Find the $L^2(-\pi,\pi)$-norm of $1, sin(x), cos(x)$
I am just wondering if I am on the right track:
$||1-cos(x)-sin(x)||_2= \bigg(\displaystyle\int_{-\pi}^{\pi} (1-cos(x)-sin(x))^2 dx\bigg)^{\displaystyle\frac{1}{2}}$
$(1-cos(x)-sin(x))^2 = sin^2(x)+2cos(x)sin(x)-2sin(x)+cos^2(x)-2cos(x)+1$
$\bigg(\displaystyle\int_{-\pi}^{\pi} (1-cos(x)-sin(x))^2 dx\bigg)^{\displaystyle\frac{1}{2}} = \bigg(\int_{-\pi}^{\pi} sin^2x dx+2\int_{-\pi}^{\pi} cos(x)sin(x)dx - 2\int_{-\pi}^{\pi} sin(x)dx + \int_{-\pi}^{\pi} cos^2(x)dx - 2\int_{-\pi}^{\pi} cos(x)dx + \int_{-\pi}^{\pi} dx \bigg)^{\displaystyle\frac{1}{2}}$
Your computation is going fine. Now some advice.
First, you usually want to work without the square roots; you calculate the norm squared, and then take the square root at the end.
Also, you can save quite a bit of effort if you already know that $1,\sin x,\cos x$ are mutually orthogonal on $L^2(-\pi,\pi)$. So you just have, for pairwise orthogonal vectors $x,y,z$, that $$ \|x+y+y\|^2=\|x\|^2+\|y\|^2+\|z\|^2.$$ Then \begin{align} \|1-\sin x-\cos x\|^2=\|1\|^2+\|\sin x\|^2+\|\cos x\|^2=2\pi+2\|\sin x\|^2. \end{align} Then the only computation one needs is $$ \int_{-\pi}^\pi \sin ^2 x\, dx=\pi. $$ Thus $$ \|1-\sin x-\cos x\|=\sqrt{4\pi}=2\sqrt\pi. $$