___cosb=1/2[sin(a+b) + sin(a-b)

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24-10-2019
Mathematics

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___cosb=1/2[sin(a+b) + sin(a-b)

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lexi4688 lexi4688 · Answer 1 · 2019-10-24T04:13:21+02:00

Use the addition and subtraction theorem:

sin (A + B) = sin A + cos B
sin (A - B) = sin A - cos B

From there it's just a matter of simplification.

sin A cos B = 1/2 [ sin (A + B) + sin (A - B) ]
sin A cos B = 1/2 [ sin A cos B + cos A sin B + sin (A - B) ]
sin A cos B = 1/2 [ sin A cos B + cos A sin B + sin A cos B - cos A sin B ]
sin A cos B = 1/2 [ 2 sin A cos B ]
sin A cos B = sin A cos B