what is (cos^4)x & (sin^4)x equal to?

Answer 1

cos^(4)x = cos²x . cos²x = cosx.cosx.cosx.cosx
sin^(4)x = sin²x.sin²x = sinx.sinx.sinx.sinx

Answer 2

You could also end up with these

cos⁴x = {cos(4x) + 4.cos(2x) + 3}/8 and
sin⁴x = {cos(4x) - 4.cos(2x) + 3}/8

by using complex numbers and binomial expansions; easy, but tedious once you know how.


The normal way at high school level is to use the formula for cos(2x)

i.e. cos 2x = 2.cos² x - 1 ............. * and rearrange it:

cos² x = ½(cos 2x + 1), then square both sides to get
cos⁴ x = ¼(cos 2x + 1)² = ¼(cos² 2x + 2.cos x +1)

Use * again with 2x instead of x cos 4x = 2.cos² 2x - 1,
so cos² 2x = ½(cos 4x + 1) ad substitute to get
cos⁴ x = ¼(cos² 2x + 2.cos x +1) = ¼{ ½(cos 4x + 1) + 2.cos x + 1} and this gives the same answer as near the top.

The sin⁴ x formula is got in a similar fashion, but using
cos 2x = 1 - 2.sin² x rearranged as sin² x = ½(1 - cos 2x) then squared and so on.

Answer 3

Depends very much on what you want to do with the expression after. You can write each term as (cos^2(X))^2 and develop cos^2 with the formula containing 2*X
cos^2(x)+sin^2(x)=1
cos^2(x)-sin^2(x)=cos(2x)
=>
cos^2(x)=(1+cos(2x))/2
cos^4(x)=((1+cos(2x))/2)^2
and so on...

Answer 4

((cos^2)x + (sin^2)x) = 1 ------> ((cos^2)x + (sin^2)x) ^ 2 = 1 =
(cos^4)x + (sin^4)x + 2*(sin^2)x*(cos^2)x ------>
(cos^4)x + (sin^4)x = 1 - 2*(sin^2)x*(cos^2)x

Answer 5

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