Answer to Question #5910 in Trigonometry for Kewal

Question

sin3A+sin2A-sinA = 4sinAcosA/2cos3A/2

Answer 1

we will use "sum to product" formula for sum of sines
sin(x)+sin(y)=2sin((x+y)/2) cos((x-y)/2)

whence sin(3A)-sin(A)=sin(3A)+sin(-A)=2sin(A)cos(2A)

hence we obtain sin3A+sin2A-sinA=2sin(A)cos(2A)+sin(2A)
Recall that sin(2A)=2sin(A)cos(A)

therefore

sin3A+sin2A-sinA=2sin(A)cos(2A)+sin(2A)=
=2sin(A)cos(2A)+2sin(A)cos(A)=2sin(A)(cos(2A)+cos(A))

there is the same "sum to product" formula for sum of cosines
cos(x)+cos(y)=2cos((x+y)/2)cos((x-y)/2)

therefore

2sin(A)(cos(2A)+cos(A))= 2sin(A)*2*cos(3*A/2)*cos(A/2)=
=4sinAcosA/2cos3A/2 and that equals to the right part of the initial equality, so that's it.

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