Question

How do you simplify 2cos^2(4θ)-1 using a double-angle formula?

Answer 1

`2 \cos^2 (4 \theta) - 1 = \cos(8 \theta)`

Explanation:

There are several double angle formulas for cosine. Usually the preferred one is the one that turns a cosine into another cosine:

`\cos 2x = 2 \cos ^2 x - 1`

We can actually take this problem in two directions. The simplest way is to say `x=4\theta` so we get

`\cos(8 \theta) = 2 \cos^2 (4 \theta) - 1`

which is pretty simplified.

The usual way to go is to get this in terms of `\cos theta`. We start by letting `x=2\theta.`

`2 \cos^2 (4 theta ) - 1`

`= 2 \cos^2 (2 (2\theta) ) - 1`

`= 2 (2\cos^2 (2 \theta) - 1)^2 - 1`

`= 2 ( 2 ( 2 \cos ^2 \theta -1)^2 -1)^2 -1`

`= 128 \cos^8 \theta - 256 \cos ^6 theta + 160 cos^4 theta - 32 cos^2 theta+ 1`

If we set `x=\cos theta` we'd have the eighth Chebyshev polynomial of the first kind, `T_8(x)`, satisfying

`cos(8x) = T_8(\cos x)`

I'm guessing the first way was probably what they're after.