Question

How do you use the double angle or half angle formulas to derive cos(4x) in terms of cos x?

Answer 1

`y^' = -16sin(x)cos^3(x) +16sin^3(x)cos(x)`

Explanation:

Knowing that
`cos(2u) = cos^2(u) - sin^2(u) = 1 - 2sin^2(u)`
`sin(2u) = 2sin(u)cos(u)`

So,

`cos(4x) = 1 - 2sin^2(2x)`
`1 - 2sin^2(2x) = 1 - 2*(2sin(x)cos(x))^2`

So `cos(4x) = 1 - 4sin^2(x)cos^2(x)`

We know that the constant is irrelevant for derivatives so we can say that for the function `y = f(x)`, we have

`y = -4sin^2(x)cos^2(x)`

So then you can differentiate in any way you want, using logarithms (after taking that minus four out of the way, remembering to tack it on later), we have

`ln(y) = 2ln(sin(x)) + 2ln(cos(x))`
`y^'/y = 2cos(x)/sin(x) + 2(-sin(x))/cos(x)`
`y^' = 4sin(x)cos^3(x) -4sin^3(x)cos(x)`

Remembering to put the -4 back in

`y^' = -16sin(x)cos^3(x) +16sin^3(x)cos(x)`

Answer 2

Refer to explanation

Explanation:

Well it is

#cos(4x) = 2cos^2(2x) - 1 = 2[2cos^2(x) - 1]^2 - 1 = 8cos^4(x) - 8cos^2(x) + 1#