we fix arbitrary constant A by normalizing wave function $\displaystyle \int_{0}^{a}|A|^2sin^2(kx)dx = 1$
by using identity $sin^2(x) = \displaystyle \frac{1}{2}-\frac{1}{2} cos{2x}$ we can rearrange integral to $|A|^2 \displaystyle \int_{0}^{a}\frac{1}{2}-\frac{1}{2} cos{2kx}dx$ = $|A|^2 \displaystyle \int_{0}^{a}\frac{1}{2}(1- cos{2kx})dx$ = $\displaystyle \frac {|A|^2}{2}\int_{0}^{a}(1- cos{2kx})dx = \displaystyle \frac {|A|^2}{2}(a-\frac{sin2ka}{2k})=1$
since normalisation constant is $\displaystyle A=\sqrt{\frac{2}{a}}$
The question is where does sine term gone?
When you found your wave functions you had two boundary conditions to satisfy, $\Psi(x) = Asin(kx) + B cos(kx)$ which were $\Psi(0) = \Psi(a) = 0$. Imposing these gave B=0 and $ka=n\pi$ so $k=\frac{n\pi}{a}$ for $n\in \mathbb{Z}$. So $sin(2ka)=sin(2n\pi)=0$, so as @achille hui said it's because of your boundary conditions.