Prove that sin^2A+sin^2(60+A)+sin^2(60-A)=3/2
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Posted by Himanshu Goyal 8 years, 3 months ago
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Samir Kumar Basu 8 years, 3 months ago
{tex}\begin{array}{l}LHS=\sin^2A+\sin^2(60^\circ+A)+\sin^2(60^\circ-A)\\=\sin^2A+\frac12\left[1-\cos2(60^\circ+A)\right]+\frac12\left[1-\cos2(60^\circ-A)\right]\\\;\;\;\;\;\;\;\;\lbrack\sin ce\;\;\sin^2\theta=\frac12(1-\cos2\theta)\;\;\rbrack\\=\sin^2A+1-\frac12\left[\cos(120^\circ+2A)+\cos(120^\circ-2A)\right]\\=\sin^2A+1-\frac12\times2\cos\frac{(120^\circ+2A)+(120^\circ-2A)}2.\cos\frac{(120^\circ+2A)-(120^\circ-2A)}2\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sin ce\;\;\cos\;C\;+\;\cos\;D=2.\cos\frac{C+D}2.\cos\frac{C-D}2\\=\sin^2A+1-\cos120^\textdegree.\cos2A\\=\sin^2A+1+\frac12.\left(1-2\sin^2A\right)\\=\sin^2A+1+\frac12-\sin^2A\\=\frac32=RHS\end{array}{/tex}
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