta $ . $ \sin \theta =\sqrt{1- \cos ^2 \theta } $ Plug $ \theta = 72 ^{\circ} $ into the above equation. $ \sin 72 ^{\circ} =\sqrt{1- \cos ^2 72 ^{\circ} } $ Substitute the value of $ \cos 72 ^{\circ} $. $ \sin 72 ^{\circ} =\sqrt{1- \left (\frac{\sqrt{5}-1}{4} \right )^2 } $ $ \sin 72 ^{\circ} =\sqrt{1- \left (\frac{6-2\sqrt{5}}{16} \right ) } $ $ \sin 72 ^{\circ} =\sqrt{ \frac{10+2\sqrt{5}}{16} } $ $ \sin 72 ^{\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4} $ By using trigonometric ratios. $ \tan \theta= \frac{\sin \theta}{\cos \theta} $ Plug $ \theta = 72 ^{\circ} $ into the above equation. $ \tan 72 ^{\circ}= \frac{\sin 72 ^{\circ}}{\cos 72 ^{\circ}} $ Substitute values. $ \tan 72 ^{\circ} =\frac{\frac{\sqrt{10+2\sqrt{5}}}{4}}{\frac{\sqrt{5}-1}{4}} $ Simplify. $ \tan 72 ^{\circ} =\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1} $ Use calculator for exact value. $ \tan 72 ^{\circ} =3.078 $
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