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Pre-Calc Exam Notes 74

# Pre-Calc Exam Notes 74 - A B C D = 180 ◦ Show that 2 sin...

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74 Chapter 3 Identities §3.2 Example3.9 Prove the following identity: sin ( A + B + C ) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C sin A sin B sin C Solution: Treat A + B + C as ( A + B ) + C and use the addition formulas three times: sin ( A + B + C ) = sin (( A + B ) + C ) = sin ( A + B ) cos C + cos ( A + B ) sin C = (sin A cos B + cos A sin B ) cos C + (cos A cos B sin A sin B ) sin C = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C sin A sin B sin C Example3.10 For any triangle ABC , show that tan A + tan B + tan C = tan A tan B tan C . Solution: Note that this is not an identity which holds for all angles; since A , B , and C are the angles of a triangle, it holds when A , B , C > 0 and A + B + C = 180 . So using C = 180 ( A + B ) and the relation tan (180 θ ) =− tan θ from Section 1.5, we get: tan A + tan B + tan C = tan A + tan B + tan (180 ( A + B )) = tan A + tan B tan ( A + B ) = tan A + tan B tan A + tan B 1 tan A tan B = (tan A + tan B ) parenleftbigg 1 1 1 tan A tan B parenrightbigg = (tan A + tan B ) parenleftbigg 1 tan A tan B 1 tan A tan B 1 1 tan A tan B parenrightbigg = (tan A + tan B ) · parenleftbigg tan A tan B 1 tan A tan B parenrightbigg = tan A tan B · parenleftbigg tan A + tan B 1 tan A tan B parenrightbigg = tan A tan B · ( tan ( A + B )) = tan A tan B · (tan (180 ( A + B ))) = tan A tan B tan
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Unformatted text preview: A + B + C + D = 180 ◦ . Show that 2 sin A sin B + sin C sin D = sin ( A + C ) sin ( B + C ) . Solution: It may be tempting to expand the right side, since it appears more complicated. However, notice that the right side has no D term. So instead, we will expand the left side, since we can eliminate the D term on that side by using D = 180 ◦ − ( A + B + C ) and the relation sin (180 ◦ − ( A + B + C )) = sin ( A + B + C ). 2 This is the “trigonometric form” of Ptolemy’s Theorem , which says that a quadrilateral can be inscribed in a circle if and only if the sum of the products of its opposite sides equals the product of its diagonals....
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