# Answer to Question #82 in Trigonometry for manank

Question #82

I don't know how to prove that:

2(sin6+cos6)-3(sin4+cos4-1)=0

2(sin6+cos6)-3(sin4+cos4-1)=0

Expert's answer

Try to use this formulas for your problem:

sin3x = 3sinx - 4(sin^3(x))

cos3x = 4(cos^3(x)) - 3cosx

sin2x = 2sinxcosx

cos2x = cos^2(x) - sin^2(x) = 2(cos^2(x)) - 1 = 1 - 2(sin^2(x))

sin3x = 3sinx - 4(sin^3(x))

cos3x = 4(cos^3(x)) - 3cosx

sin2x = 2sinxcosx

cos2x = cos^2(x) - sin^2(x) = 2(cos^2(x)) - 1 = 1 - 2(sin^2(x))

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