<s>[INST] 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.
 [/INST] 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.
</s>