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.

  • $\begingroup$ You mean $\sin 75 = \sin(30+45)$. $\endgroup$
    – KM101
    Dec 23, 2018 at 10:09
  • $\begingroup$ yes you are right $\endgroup$ Dec 23, 2018 at 10:14
  • 3
    $\begingroup$ Before claiming values are different, you should try calculating them with a calculator, notice they match and then think about an algebraic manipulation. $\endgroup$
    – zwim
    Dec 23, 2018 at 10:49
  • $\begingroup$ 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 $\endgroup$ Dec 23, 2018 at 22:23

1 Answer 1


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.


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