Find an identity for cos (4t) in terms of cos (t)

So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.

Cos (0) = 1

Cos (pi/2) = 0

Cos (pi) = - 1

Cos (3pi/2) = 0

Cos (2pi) = 1

Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).

But you can always reduce it.

Cos (0) = 1

Cos (4pi/2) = cos (2pi) = 1

Cos (4pi) = Cos (2pi) = 1 (Any multiple of 2pi = = 1)

etc ...

the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.

Now that you know that, the identities of Cosine are another beast, but mathematically.

You have.

Cos (2*2t) = Cos^2 (2t) - Sin^2 (2t)

Sin^2 (t) = - Cos^2 (t) + 1 ... (all A^2+B^2=C^2)

Cos (2*2t) = Cos^2 (t) - (-Cos^2 (t) + 1)

Cos (2*2t) = 2Cos^2 (2t) - 1

2Cos^2 (2t) - 1 = 2 (Cos^2 (t) - Sin^2 (t)) ^2 - 1

(same thing as above but done twice because it's cos ^2 now)

convert sin^2

2Cos^2 (2t) - 1 = 2 (Cos^2 (t) + Cos^2 (t) - 1) ^2 - 1

2 (2Cos^2 (t) - 1) ^2 - 1

2 (2Cos^2 (t) - 1) (2Cos^2 (t) - 1) - 1

2 (4Cos^4 (t) - 2 (2Cos^2 (t)) + 1) - 1

Distribute

8Cos^4 (t) - 8Cos^2 (t) + 1

Cos (4t) = 8Cos^4-8Cos^2 (t) + - 1