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!
 A: 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.
