<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] 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 $\sin 81^\circ$).
</s>