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Callisto
Group Title
Trigo. question #2
Suppose in ΔABC, a = ksinA, b=ksinB and c=ksinC where k is a real constant with k>0. By using compound angle formulas, prove that \[a^2 = b^2 + c^2  2bc \ cosA\]
 2 years ago
 2 years ago
Callisto Group Title
Trigo. question #2 Suppose in ΔABC, a = ksinA, b=ksinB and c=ksinC where k is a real constant with k>0. By using compound angle formulas, prove that \[a^2 = b^2 + c^2  2bc \ cosA\]
 2 years ago
 2 years ago

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FoolAroundMath Group TitleBest ResponseYou've already chosen the best response.1
\(C = \pi  A  B \Rightarrow sinC = sin(A+B)\)\[\frac{b^{2}+c^{2}a^{2}}{2bc} = \frac{k^{2}(sin^{2}B+sin^{2}C sin^{2}A)}{2k^{2}sinBsinC}\] \[=\frac{sin^{2}B+sin(C+A)sin(CA)}{2sinBsinC}\] \(C+A = \pi  B \Rightarrow sin(C+A)=sinB\)\[=\frac{sinB(sinB + sin(CA))}{2sinBsinC} = \frac{2sin((B+CA)/2)cos((BC+A)/2)}{2sinC}\] \[=\frac{cosAsinC}{sinC} = cosA\]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
I'm not sure if it is okay to do it in this way... But thanks for your ideas!!
 2 years ago
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