Simplify into first power of cosine: cos^2(x)*sin^4(x)?

cos^2(x)sin^4(x)

= cos^2(x)[sin^2(x)]^2

= cos^2(x)[1 - cos^2(x)]^2

= cos^2(x)[1 - 2cos^2(x) + cos^4(x)]

= cos^2(x) - 2cos^4(x) + cos^6(x) .......... (1)

Now we need to do a bit of work with cos(2x), cos(4x) and cos(6x) :

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

Therefore, cos^2(x) = [cos(2x) + 1]/2 .......... (2)

cos(4x) = cos[2(2x)]

= 2cos^2(2x) - 1

= 2[2cos^2(x) - 1]^2 - 1

= 2[4cos^4(x) - 4cos^2(x) + 1] - 1

= 8cos^4(x) - 8cos^2(x) + 1

Substitute from (2) :

cos(4x) = 8cos^4(x) - 8[cos(2x) + 1]/2 + 1

= 8cos^4(x) - 4cos(2x) - 3

Therefore, cos^4(x) = [cos(4x) + 4cos(2x) + 3]/8 .......... (3)

cos(3x) = cos(2x + x)

= cos(2x)cos(x) - sin(2x)sin(x)

= [2cos^2(x) - 1]cos(x) - 2sin^2(x)cos(x)

= 2cos^3(x) - cos(x) - 2[1 - cos^2(x)]cos(x)

= 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)

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

Now, cos(6x) = cos[2(3x)]

= 2cos^2(3x) - 1

= 2[4cos^3(x) - 3cos(x)]^2 - 1

= 2[16 cos^6(x) - 24cos^4(x) + 9cos^2(x)] - 1

= 32cos^6(x) - 48cos^4(x) + 18cos^2(x) - 1

Substitute from (2) and (3) :

cos(6x) = 32cos^6(x) - 48[cos(4x) + 4cos(2x) + 3]/8 + 18[cos(2x) + 1]/2 - 1

= 32cos^6(x) - 6cos(4x) - 24cos(2x) - 18 + 9cos(2x) + 9 - 1

= 32cos^6(x) - 6cos(4x) - 15cos(2x) - 10

Therefore, cos^6(x) = [cos(6x) + 6cos(4x) + 15cos(2x) +10]/32 .......... (4)

Now substitute (2), (3) and (4) into (1) :

cos^2(x)sin^4(x)

= [cos(2x) + 1]/2 - 2[cos(4x) + 4cos(2x) + 3]/8 +

[cos(6x) + 6cos(4x) + 15cos(2x) + 10]/32

= (1/32)[16cos(2x) + 16 - 8cos(4x) - 32cos(2x) - 24 +

cos(6x) + 6cos(4x) + 15cos(2x) + 10]

= (1/32)[cos(6x) - 2cos(4x) - cos(2x) + 2] (= final answer)

cos(2x) = 2cos^2(x) - 1 = (1 - 2sin^2x). So:

cos^2(x) = 1/2 (cos(2x) + 1)

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

Now just plug in:

= 1/8 (cos(2x) + 1)(cos(2x) - 1)^2

You can further simplify this - as you can see you'll end up with a cos^3(2x) term... and you'll get a nasty long expression.