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## Callisto 4 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

$$C = \pi - A - B \Rightarrow sinC = sin(A+B)$$$\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

I'm not sure if it is okay to do it in this way... But thanks for your ideas!!

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