Trigonometry Question ABCD is an isosceles trapezoid.
$$CAB = \alpha,\quad CAD = \beta,\quad AC = m$$
a) Find $AB$ and $DC$.
b) $\beta = 2\alpha$, $\frac{AB}{DC}$ = $\frac{1}{2}$, Find $a$ without using $β$ or $m$.
c) Is it possible to inscribe a circle in $ABCD$? Why?
Answers:
a) $AB = \frac{m\sin(2\alpha+\beta)}{\sin(\alpha+\beta)}$, $DC = \frac{m\sin(\beta)}{\sin(\alpha+\beta)}$
b) $\alpha = 37.76^\circ$
c) No.
I've managed to solve a, but I'm having trouble solving b.

 A: Part (b) follows directly from part (a). We have $\dfrac{AB}{DC}=\dfrac{1}{2}$.
Using the expressions you got for $AB$ and $CD$, putting $\beta=2\alpha$, and simplifying a bit, we find that 
$$\frac{1}{2}=\frac{AB}{CD}=\frac{\sin 4\alpha}{\sin 2\alpha}$$
(there is a fair amount of cancellation.)
Now use the double-angle identity $\sin 4\alpha=2(\sin 2\alpha)(\cos 2\alpha)$. We find that
$$\cos 2\alpha =\frac{1}{4}.$$
The rest is a job for the calculator, use the $\cos^{-1}$ button.
If we feel like it we can get an explicit expression for $\cos \alpha$ or $\sin \alpha$, from the fact that $\cos 2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha$, but that is not necessary.
As to part (c), one cannot inscribe a circle in any trapezoid that satisfies the conditions of (b). First  draw an isosceles trapezoid with an inscribed circle. Suppose that the two tangents to a circle from an external point $P$, meet the circle at $M$ and $N$. Then $PM=PN$.  So in any isosceles trapezoid with an inscribed circle, each "slant" side has length equal to half the sum of the two parallel sides.
In our case, we can take $AB=2$ and $CD=4$. So if our trapezoid had an inscribed circle, then each slant side would have length $3$. But then the height of the trapezoid would be $\sqrt{3^2-1^2}$, that is, $2\sqrt{2}$. That makes $\tan\alpha =\frac{2\sqrt{2}}{3}$. But then $\angle \alpha\approx 43.3$ degrees, which is not what we found in part (b).
