Question

cos4x= 1/3 then cos^4x+sin^4x=?

Answer 1

Is this:
\[\cos^4(x) = 1/3\textrm{ ?}\]

Answer 2

no cos (4x)= 1/3

Answer 3

for the 1st one and the other two are to the power of

Answer 4

Solve for x then:
\[\cos(4x)=1/3 \implies x=\frac{\arccos(1/3)}{4}\approx 17.63^0\]
Then you just need to evaluate sin^4(17.63) and cos^4(17.63). Make sure your calculator is in degree mode.

Answer 5

mk ty

Answer 6

Woahhh!! your problem!!! but I got it
cos (4x) = 1-2sin^2(2x) =1/3
therefore 1-1/3 = 2sin^2(2x)
or sin^2(2x) = 1/3
moreover \(sin^2(2x) = 4sin^2xcos^2x = \dfrac{1}{3}\) so, I have \(2sin^2x cos^2x = \dfrac{1}{6}\) (a) Let it here

Answer 7

and \((sin^2x)^2\) = sin^4 x, right?

Answer 8

so, \(sin^4x + cos^4x = (sin^2x)^2 +(cos^2x)^2 +2sin^2xcos^2x-2sin^2xcos^2x\)
the first 3 terms forms the form of \( (sin^2x +cos^2x)^2 \) = 1
so, it is \(1 - 2 sin^2x cos^2x\) got me so far?

Answer 9

yep

Answer 10

be back to my previous comment, when I calculate 2sin^2 cos^2 = 1/6 (a)

Answer 11

so, sin^4 +cos^4 = 1-1/6 = 5/6

Answer 12

lalalalala.... hehehehe life is beaaauuutiful

Answer 13

oh tytyty lemme reread everything thankss this made more sense