<s>[INST] Find $\sin81^\circ$ given $\sin18^\circ=\dfrac{\sqrt{5}-1}{4}$ 
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 ?
 [/INST] $\sin81^\circ = \sin(45 + 36)\\
\sin45\cos 36 + \cos 45\sin 36$
$\cos 36 = 1 - 2\sin^2 18 = \frac {1+\sqrt 5}{4}$
$\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.
$3+\sqrt{5} = \frac 12 (1+\sqrt{5})^2\\
\frac {\sqrt{3+\sqrt{5}}}{4} = \frac {1}{\sqrt 2} \frac {1+\sqrt{5}}{4} = \sin 45\cos 36$
</s>