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 243: 53


By multiplying both numerator and denominator by $(1-\cos A)$, we can derive the equivalent identity as shown below.

Work Step by Step

$$\tan\frac{A}{2}=\frac{\sin A}{1+\cos A}$$ The job is to derive the equivalent identity: $$\tan\frac{A}{2}=\frac{1-\cos A}{\sin A}$$ As hinted by the exercise, we multiply both numerator and denominator of the original identity by $(1-\cos A)$ $$\tan\frac{A}{2}=\frac{\sin A(1-\cos A)}{(1+\cos A)(1-\cos A)}$$ - Denominator: $$(1+\cos A)(1-\cos A)=1-\cos^2A$$ (as $(A+B)(A-B)=A^2-B^2$) $$(1+\cos A)(1-\cos A)=\sin^2A$$ (Pythagorean Identity) Therefore, $$\tan\frac{A}{2}=\frac{\sin A(1-\cos A)}{\sin^2 A}$$ $$\tan\frac{A}{2}=\frac{1-\cos A}{\sin A}$$ which is the equivalent identity. Thus, we have completed deriving the equivalent identity as requested.
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