Expected value of sin of sum of n random angles Consider the following problem.
Let $θ_1,θ_2,...θ_n ∈[0, \frac{π}{2}]$ be independent and uniformly distributed variables. Find $E[sin(θ_1 + ... + θ_n)].$
I was able to solve for $n=1$ (of course), but I'm not very sure how to go from there. I was thinking of using the linearity of expectation and doing induction on n (basically split up the $sin$ into two $sin*cos$ parts), but I'm not so sure if that works with deterministic functions of variables. Any help is welcome, thank you!

This is made easy when one uses complex numbers (in particular the formula for the sine), which converts the summation to a product and makes the process of separating the independent parts easier.
Note that $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$ for all $x$ real.
Therefore, we get $$
\sin(\theta_1+\ldots+\theta_n) =  \frac{e^{i(\theta_1+\ldots+\theta_n)} + e^{-i(\theta_1+\ldots+\theta_n)}}{2i}= \frac{\prod_{i=1}^ne^{i\theta_i} - \prod_{i=1}^n e^{-i\theta_i}}{2i}
$$
In particular, extending the definition of the expectation to complex-valued random variables using the obvious $\mathbb E[U+iV] = \mathbb E[U] + i\mathbb E[V]$  and noting that properties of the real-valued expectation carry over,
$$
\mathbb E[\sin(\theta_1+\ldots+ \theta_n)] =\frac 1{2i} \left(\mathbb E\left[\prod_{j=1}^n e^{i\theta_j}\right] - \mathbb E\left[\prod_{j=1}^n e^{-i\theta_j}\right] \right) 
$$
Now, if $\theta_j$ are independent, then so are the collections $\{e^{i \theta_j}\}$ and $\{e^{-i \theta_j}\}$ (this is easily seen using an argument analogous to the real number situation). In particular, we get $$
\mathbb E\left[\prod_{j=1}^n e^{i\theta_j}\right] = \prod_{i=1}^n \mathbb E[e^{i \theta_j}] = (\mathbb E[\sin \theta] + i \mathbb E[\cos \theta])^n
$$
Likewise
$$
\mathbb E\left[\prod_{j=1}^n e^{-i\theta_j}\right] = \prod_{i=1}^n \mathbb E[e^{-i \theta_j}] = (-\mathbb E[\sin \theta] + i \mathbb E[\cos \theta])^n
$$
which upon substitution give the answer.

Alternately, if you wish to stay in the realm of real numbers for a longer time (an eventual formula will involve complex numbers in some form), define the sequences $a_n, b_n$ for $n \geq 1$ by $$
a_n = \mathbb E[\sin(\theta_1+\ldots+\theta_n)] \\
b_n = \mathbb E[\cos(\theta_1+\ldots+\theta_n)] 
$$
Then, the formulas $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B + \sin A \sin B$, along with the natural split $$
\theta_1+\ldots+\theta_n = \underbrace{\theta_n}_{A} + \underbrace{\theta_1+\ldots+\theta_{n-1}}_{B}
$$
and independence, lead instantly to the two recursion formulas$$
a_{n} = b_{n-1}a_1+b_1a_{n-1} \quad ; \quad b_n = b_{n-1}b_1 - a_{n-1}a_1
$$
That can be written in matrix form as $$
\begin{pmatrix}
a_n \\ b_n
\end{pmatrix} = \begin{pmatrix} b_1 & a_1 \\ -a_1 & b_1 \end{pmatrix} \begin{pmatrix} a_{n-1} \\b_{n-1}\end{pmatrix}
$$
which has the solution $$\begin{pmatrix}
a_n \\ b_n
\end{pmatrix} = \begin{pmatrix} b_1 & a_1 \\ -a_1 & b_1 \end{pmatrix}^{n-1} \begin{pmatrix} a_{1} \\b_{1}\end{pmatrix}
$$
$a_1,b_1$ can be obtained straightforwardly, while $a_n,b_n$ can be obtained by diagonalizing the above matrix to find an explicit formula for the $n-1$th power. This will involve complex numbers : for that formula, see here. Substitution provides an answer.
