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

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

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))

Answer 2

Answer:
`((1)/(8))`)(3 - 4cos 2x + cos 4x)

Explanation:

sin⁴ x

= sin²x * sin²x

=
`(1 - cos 2x)/(2) * (1 - cos 2x)/(2)`

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

=
`((1)/(4))`(1 - 2cos 2x + cos²2x)

=
`((1)/(4))`(1 - 2cos 2x +
`(1 + cos 4x)/(2)`)

=
`((1)/(4))`(1 - 2cos 2x +
`(1)/(2)` +
`(cos 4x)/(2)`)

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

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