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Callisto

  • 3 years 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 \[a^2 = b^2 + c^2 - 2bc \ cosA\]

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  1. FoolAroundMath
    • 3 years ago
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    \(C = \pi - A - B \Rightarrow sinC = sin(A+B)\)\[\frac{b^{2}+c^{2}-a^{2}}{2bc} = \frac{k^{2}(sin^{2}B+sin^{2}C -sin^{2}A)}{2k^{2}sinBsinC}\] \[=\frac{sin^{2}B+sin(C+A)sin(C-A)}{2sinBsinC}\] \(C+A = \pi - B \Rightarrow sin(C+A)=sinB\)\[=\frac{sinB(sinB + sin(C-A))}{2sinBsinC} = \frac{2sin((B+C-A)/2)cos((B-C+A)/2)}{2sinC}\] \[=\frac{cosAsinC}{sinC} = cosA\]

  2. Callisto
    • 3 years ago
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    I'm not sure if it is okay to do it in this way... But thanks for your ideas!!

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