Deriving what $\cos(A+2B)$ is We know that $\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ , but I don't understand with the $2B$. Would it just become $\cos(A+2B) = \cos(A)\cos(2B) - \sin(A)\sin(2B)$?
Thank you for any help you can give me.

Further expansion is possible with repeated use of the angle addition identities.
$$\begin{align}
\cos (a + 2b) &= \cos a \cos 2b - \sin a \sin 2b \\
&= \cos a (\cos b \cos b - \sin b \sin b) - \sin a (\sin b \cos b + \cos b \sin b) \\
&= \cos a (\cos^2 b - \sin^2 b) - \sin a (2 \sin b \cos b) \\
&= \cos a \cos^2 b - \cos a \sin^2 b - 2 \sin a \sin b \cos b.
\end{align}$$
