Let $X$, $Y$ be independent and exponentially distributed with mean $1$. Find $\mathbb E(\cos(X+Y)\mid X)$.
What I did: \begin{align*} \mathbb E(\cos(X+Y)\mid X) &= \mathbb E(\cos X \cos Y-\sin X\sin Y|X) \\&= \mathbb E(\cos X \cos Y|X)-\mathbb E(\sin X \sin Y|X) \\&= \cos X \mathbb E(\cos Y|X)-\sin X \mathbb E(\sin Y|X) \end{align*} Is this correct? I am not sure we can open it like this or not. Does this imply that $\cos X \cos Y$ and $\sin X \sin Y$ are independent?
What you did so far is correct: the first equality is just $\cos(a+b)=\cos a\cos b-\sin a\sin b$, the second one the linearity of conditional expectation and the third one is the pull-out property, since $\cos X$ is $\sigma(X)$-measurable as well as $\sin(X)$.
Then $\mathbb E(\cos(Y)\mid X)$ can be computed thanks to independence: this is simply $\mathbb E(\cos(Y))$, and similarly, $\mathbb E(\sin(Y)\mid X)=\mathbb E(\sin(Y))$ and these expectations can be computed explicitely.
I do not see why what you did implies that $\cos \cos $ and $\sin \sin $ are independent.