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Question no. 12 please


12 .       sin 2   160 °   +   sin 2   140 °   +   sin 2   100 °   =   ( a )     1 2                                                   ( b )     3 2                                           ( c )     5 2                                                             ( d )     7 2

Hi, we know that sin180-θ = sinθsin2160+sin2140+sin2100=sin2180-20+sin2180-40+sin2180-80=sin220+sin240+sin280=sin220+sin260-20+sin260+20=sin220+sin60cos20-cos60sin202+sin60cos20+cos60sin202=sin220+sin260cos220+cos260sin220-2sin60cos20cos60sin20+sin260cos220+cos260sin220+2sin60cos20cos60sin20=sin220+2sin260cos220+2cos260sin220=sin220+2×34cos220+2×14sin220=sin220+32cos220+12sin220= 32cos220+sin220+12sin220=32cos220+32sin220=32sin220+cos220=32×1= 32

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1/2
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We know that sin2(180°-θ)= sin2θ
So, sin2160°+ sin2140°+ sin2100°= sin220°+ sin240°+ sin280°= sin220°+ sin2(60°- 20°)+ sin2(60°+20°)= 3/2

[(sin2θ+ sin2(60°-θ)+ sin2(60°+θ)= 3/2 irrespective of θ, as can easily be proved by expanding sin2(60°-θ and sin2(60°+θ) using the formula for sin(A+B) and sin(A-B).]
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