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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 \[a^2 = b^2 + c^2  2bc \ cosA\]
 one year ago
 one year ago
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 \[a^2 = b^2 + c^2  2bc \ cosA\]
 one year ago
 one year ago

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

CallistoBest 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!!
 one year ago
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