Question

Verify the identity.

cos 4u = cos^2(2u) - sin^2(2u)

Answer 1

Can you verify it using the Double Angle Formula for Cosine?
If so then we really only have one small step to apply.

Answer 2

What is the formula? @zepdrix

Answer 3

\[\large \cos 2\theta=\cos^2 \theta-\sin^2 \theta\]

Answer 4

ok so how would I use that exactly?

Answer 5

\[\large \cos (4u)=\cos (2\cdot\color{orangered}{2u})\]Let,\[\large \color{orangered}{2u=\theta}\]\[\large \cos(2\cdot \color{orangered}{2u})=\cos(2 \color{orangered}{\theta})\]Using the Double Angle Formula for Cosine gives us,\[\large \cos(2 \color{orangered}{\theta})=\cos^2 \color{orangered}{\theta}-\sin^2 \color{orangered}{\theta}\]Plugging back in our original value gives us,\[\large \color{black}{\cos^2 2u-\sin^2 2u}\]

Maybe something like that? :o

Answer 6

cos(4 u) = cos(2 u)^2-sin(2 u)^2

cos(4 u) = cos(2 u)^2-(1-cos(4 u))/2

cos(4 u) = cos(2 u)^2-1/2-(cos(4 u))/2

cos(4 u) = (1+cos(4 u))/2-(1/2-1/2 cos(4 u))

cos(4 u) = 1/2+(cos(4 u))/2-(1/2-1/2 cos(4 u))

cos(4 u) = 1/2+(cos(4 u))/2+1/2 cos(4 u)-1/2

cos(4 u) = cos(4 u)

Does this work?

Answer 7

So you went in the other direction using the Half Angle Identity for Sine and Cosine?
Haha that's clever :) Yah I suppose that works!