I am reading the book Applied Fourier Analysis by Tim Olson. There I got introduced the Fourier Series and we have derived the coefficients $a_k$ and $b_k$ for the Fourier Series on $L^2[-\pi,\pi]$. The author is now setting up to introduce orthonormal expansions.
I could recreate the norm of $\sin$ and $\cos$ in $L^2[-\pi,\pi]$ as follows:
For $\cos$ $$\begin{align} \lvert\lvert \cos \rvert\rvert &= \sqrt{ \int _{-\pi}^\pi \cos^2(t) \, dt} \\ &= \sqrt{ 2\int _{0}^\pi \cos^2(t) \, dt} \tag{$\cos$ even function}\\ &= \sqrt{ 2\left[\frac{1}{2}(\cos(t)\sin(t) + t)\right]_{0}^\pi} \\ &= \sqrt{ \cos(\pi)\sin(\pi) + \pi - (\cos(0)\sin(0)) } \, \\ &= \sqrt{ (-1)\cdot0 + \pi } \\ &= \sqrt{ \pi} \, \\ \end{align}$$ For $\sin$ $$\begin{align} \lvert\lvert \sin \rvert\rvert &= \sqrt{ \int_{-\pi}^\pi \sin^2(t) \, dt} \\ &= \sqrt{ \int_{-\pi}^\pi \frac{1 - \cos(2t)}{2} \, dt} \\ &= \sqrt{ \frac{1}{2} \int_{-\pi}^\pi (1 - \cos(2t)) \, dt }\\ &= \sqrt{ \frac{1}{2} \left[t - \frac{\sin(2t)}{2}\right]_{-\pi}^\pi }\\ &= \sqrt{ \frac{1}{2} \left(\pi - (-\pi) - \frac{\sin(2\pi)}{2} + \frac{\sin(-2\pi)}{2}\right)} \\ &= \sqrt{ \frac{1}{2} \left(2\pi \right) }\\ &=\sqrt{ \pi} \end{align}$$
When continuing to read the author uses $\sqrt{T}$as normalization for $\sin$ and $\cos$ in $L^2[-T,T]$. I could recreate this for $\sin$ but I am stuck with $\cos$. My calculations are the following: For $\sin$ $$\begin{align} \lvert\lvert \sin \rvert\rvert &= \sqrt{ \int _{-T}^T \lvert \sin(t) \rvert^2 \, dt } \\ &= \sqrt{ \int _{-T}^T \sin^2(t) \, dt } \\ &= \sqrt{ \int_{-T}^T \frac{1 - \cos(2t)}{2} \, dt }\\ &= \sqrt{\frac{1}{2} \int_{-T}^T 1-\cos(2t) \, dt } \\ &= \sqrt{ \frac{1}{2} \left[t-\frac{\sin(2t)}{2}\right]_{-T}^T } \\ &= \sqrt{ \frac{1}{2} \left(T-\frac{\sin(2T)}{2} - \left( -T-\frac{\sin(-2T)}{2} \right)\right) }\\ &= \sqrt{ \frac{1}{2} \left(T - \frac{\sin(2T)}{2} + T + \frac{\sin(2T)}{2} \right) } \\ &= \sqrt{ \frac{1}{2} \left(2T\right) } \\ &= \sqrt{ T } \\ \end{align}$$
For $\cos$ $$\begin{align} \lvert\lvert \cos \rvert\rvert &= \sqrt{ \int _{-T}^T \lvert \cos(t) \rvert^2 \, dt } \\ &= \sqrt{ \int _{-T}^T \cos^2(t) \, dt } \\ &= \sqrt{ 2\int _{0}^T \cos^2(t) \, dt }\\ &= \sqrt{ 2\left[\frac{1}{2}(\cos(t)\sin(t) + t)\right]_{0}^T } \\ &= \sqrt{ \cos(T)\sin(T) + T } \\ \end{align}$$
First of all, I need some help with $\cos$, because this is not the expected result I have. Additionally the book will continue to work on $L^2[a,b]$. Is there any trick I need to keep in mind going on also to show the normalized versions in $L^2[a,b]$?
Also, if there are better books you can recommend, I am happy to take recommendations.
Since the subject is Fourier series you should probably compute the norm of the trigonometric polynomials which are periodic of period $P$. Your author uses the notation where the period is $P=2T$
This means you should consider the norm of the functions $\cos(\omega n x)$, $\sin(\omega n x)$ where $n$ is an integer and $\omega=2\pi/P=\pi/T$.
After a simple change of variable the computation is reduced to the case where the period is $2\pi$.
The trick to integrate is either use bisection formulae:
\begin{align} \cos(y)^2 &= \frac{1+\cos(2y)}{2} \\ \sin(y)^2 &= \frac{1-\cos(2y)}{2} \, , \\ \end{align}
as you are doing, or integrate by parts.
Finally, Fourier series become much simpler in the complex case, where the orthonormal basis is simply $e^{ikx}/\sqrt{2\pi}$.