Represent `cos 6 theta` in terms of `cos theta`.

Answer

`32 cos^(6) theta - 48 cos^(4) theta +18 cos^(2) theta - 1`

Explanation

`cos 6 theta = "Real part of"(cos theta + i sin theta)^(6)`
`= .^(6)C_(0)cos^(6)theta - .^(6)C_(2)cos^(4) theta sin^(2) theta + .^(6)C_(4) cos^(2)theta sin^(4) theta-.^(6)C_(6)sin^(6) theta`
`= cos^(6)theta - 15cos^(4)theta(1-cos^(2)theta)+15cos^(2)theta(1-cos^(2)theta)^(2)-(1-cos^(2)theta)^(3)`
`= cos^(6)theta - 15 cos^(4)theta(1-cos^(2)theta)+15cos^(2)theta(1-2cos^(2)theta+cos^(4)theta)-(1-3cos^(2)theta+3cos^(4)theta-cos^(6)theta)`
`= 32 cos^(6) theta - 48 cos^(4)theta + 18 cos^(2) theta - 1`