Proof of inner product Let $V$ be a vector space with inner product $\langle\,,\,\rangle$ and $\{v_1,v_2,\dots,v_n\}$ an orthonormal basis of $V$. Prove that $\sum_{i=1}^n \langle x,v_i\rangle^2=\|x\|^2$, for $x\in V$.

I´m not very fond of proofs but this is what I have:

\begin{align}
\sum_{i=1}^n \langle x,v_i\rangle^2
&amp;=\sum_{i=1}^n (\|x\|\,\|v_i\| \cos(\theta))^2\\
&amp;=\sum_{i=1}^n \|x\|^2 \|v_i\|^2 \cos^2(\theta)\\
&amp;=\sum_{i=1}^n (\|x\|^2\cos(\theta)) (\|v_i\|^2 \cos(\theta))\\
&amp;=\sum_{i=1}^n \langle x,x\rangle \langle v_i,v_i\rangle\\
&amp;=\langle x,x\rangle\sum_{i=1}^n \langle v_i,v_i\rangle\\
&amp;=\|x\|^2 n
\end{align}

What am I wrong? Any suggestions to improve my proofs would be helpful. Thanks.
Your mistake is this:
$$\|x\|^2\cos\theta \ \ne\ \langle x,x\rangle$$
Instead, we have $\langle x,x\rangle = \|x\|^2$ without the $\cos$ term.

And, hint for the solution: write $x=\xi_1 v_1+\xi_2 v_2+\dots$, then calculate $\|x\|^2=\langle x,x\rangle$ by substituting this into the right argument.
$$x=\sum_{i=1}^n v_ix_i,$$ where $x_i=\langle x,v_i\rangle$ are scalars. 

Indeed, by linearity and orthonormality, $$\langle x,v_j\rangle=\sum_{i=1}^n \langle v_j,v_i\rangle x_i=\sum_{i=1}^n \delta_{ij} x_i=x_j.$$

Then 

$$\|x\|^2=x^2=\sum_{i=1}^n\sum_{j=1}^n \langle v_i,v_j\rangle x_i x_j=\sum_{i=1}^n\sum_{j=1}^n \delta_{ij} x_i x_j=\sum_{i=1}^n x_i^2.$$