Show that one equation equals another (trigonometry)

Question

I'm studying for a test and when going through old exams I find this one which I'm not able to solve.

Show that

$$a^2 = (b−c)^2 + 4bc \sin^2 \left(\frac A2\right)$$

equals

$$a^2 = b^2 + c^2 - 2bc \cos(A)$$

Hint:

$$\cos(θ) = \frac{e^{iθ} + e^{-iθ}}2 \quad\text{and}\quad \sin(θ) = \frac{e^{iθ} - e^{-iθ}}{2i}$$

Answer 1

$$a^2 = (b−c)^2 + 4bc \sin^2\frac{A}{2} = b^2+c^2-2bc+4bc \sin^2\frac{A}{2} = \\ = b^2+c^2-2bc(1-2\sin^2\frac{A}{2})=b^2+c^2-2bc\ cosA$$ where in the last line, the well known half-angle identity: $$1-2\sin^2\frac{A}{2}=cosA$$ has been used.