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Question

If f(θ)=sin2θ+sin2(θ+2π3)+sin2(θ+4π3),thenf(π15)


A

23

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B

32

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C

12

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D

13

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Solution

The correct option is B

32


f(θ)=sin2θ+sin2(θ+2π3)+sin2(θ+4π3)=1cos2θ2+1cos(2θ+4π3)2+1cos(2θ+8π3)2=12(3cos2θcos(4π3+2θ)cos(8π3+2θ))=12(3cos2θ2cos(2π+2θ)cos2π3)=12(3cos2θ+cos2θ)f(θ)=32f(π15)=32


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