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Prove that the area of a triangle with one of its angles $\frac{\pi}4$ rad, and the side opposite to this angle is $2$cm equals $\sin(2\theta)-\cos(2\theta)+1$, where $\theta$ is the angle adjacent to the $2$cm side.
and here is a picture to make things cleaner:
Let $a = 2, A = \frac {\pi} 4, B = \theta$
Then,
$$ \frac{b}{\sin \theta} = \frac{2}{\frac{\sqrt{2}}2} \implies b = 2\sqrt 2 \sin \theta $$
Then,
$$ \sin C = \sin(A+B) = \frac {\sqrt{2}}2 (\sin \theta + \cos \theta) $$
Then,
$$ S = \frac 12 ab\sin C =\frac 12 \times 2\times 2\sqrt 2\sin \theta\times \frac {\sqrt{2}}2 (\sin \theta + \cos \theta)\\ = 2\sin \theta (\sin \theta + \cos \theta)=2\sin^2\theta+2\sin \theta \cos \theta = 1-\cos 2\theta + \sin 2\theta $$
