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verify algebraically these two equations are equivalent: (sinx+sin3x)2sin2x = cosx
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Sorry, (sinx+sin3x)/(2sin2x) = cosx
(sinx + sin(2x + x)) = 2(cosx)sin2x sinx + sin2xcosx + cos2xsinx = 2cosxsin2x sinx + cos2xsinx = cosxsin2x cos2x = (cosx)^2 - (sinx)^2 <- identity sin2x = 2sinxcosx <- identity sinx + sinx(cosx)^2 - (sinx)^3 = 2sinx(cosx)^2 1 + (cosx)^2 - (sinx)^2 = 2(cosx)^2 1-(sinx)^2 = (cosx)^2 <- identity so : (cosx)^2 + (cosx)^2 = 2(cosx)^2 maybe there is a shorter way
Thank You So Much. I'm staring at my double angle identities right now and was just having a hard time make it through line by line.
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