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Question

The value of 13k=11sin(π4+(k1)π6)sin(π4+kπ6) is equal to

A
33
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B
2(33)
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C
2(31)
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D
2(2+3)
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Solution

The correct option is C 2(31)
Let
S=13k=11sin(π4+(k1)π6)sin(π4+kπ6)

Assuming π4=θ,α=π6
=13k=11sin(θ+kα)sin(θ+(k1)α)=1sinα13k=1sinαsin(θ+kα)sin(θ+(k1)α)=1sinα13k=1sin[(θ+kα)(θ+(k1)α)]sin(θ+kα)sin(θ+(k1)α)=1sinπ613k=1cot(θ+(k1)α)cot(θ+kα)=213k=1cot(θ+(k1)α)cot(θ+kα)=2[cotθcot(θ+13α)]=2[cotπ4cot(π4+13π6)]=2[1cot(5π12)]=2[1cot75]=2[1(23)]=2(31)

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