20 points precalc plz meh rewrite sin^4 x as a sum of first powers of the cosines of multiple-angles. idk what its asking

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

Step-by-step explanation:

sin⁴ x

= sin²x * sin²x

=

=

= (1 - 2cos 2x + cos²2x)

= (1 - 2cos 2x + )

= (1 - 2cos 2x + + )

= ( - + )

= (3 - 4cos 2x + cos 4x)

(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))

6, 10

9, -13

12, -21

18, -23

step-by-step explanation:

you just add 6 to the x-coordinate and subtract 10 from the y-coordinate.