21(SIN2θ+SIN4θ)3+2 SIN4θ-2COS2θ=(COSθ+COS2θ)2+2COS2θ-2COS2θ=13=1(REPLACING EARLIER VALUES) I COULDN'T FIND A BRACKET IN YOUR OPTIONS TO MEAN WHOLE CUBE- SIN4θ+SIN2θ WHOLE CUBE. & "COSθ3" WAS A PRINTING ERROR.
If cos θ + cos2 θ = 1, then
sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ – 2 =
(A) 0
(B) 1
(C) 2
(D) 3
Thanx in advance...
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UP 0 DOWN 0 1 3
3 Answers
21cosθ+cos2θ=1
→1-cos2θ=cosθ
→sin2θ=cosθ
→sin4θ=cos2θ
sin12θ+3sin10θ+3sin8θ+sin6θ+2sin4θ+2sin2θ-2=sin4θ+sin2θ3 +2sin4θ-2cos2θ=cos2θ+cosθ3+2cos2θ-2cos2θ=13=1
337Thank you for your reply. But I could not get the folloeing:
"sin2θ3" and "cosθ3"
Thanx...
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