I calculated $\sin 75^\circ$ as $\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$, but the answer is $\frac{\sqrt{2}+\sqrt{6}}{4}$. What went wrong? I calculated the exact value of $\sin 75^\circ$ as follows:    
$$\begin{align}
\sin 75^\circ &= \sin(30^\circ + 45^\circ)  \\
&=\sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ \\  
&=\frac12\cdot\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}} \\  
&= \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}
\end{align}$$
The actual answer is
$$\frac{\sqrt{2} + \sqrt{6}}{4}$$
My main confusion is how the textbook answer is completely different from mine, even though if I compute $\sin 30^\circ \cos45^\circ + \cos 30^\circ \sin 45^\circ$, it will be approximately the same value of $\sin 75^\circ$. 
I think I'm having difficulty adding and subtracting the radicals. So, if someone can demonstrate to me how they got that answer, it will be helpful. Thanks.

They’re the same value. Multiply the numerator and denominator of your answer by $\sqrt 2$ to see why.
$$\frac{1+\sqrt 3}{2\sqrt 2} = \frac{\sqrt 2}{\sqrt 2}\cdot\frac{1+\sqrt 3}{2\sqrt 2} = \frac{\sqrt 2+\sqrt 6}{4}$$
You can also use $\sin 45 = \cos 45 = \frac{\sqrt 2}{2}$ (rationalizing $\frac{1}{\sqrt 2}$) to get the answer more easily.
