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The expression 3 sin43π2−α+sin4(3π+α)−2sin6π2+α+sin6(5π−α)=

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a

0

b

1

c

3

d

none of these

answer is B.

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Detailed Solution

sin43π2−α=cos4αsin4(3π+α)=sin4αsin6π2+α=cos6α,sin6(5π−α)=sin6αTherefore, the given quantity is =3cos4α+sin4α−2cos6α+sin6α=31−2sin2αcos2α−21−3sin2αcos2α=1

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