# Help with Proof!Watch

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#1
Anyone help me prove this identity?

sinx - 2sin3x + sin5x = 2sinx(cos4x - cos2x)

Thanks
0
15 years ago
#2
(Original post by Hoofbeat)
Anyone help me prove this identity?

sinx - 2sin3x + sin5x = 2sinx(cos4x - cos2x)

Thanks
sinx - 2sin3x + sin5x = sinx - sin3x + sin5x - sin3x.

Now sinA - sinB = 2sin((A-B)/2)cos((A+B)/2).

Using this twice on the above gives:

2sin(-x)cos(2x) + 2sin(x)cos(4x).

Now use sin(-x) = -sin(x) and factorise to get 2sin(x)(cos(4x)-cos(2x)).

Chris
0
15 years ago
#3
sinx - 2sin3x + sin5x = 2sinx(cos4x - cos2x)
l.h.s = sinx - 2(sin2xcosx + cos2xsinx) + sin4xcosx + cos4xsinx
sinx - 2.2sinxcos^2x - 2sinxcos2x + 2sin2xcos2xcosx + cos4xsinx
sinx - 4sinxcos^2x - 2sinxcos2x +4sinxcos^2xcos2x + cos4xsinx
sinx(1 - 4cos^2x + 4cos^2xcos2x) + sinxcos4x - 2sinxcos2x
sinx(1 - 4cos^2x(1 - cos2x)) + sinxcos4x - 2sinxcos2x
sinx(1-8sin^2xcos^2x) + sinxcos4x - 2sinxcos2x
sinx(1 - 2(2sinxcosx)^2) + sinxcos4x - 2sinxcos2x
sinx(1 - 2sin^2(2x)) + sinxcos4x - 2sinxcos2x
sinxcos4x + sinxcos4x - 2sinxcos2x
2sinxcos4x - 2sinxcos2x
2sinx(cos4x - cos2x) = r.h.s
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