The diagram below shows an isosceles triangle $ABC$, where $AB = AC = 2\text{ cm}$, and a square $BCDE$. $M$ is the midpoint of $BC$. Angle $BAM = \theta$ and can vary.

1. Find the lengths of $BM$ and $AM$ in terms of $\theta$.

In right-angled $\triangle ABM$,
$\sin\theta = \frac{BM}{AB} \implies BM = 2\sin\theta$
$\cos\theta = \frac{AM}{AB} \implies AM = 2\cos\theta$

2. Find the length of $BC$ and the area of the square $BCDE$.

Since $M$ is the midpoint of $BC$,
$BC = 2BM = 4\sin\theta$
Area of square $BCDE = BC^2 = (4\sin\theta)^2 = 16\sin^2\theta$

3. Find the area of $\triangle ABC$ and the total area.

Area of $\triangle ABC = \frac{1}{2} \times BC \times AM$
$= \frac{1}{2}(4\sin\theta)(2\cos\theta) = 4\sin\theta\cos\theta$
Total area = Area of square + Area of $\triangle ABC$
$= 16\sin^2\theta + 4\sin\theta\cos\theta$ (shown)

1. Apply the double angle identity for sine to the term $4\sin\theta\cos\theta$.

$4\sin\theta\cos\theta = 2(2\sin\theta\cos\theta) = 2\sin 2\theta$

2. Apply the double angle identity for cosine to the term $16\sin^{2}\theta$.

$\cos 2\theta = 1 - 2\sin^{2}\theta \implies 2\sin^{2}\theta = 1 - \cos 2\theta$
$16\sin^{2}\theta = 8(2\sin^{2}\theta) = 8(1 - \cos 2\theta) = 8 - 8\cos 2\theta$

3. Combine the terms to show the final expression.

$16\sin^{2}\theta + 4\sin\theta\cos\theta = (8 - 8\cos 2\theta) + 2\sin 2\theta$
$= 2\sin 2\theta - 8\cos 2\theta + 8$ (shown)

1. Expand the target form and compare coefficients to find equations for $R$ and $\alpha$

$R\sin(2\theta - \alpha) = R\sin 2\theta \cos \alpha - R\cos 2\theta \sin \alpha$
Comparing with $2\sin 2\theta - 8\cos 2\theta$:
$R\cos \alpha = 2$
$R\sin \alpha = 8$

2. Calculate the value of $R$

$R = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68}$ (or $2\sqrt{17}$)

3. Calculate the value of $\alpha$ and write the final expression

$\tan \alpha = \frac{8}{2} = 4$
$\alpha = \tan^{-1}(4) \approx 75.964^\circ \approx 76.0^\circ$ (to 1 d.p.)
Expression is $\sqrt{68}\sin(2\theta - 76.0^\circ)$