If sinθ+cosθ=12, then 16(sin(2θ)+cos(4θ)+sin(6θ)) is equal to
A
27
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B
−23
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C
−27
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D
23
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Solution
The correct option is B−23 Given: sinθ+cosθ=12
squaring on both sides, we get sin2θ=14−1 ∴sin2θ=−34
and cos4θ=1−2(sin2θ)2 ⇒cos4θ=1−98 ∴cos4θ=−18
and sin6θ=3sin2θ−4sin32θ ⇒sin6θ=−94+2716 ∴sin6θ=−916
Now, 16(sin(2θ)+cos(4θ)+sin(6θ)) =16(−34−18−916) =−23