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: 6


$$\tan15^\circ=2-\sqrt3$$ 6 goes with A.

Work Step by Step

$$\tan15^\circ$$ $15^\circ$ can be written as half of the angle $30^\circ$. $$15^\circ=\frac{30^\circ}{2}$$ Thus, $$\tan15^\circ=\tan\Big(\frac{30^\circ}{2}\Big)$$ We can apply here the Half-Angle Identity for tangent, which states $$\tan\Big(\frac{A}{2}\Big)=\frac{\sin A}{1+\cos A}$$ with $A=30^\circ$: $$\tan15^\circ=\frac{\sin30^\circ}{1+\cos30^\circ}$$ $$\tan15^\circ=\frac{\frac{1}{2}}{1+\frac{\sqrt3}{2}}$$ $$\tan15^\circ=\frac{\frac{1}{2}}{\frac{2+\sqrt3}{2}}$$ Two similar denominators would eliminate each other. $$\tan15^\circ=\frac{1}{2+\sqrt3}$$ Now multiply both numerator and denominator with $(2-\sqrt3)$. - Numerator: $1\times(2-\sqrt3)=2-\sqrt3$ - Denominator: $(2+\sqrt3)(2-\sqrt3)=4-3=1$ (since $(A-B)(A+B)=A^2-B^2$) That means, $$\tan15^\circ=\frac{2-\sqrt3}{1}$$ $$\tan15^\circ=2-\sqrt3$$ 6 goes with A.
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