Hm, (1-cos^1x*cos^1x)*cos^1x*cos^1x*cos^1x*cos^1x ? What's the chapter about?

OK, let's call sin^2 x "s^2" formula: s^2 = (1-c2)/2 formula: c^2 = (1+c2)/2 we have: (s^2)(c^2)(c^2) = (1-c2)(1+c2)(1+c2)/2*2 and just multiply it out?

We don't do other peoples homework on Answerbag

We are forewarned, so let's mult by 32 32*y = 8*(1-c2)(1+c2)(1+c2) let's do (1+c2)(1+c2) = 1+2c2+cc2 let's call cos^2 "cc" = 1+2c2+(1+c2*2)/2 = (2+4c2+1+c4)/2 = (3+4c2+c4)/2 so 32*y = 4*(1-c2)(3+4c2+c4) let's do c2(3+4c2+c4) = (3c2+4cc2+c2c4) = (3c2+4(1+c2)/2+c2c4) = (3c2+2(1+c2)+c2c4) = (3c2+2+2c2+c2c4) ...

sin^2(x)cos^4(x) = A * B A = sin^2(x) = [1 - cos(2x)]/2 B = cos^4(x) = [3 + 4cos(2x) + cos(4x)]/8 A * B = {[3 + 4cos(2x) + cos(4x)] - [3cos(2x) + 4cos^2(2x) + cos(2x)cos(4x)]}/16 cos^2(2x) = [1 + cos(4x)]/2 cos(2x)cos(4x) = C * D C = cos(2x) = [exp^{j2x} + exp^{-j2x}]/2 D = cos(4x) = [exp^{j4x} + exp^{-j4x}]/2 C * D = [1/2][exp^{j(2x+4x)} + exp^{j(2x-4x)} + exp^{j(-2x+4x)} + exp^{j(-2x-4x)}]/2 C * D = [1/2][exp^{j6x} + exp^{-j2x} + exp^{j2x} + exp {-j6x}]/2 C * D = [1/2] [ <[exp^{j6x} + exp{-j6x}]/2> + <[exp^{j2x} + exp^{-j2x}]/2> ] C * D = [ cos(6x) + cos(2x) ]/2 A * B = { [3 + 4cos(2x) + cos(4x)] - [ 3cos(2x) + 2{1 + cos(4x)} + cos(6x)/2 + cos(2x)/2] }/16 A * B = { 3 + 4cos(2x) + cos(4x) -3cos(2x) - 2{1 + cos(4x)} - cos(6x)/2 - cos(2x)/2 }/16 A * B = { 3 + 4cos(2x) + cos(4x) -3cos(2x) - 2 -2cos(4x)} - cos(6x)/2 - cos(2x)/2] }/16 A * B = { [3-2] + [4 - 3 - (1/2)]cos(2x) + [1 - 2]cos(4x) - cos(6x)/2 } / 16 A * B = { 1 + cos(2x)/2 - cos(4x) - cos(6x)/2 } / 16 A * B = { 2 + cos(2x) - 2cos(4x) - cos(6x) } / 32 It is wrong to give H.S. students this problem!!!!!! It uses Euler's identity and this is college material.