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: 58

Answer

$1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = cos~2x$
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Work Step by Step

$1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2}$ When we graph this function, we can see that it looks like the graph of $~~cos~2x$ Note that: $~sin~2a = 2~sin~a~cos~a$ Also: $~~cos~2b = 1-2~sin^2~b$ We can verify this algebraically: $1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = 1 - 2~(4~sin^2~\frac{x}{2}~cos^2~\frac{x}{2})$ $1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = 1 - 2~(2~sin~\frac{x}{2}~cos~\frac{x}{2})^2$ $1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = 1 - 2~(sin~x)^2$ $1 - 8 ~sin^2~\frac{x}{2}~cos^2~\frac{x}{2} = cos~2x$
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