Prove that sin^2A+sin^2(60+A)+sin^2(60-A)=3/2
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Posted by Himanshu Goyal 7 years, 10 months ago
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Samir Kumar Basu 7 years, 10 months ago
LHS=sin2A+sin2(60∘+A)+sin2(60∘−A)=sin2A+12[1−cos2(60∘+A)]+12[1−cos2(60∘−A)][sincesin2θ=12(1−cos2θ)]=sin2A+1−12[cos(120∘+2A)+cos(120∘−2A)]=sin2A+1−12×2cos(120∘+2A)+(120∘−2A)2.cos(120∘+2A)−(120∘−2A)2sincecosC+cosD=2.cosC+D2.cosC−D2=sin2A+1−cos120\textdegree.cos2A=sin2A+1+12.(1−2sin2A)=sin2A+1+12−sin2A=32=RHS
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