Question

20 POINTS PRECALC PLZ HELP MEH

Rewrite sin^4 x as a sum of first powers of the cosines of multiple-angles.

idk what its asking for...

Answer 1

Answer: $(\frac{1}{8})$)(3 - 4cos 2x + cos 4x)

Step-by-step explanation:

sin⁴ x

= sin²x * sin²x

= $\frac{1 - cos 2x}{2} * \frac{1 - cos 2x}{2}$

= $\frac{1 - 2cos 2x + cos^{2}2x} {4}$

= $(\frac{1}{4})$(1 - 2cos 2x + cos²2x)

= $(\frac{1}{4})$(1 - 2cos 2x + $\frac{1 + cos 4x}{2}$)

= $(\frac{1}{4})$(1 - 2cos 2x + $\frac{1}{2}$ + $\frac{cos 4x}{2}$)

= $(\frac{1}{4})$($\frac{3}{2}$ - $\frac{4cos 2x}{2}$ + $\frac{cos 4x}{2}$)

= $(\frac{1}{8})$(3 - 4cos 2x + cos 4x)

Answer 2

Answer:

(3 - 4cos(2x) + cos(4x))/8

Step-by-step explanation:

You know that ...

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

Then

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

Squaring this, we get ...

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

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

Using the above relation for sin^2, we have ...

... sin(x)^4 = (1/4)(2 - 2cos(2x) - (1-cos(4x))/2)

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