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Sorry - not true.
sin2X = 2sinXcosX, so sin2X - sinXcosX = sinXcosX, which is not the same thing as cosX.
sin(2x) = 2*sin(x)*cos(x), therefore:
sin(2x) - cos(x)*sin(x) = sin(x)*cos(x)
What you get is a simple equation:
sin(x)*cos(x) = cos(x) or
cos(x)*(sin(x)-1)=0
This equation holds if:
(a) cos(x)=0 => x = pi/2 + pi*n; n is any whole number, or
(b) sin(x)=1 => x = pi/2 + 2*pi*k; k is any whole number
The answer (b) is a subset of and contained in the answer (a), therefore just answer (a) would be sufficient. Hope this helps.
Yes, the answer would depend on what the question asks.
If it's a proof the identify is false, and if it's an equation then the post of harum is the way to go.
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