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Question

Evaluate sin263o+sin227ocos217o+cos273o.

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Solution

We know that sin2θ+cos2θ=1
& cos2Asin2B=cos(A+B)cos(AB)
Now, sin263o=1cos263o
& cos273o=1sin273o
So,
1cos263o+sin227ocos217o+1sin273o
=1(cos263osin227o)1+(cos217osin273o)
=1cos(63+27)cos(6327)o1+cos(17+73)ocos(1773)o
=1cos(90)ocos(36)o1+cos(90)ocos(56)o
But cos(90o)=0
So,
sin263o+sin227ocos217o+cos273o=1


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