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How to develop a formula for cos3x?

Develop a formula for cos3x in terms of cosx?

  • Hello, I need some help on how to express cos 3x in terms of cos x. I realize that you need to use double angle formulas, but I'm not able to group the terms properly in the second step and need some help in finding the final answer. Thanks in advance!

  • Answer:

    cos3x = cos (2x+ x) =cos2x*cosx - sin2x*sinx =(2*cos^2 (x) - 1) *cosx - 2sinx*cosx*sinx { since cos2x = 2cos^2 (X) -1 , sin2x = 2*sinx*cosx } =2 cos^3 (x) - cos x - 2 cosx *sin^2 (x) = 2 cos^3 (x) - cos x (1 + 2 sin^2 (x) ) = 2 cos^3 (x) - cos x (1 + 2 ( 1 - cos^2 (x) )) = 2 cos^3 (x) - cos x ( 1+ 2 - 2*cos^2(x) ) =2 cos^3 (x) - cos x (3 -2*cos^2(x) ) =2 cos^3 (x) - 3cos x + 2*cos^3(x) =4 cos^3 (x) - 3cos x

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cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x) Now figure out what cos(2x) and sin(2x) are equal to cos(2x) = cos^2(x)-sin^2(x) sin(2x) = 2sin(x)cos(x) Put those into our equation, and we have: (cos^2(x)-sin^2(x))*cos(x) - (2sin(x)cos(x))*sin(x) We know that sin^2(x) = 1-cos^2(x) so plug that in everywhere possible (2cos^2(x)-1)*cos(x) - (2*(1-cos^2(x))*cos(x) cos(x)*(2cos^2(x)-1 - (1-cos^2(x))) cos(x)*(3cos^2(x)-2) 3cos^3(x)-2cos(x) Umm I hope I did that right.. lol

Andrew

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