if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then... 
if $a,b,c,d$ be the first positive solutions of $\sin x=\frac{1}{4}$ then find $$\sin (d/2)+2\sin (c/2)+3\sin (b/2)+4\sin (a/2)$$.$(a<b<c<d)$

Attempts
$$\sin x=1/4$$,$$4\sin^2(x/2)(1-\sin^2(x/2))=\frac{1}{16}$$
Let $\sin^2 (x/2)=t$:it remains to solve $$t-t^2=\frac{1}{64}$$ .The rest is just brute force i.e we calculate  the 4 values $\sin(x/2)$ can take , make sure the angles are arranged in ascending order and substitute the roots.As far as i can see i a getting irrational roots.

My question is does anyone spot a trick, that can avoid all this work.!
 A: Hint: Since $\sin a$ takes a positive value and $a$ is the smallest solution of $\sin x = \frac{1}{4}$, it follows that angle $a$ is between $0$ and $\frac{π}{2}$. This leads to the conclusion that $b=π−a, c=2π+a$ and $d=3π−a$.
Substituting these values in the original equation and using some basic trigonometric identities gives $2\sqrt{1+\sin a}.$
A: The problem can be simplified somewhat by looking at the geometry of the exercise with respect to the unit circle. The four half angles are distributed as in the diagram:

We see that $\sin(a/2)+sin(c/2)=0$ and $\sin(b/2)+\sin(d/2)=0$ which results in
$$\sin (d/2)+2\sin (c/2)+3\sin (b/2)+4\sin (a/2)=2\sin(a/2)+2\sin(b/2)$$
with much less hassle.
A: So a is in the first quadrant
$b = \pi-a$
$c = a + 2\pi$
$d = 3\pi-a$
$\sin(\frac{b}{2}) = \sin(\frac{\pi}{2}-\frac{a}{2}) = \sin(\frac{\pi}{2})\cos(-\frac{a}{2})+\cos(\frac{\pi}{2})\sin(-\frac{a}{2}) = \cos(\frac{a}{2})$
$\sin(\frac{c}{2}) = \sin(\frac{a}{2}+\pi) = \sin(\frac{a}{2})\cos(\pi)+\cos(\frac{a}{2})\sin(\pi) = -\sin(\frac{a}{2})$
$\sin(\frac{d}{2}) = \sin(\frac{3\pi}{2}-\frac{a}{2}) = \sin(\frac{3\pi}{2})\cos(-\frac{a}{2})+\cos(\frac{3\pi}{2})\sin(-\frac{a}{2}) = -\cos(\frac{a}{2})$
So $\sin(\frac{d}{2}) + 2\sin(\frac{c}{2}) + 3\sin(\frac{b}{2}) + 4\sin(\frac{a}{2})$
$=2\sin(\frac{a}{2}) + 2\cos(\frac{a}{2})$
We know $\cos(a) = \sqrt{1-\frac{1}{16}} = \frac{\sqrt{15}}{4}$
Using the half-angle formla
$\sin(\frac{a}{2}) = \sqrt{0.5(1-\cos(a))}$
$\sin(\frac{a}{2}) = \sqrt{\frac{4-\sqrt{15}}{8}}$
similarly
$\cos(\frac{a}{2}) = \sqrt{\frac{4+\sqrt{15}}{8}}$
So
$2\sin(\frac{a}{2}) + 2\cos(\frac{a}{2}) = \sqrt{\frac{4-\sqrt{15}}{2}}+\sqrt{\frac{4+\sqrt{15}}{2}}$
ok. simplifying the right side by squaring (comes out to 5) and then taking the square root, I get $\sqrt{5}$.
