Question 188285
Verify that the equation is an identity.
{{{sin2x/2sinx = cos^2 x/2 - sin^2 x/2}}}
I tried solving on the right side
 = [(1 + cos x) / 2] - [(1 - cos x) / 2]
 = [1 + cosx - 1 + cos x]/2
 = 2cosx/2
 = cosx 
I have no clue how to get sin2x over 2sinx from this and I don't know what I did wrong.
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Picking up where you left off:
sin(2x) = 2sin(x)cos(x)
Dividing that by 2sin(x) --> cos(x)