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Question

If sinθsin2(π8+θ2)sin2(π8θ2)=k (θ2nπ), then value of 2k2 is

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Solution

sinθsin2(π8+θ2)sin2(π8θ2)=kk=sinθsin(π8+θ2+π8θ2)sin(π8+θ2π8+θ2)k=sinθsinπ4sinθ=22k2=4

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