What is 4 cos^4(x) + 4 sin^4(x) - cos(4x) after simplification?

?2017-12-12T03:17:17Z

Favorite Answer

4 cos^4(x) + 4 sin^4(x) - cos(4x) =
4 cos^4(x) + 4 sin^4(x) - [2cos^2(2x)-1] =
1+4 cos^4(x) + 4 sin^4(x) - 2cos^2(2x) =
1+4 cos^4(x) + 4 sin^4(x) - 2cos(2x)*cos(2x) =
1+4 cos^4(x) + 4 sin^4(x) - 2 [2cos^2(x)-1][2cos^2(x)-1] =
1+4 cos^4(x) + 4 sin^4(x) - 2 [4cos^4(x)-4cos^2(x)+1] =
-1 + 4 sin^4(x) - 4 cos^4(x)+8 cos^2(x) =
-1 + 4 [sin^4(x) - cos^4(x)]+8 cos^2(x) =
-1 + 4 [sin^2(x)+cos^2(x)][sin^2(x-cos^2(x)]+8 cos^2(x) =
-1 + 4 [sin^2(x)-cos^2(x)]+8 cos^2(x) =
-1 + 4 [sin^2(x)+cos^2(x)] =
-1 + 4 = 3

Captain Matticus, LandPiratesInc2017-12-12T03:45:09Z

4 * (cos(x)^4+ sin(x)^4) - cos(4x) =>
4 * (((1/2) * (1 + cos(2x)))^2 + ((1/2) * (1 - cos(2x)))^2) - cos(4x) =>
4 * ((1/2)^2 * (1 + cos(2x))^2 + (1/2)^2 * (1 - cos(2x))^2) - cos(4x) =>
4 * (1/4) * (1 + 2 * cos(2x) + cos(2x)^2 + 1 - 2 * cos(2x) + cos(2x)^2) - cos(4x) =>
1 * (2 + 2 * cos(2x)^2) - cos(4x) =>
2 + 2 * cos(2x)^2 - cos(4x) =>
2 + 2 * (1/2) * (1 + cos(4x)) - cos(4x) =>
2 + 1 + cos(4x) - cos(4x) =>
3

L. E. Gant2017-12-12T02:15:33Z

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