Find $\sin81^\circ$ given $\sin18^\circ=\dfrac{\sqrt{5}-1}{4}$

Question

If $\sin18^\circ=\dfrac{\sqrt{5}-1}{4}$, then $\sin81^\circ$ is equal to \begin{align} &a)\quad\frac{\sqrt{5}+1}{4}\\ &b)\quad\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}\\ &c)\quad\frac{\sqrt{10+2\sqrt{5}}}{4} \end{align}

$$ \cos18^\circ=\sqrt{1-\frac{6-2\sqrt{5}}{16}}=\frac{\sqrt{10+2\sqrt{5}}}{4} $$ $$ \sin81^\circ=\cos9^\circ=\sqrt{\frac{1+\cos18^\circ}{2}}=\sqrt{\frac{4+\sqrt{10+2\sqrt{5}}}{8}}=\frac{\sqrt{8+2\sqrt{10+2\sqrt{5}}}}{4} $$ How do I proceed further and find the solution ?

Or from here, at least can I just identify the solution as $\sin81^\circ=\dfrac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$ from the options given ?

Answer 1

If you suspect (b) is the answer, just square it. You'll get $$ \frac1{16}\left(3 +\sqrt 5+2\sqrt{(3+\sqrt 5)(5-\sqrt 5)} + 5 - \sqrt 5\right) $$ which simplifies to $$ \frac1{16}\left(8 + 2\sqrt{10+2\sqrt 5}\right), $$ the square of your answer. Your answer must then agree with (b), since both answers are positive (as is $\sin 81^\circ$).

Answer 2

$\sin81^\circ = \sin(45 + 36)\\ \sin45\cos 36 + \cos 45\sin 36$

$\cos 36 = 1 - 2\sin^2 18 = \frac {1+\sqrt 5}{4}$

$\sin 36 =$$ \sqrt {\frac {1-\cos 72}{2}}\\ \sqrt {\frac {1-\sin 18}{2}}\\ \sqrt {\frac {5-\sqrt 5}{8}}\\ \cos 45\sin 36 = \frac {\sqrt{5-\sqrt5}}{4}$

$\sin 81 = \frac {\sqrt 2 + \sqrt 10}{8} + \frac {\sqrt{5-\sqrt5}}{4}$

But they have pulled a dirty trick on you.

$3+\sqrt{5} = \frac 12 (1+\sqrt{5})^2\\ \frac {\sqrt{3+\sqrt{5}}}{4} = \frac {1}{\sqrt 2} \frac {1+\sqrt{5}}{4} = \sin 45\cos 36$