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You may solve the problem either transforming the product`sin theta*cos theta` into a sum of two like trigonometric function or writing the double of the angle `theta` as the sum `theta + theta` .
I'll focus on the first solving strategy such that:
`sin theta*cos theta = (1/2)[sin(theta - theta) + sin (theta + theta)]`
`sin theta*cos theta = (1/2)[sin(0) + sin (2theta)]`
Since sin 0 = 0 => `2sin theta*cos theta = sin (2theta)`
This last line proves the given identity.
the sine addition formula is
sin(A+B) = sin(A)cos(B) + sin(B)cos(A)
if A = B then
sin(A+A) = sin(A)cos(A) + sin(A)cos(A)
this simplifies to
sin(2A) = 2 sin(A)cos(A) which proves your question.
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