SOLUTION: prove the following identity sin(x)cos(y)=1/2[sin(x-y)-sin(x+y)] Thank you to the one that helps out.

Question 22705: prove the following identity
sin(x)cos(y)=1/2[sin(x-y)-sin(x+y)]
Thank you to the one that helps out.

Found 2 solutions by khwang, mahan6221:
Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
prove the following identity
sin(x)cos(y)=1/2[sin(x-y)+ sin(x+y)]
THE MIDDLE SIGN SHOULD BE +
-------
Use sin(x+y) = sin x cos y + cos x sin y
and sin(x-y) = sin x cos y - cos x sin y

Kenny
Answer by mahan6221(2) (Show Source): You can put this solution on YOUR website!
This relation is not true because: X=Y=(Pi/4)
(1/2) [ sin(x-y) - sin(x+y)] = (1/2)[sin(0)-sin(Pi/2)] = -(1/2)
sin(x)cos(y) = sin(Pi/2)cos(Pi/2) = (1/2)
true is :-(1/2) [SIN(X-Y)-SIN(X+Y)]= COS(X)SIN(Y)
solve:
SIN(X-Y) � SIN(X+Y) =
=SIN(X)COS(Y)-COS(X)SIN(Y) � ( SIN(X)COS(Y)+COS(X)SIN(Y))
= -2COS(X)SIN(Y)
=> SIN(X-Y)-SIN(X+Y)= -2COS(X)SIN(Y)
=> -(1/2) [SIN(X-Y)-SIN(X+Y)]= COS(X)SIN(Y)

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