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

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Answer 1 · 2018-04-21T09:00:34

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

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.