prove that sin6x+sin2x+2sin4x/sin7x+sin3x+2sin5x=sin4x/sin5x
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Keep using the identity sin(A + B) = sinAcosB + cosAsinB and sin(A - B) = sinAcosB - cosAsinB.
sin(x) + sin(3x) + sin(5x) + sin(7x)
= sin(2x-x) + sin(2x+x) + sin(6x-x) + sin(6x+x)
= sin(2x)cos(x) - cos(2x)sin(x) + sin(2x)cos(x) + cos(2x)sin(x) + sin(6x)cos(x) - cos(6x)sin(x) + sin(6x)cos(x) + cos(6x)sin(x)
= 2cosx[sin(2x) + sin(6x)]
=2cosx[sin(4x-2x) + sin(4x+2x)]
=2cosx[sin(4x)cos(2x) - cos(4x)sin(2x) + sin(4x)cos(2x) + sin(4x)cos(2x)]
=4cos(x)cos(2x)sin(4x)
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