Link to original question: https://web2.0calc.com/questions/simplify-each-expression_1
CPhill answered but he dropped the 2 on the -2cos4x.
Simplify: Sorry Adam, no LaTex !!
sin(x)^4 - sin(x)^2 cos(x)^2 - 2 cos(x)^4 =
Express sin(x)^2 in terms of cosine via the Pythagorean identity.
sin(x)^2 = 1 - cos(x)^2:
-2 cos(x)^4 - cos(x)^2 1 - cos(x)^2 + sin(x)^4 =
Expand -cos(x)^2 (1 - cos(x)^2).
-cos(x)^2 (1 - cos(x)^2) = cos(x)^4 - cos(x)^2:
-2 cos(x)^4 + cos(x)^4 - cos(x)^2 + sin(x)^4 =
Express sin(x)^4 in terms of cosine via the Pythagorean identity.
sin(x)^4 = (sin(x)^2)^2 = (1 - cos(x)^2)^2:
-2 cos(x)^4 - cos(x)^2 + cos(x)^4 + (1 - cos(x)^2)^2 =
Expand (1 - cos(x)^2)^2.
(1 - cos(x)^2)^2 = 1 - 2 cos(x)^2 + cos(x)^4:
-2 cos(x)^4 - cos(x)^2 + cos(x)^4 + 1 - 2 cos(x)^2 + cos(x)^4 =
Evaluate -2 cos(x)^4 - cos(x)^2 + cos(x)^4 + 1 - 2 cos(x)^2 + cos(x)^4.
-2 cos(x)^4 - cos(x)^2 + cos(x)^4 + 1 - 2 cos(x)^2 + cos(x)^4
= 1 - 3cos(x)^2
Sorry for the earlier error!!!
This becomes a little easier if we factor up front
sin^4x -sin^2xcos^2x -2cos^4x factor as
(sin^2x - 2cos^2x) (sin^2x + cos^2x)
(sin^2x - 2cos^2x) (1)
sin^2x - 2cos^2x
sin^2x - 2 (1 - sin^2x)
3sin^2x - 2 or
3(1 - cos^2x) - 2
3 - 3cos^2x - 2
1 - 3cos^2x