# 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.

• You mean $\sin 75 = \sin(30+45)$. – KM101 Dec 23 '18 at 10:09
• yes you are right – DreamVision2017 Dec 23 '18 at 10:14
• Before claiming values are different, you should try calculating them with a calculator, notice they match and then think about an algebraic manipulation. – zwim Dec 23 '18 at 10:49
• Sorry I didn't think I clarified this in my question, I knew that my answer was equivalent but just didn't know how the textbook simplified it like that algebraically – DreamVision2017 Dec 23 '18 at 22:23

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.