Pre-Calc Exam Notes 84

# Pre-Calc Exam Notes 84 - 84 Chapter 3 Identities 3.4...

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84 Chapter 3 Identities §3.4 Example 3.18 For any triangle ABC , show that cos A + cos B + cos C = 1 + 4 sin 1 2 A sin 1 2 B sin 1 2 C . Solution: Since cos ( A + B + C ) = cos 180 =− 1, we can rewrite the left side as cos A + cos B + cos C = 1 + (cos ( A + B + C ) + cos C ) + (cos A + cos B ) , so by formula (3.43) = 1 + 2 cos 1 2 ( A + B + 2 C ) cos 1 2 ( A + B ) + 2 cos 1 2 ( A + B ) cos 1 2 ( A B ) = 1 + 2 cos 1 2 ( A + B ) ( cos 1 2 ( A + B + 2 C ) + cos 1 2 ( A B ) ) , so = 1 + 2 cos 1 2 ( A + B ) · 2 cos 1 2 ( A + C ) cos 1 2 ( B + C ) by formula (3.43), since 1 2 ( 1 2 ( A + B + 2 C ) + 1 2 ( A B ) ) = 1 2 ( A + C ) and 1 2 ( 1 2 ( A + B + 2 C ) 1 2 ( A B ) ) = 1 2 ( B + C ). Thus, cos A + cos B + cos C = 1 + 4 cos (90 1 2 C ) cos (90 1 2 B ) cos (90 1 2 A ) = 1 + 4 sin 1 2 C sin 1 2 B sin 1 2 A , so rearranging the order gives = 1 + 4 sin 1 2 A sin 1 2 B sin 1 2 C . Example 3.19
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## This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.

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