The estimable @solvemymaths tweeted, some time back:

hmm, perhaps I'll keep this one as "sin(22)" pic.twitter.com/cT5IHonoyb

— solve my maths (@solvemymaths) January 16, 2016

A sensible option? Perhaps. But Wolfram Alpha is being a bit odd here: that's something that can be simplified significantly. (One aside: I'm not convinced that actually *is* $\sin(22º)$, because I get a different answer in the first line, and WA doesn't give an exact answer for it -- but I'm willing to be corrected! So I'm going to call it $S$ instead of $\sin(22º)$)

Before I do anything, I'll reiterate my Key Rule for simplifying anything: take the ugliest thing and make it nicer.

Let's start from what they have:

$S = \frac{ - \frac 18 \sqrt{3} \left( -1 - \sqrt{5} \right) - \frac{1}{4}\sqrt{\frac 12 (5 - \sqrt{5}) } } {\sqrt{2}} - \frac{ - \frac 18 \left( -1 - \sqrt{5} \right) - \frac{1}{4}\sqrt{\frac 32 (5 - \sqrt{5}) } } {\sqrt{2}}$

Honestly. If a student of mine turned that in, I'd roll my eyes at them. For a start, we've got two fractions with the same denominator. Let's combine them, being very careful with our minus signs:

$S = \frac{

-\frac 18 \sqrt{3} \left( -1 - \sqrt{5} \right)

- \frac 14 \sqrt{\frac 12 (5 - \sqrt{5}) } + \frac 18 \left( -1 - \sqrt{5} \right) + \frac{1}{4}\sqrt{\frac 32 (5 - \sqrt{5}) } } {\sqrt{2}}$

Next, the ugliest thing is the stacked fractions. Let's multiply top and bottom by 8:

$S = \frac{ - \sqrt{3} \left( -1 - \sqrt{5} \right) - 2 \sqrt{\frac 12 (5 - \sqrt{5}) } + \left( -1 - \sqrt{5} \right) + 2 \sqrt{\frac 32 (5 - \sqrt{5}) } } {8\sqrt{2}}$

There are minus signs flying around like squirrels. This can be improved:

$S = \frac{ \sqrt{3} \left( 1 + \sqrt{5} \right) - 2 \sqrt{\frac 12 (5 - \sqrt{5}) } - \left( 1 + \sqrt{5} \right) + 2 \sqrt{\frac 32 (5 - \sqrt{5}) } } {8\sqrt{2}}$

I don't like the fractions inside the square roots, either.

$S = \frac{ \sqrt{3} \left( 1 + \sqrt{5} \right) - \sqrt{2} \sqrt{ (5 - \sqrt{5}) } - \left( 1 + \sqrt{5} \right) + \sqrt{3}\sqrt{2} \sqrt{5 - \sqrt{5} } } {8\sqrt{2}}$

There's a reason I didn't make that $\sqrt{6}$, by the way. There's now some obvious factorising:

$S = \frac{

\left(\sqrt{3} - 1\right)

\left( 1 + \sqrt{5} \right)

+ \sqrt{2} \left( \sqrt{3}- \sqrt{1} \right)

\sqrt{ (5 - \sqrt{5}) } } {8\sqrt{2}}$

And the $\left(\sqrt{3} -1\right)$ comes out as well:

$S = \frac{

\left(\sqrt{3} - 1\right)

\left[

\left( 1 + \sqrt{5} \right)

+\sqrt{2} \sqrt{ 5 - \sqrt{5}} \right] } {8\sqrt{2}}$

I'm not going to claim that's *nice*, but it's definitely nicer than it was!