Pre-Calc Exam Notes 83

Pre-Calc Exam Notes 83 - C − A tan 1 2 C A We need only...

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Other Identities Section 3.4 83 Example 3.16 We are now in a position to prove Mollweide’s equations, which we introduced in Section 2.3: For any triangle ABC , a b c = sin 1 2 ( A B ) cos 1 2 C and a + b c = cos 1 2 ( A B ) sin 1 2 C . First, since C = 2 · 1 2 C , by the double-angle formula we have sin C = 2 sin 1 2 C cos 1 2 C . Thus, a b c = a c b c = sin A sin C sin B sin C (by the Law of Sines) = sin A sin B sin C = sin A sin B 2 sin 1 2 C cos 1 2 C = 2 cos 1 2 ( A + B ) sin 1 2 ( A B ) 2 sin 1 2 C cos 1 2 C (by formula (3.42)) = cos 1 2 (180 C ) sin 1 2 ( A B ) sin 1 2 C cos 1 2 C (since A + B = 180 C ) = cos (90 1 2 C ) sin 1 2 ( A B ) ✟✟ sin 1 2 C cos 1 2 C = sin 1 2 ( A B ) cos 1 2 C (since cos (90 1 2 C ) = sin 1 2 C ) . This proves the ±rst equation. The proof of the other equation is similar (see Exercise 7). Example 3.17 Using Mollweide’s equations, we can prove the Law of Tangents: For any triangle ABC , a b a + b = tan 1 2 ( A B ) tan 1 2 ( A + B ) , b c b + c = tan 1 2 ( B C ) tan 1 2 ( B + C ) , c a c + a = tan 1 2
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Unformatted text preview: ( C − A ) tan 1 2 ( C + A ) . We need only prove the ±rst equation; the other two are obtained by cycling through the letters. We see that a − b a + b = a − b c a + b c = sin 1 2 ( A − B ) cos 1 2 C cos 1 2 ( A − B ) sin 1 2 C (by Mollweide’s equations) = sin 1 2 ( A − B ) cos 1 2 ( A − B ) · sin 1 2 C cos 1 2 C = tan 1 2 ( A − B ) · tan 1 2 C = tan 1 2 ( A − B ) · tan (90 ◦ − 1 2 ( A + B )) (since C = 180 ◦ − ( A + B )) = tan 1 2 ( A − B ) · cot 1 2 ( A + B ) (since tan (90 ◦ − 1 2 ( A + B )) = cot 1 2 ( A + B ), see Section 1.5) = tan 1 2 ( A − B ) tan 1 2 ( A + B ) . QED...
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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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