$\sin2\alpha$ in terms of $\tan\alpha$

Question

I am solving a geometry problem and I arrived at $$QD^2=54+36\sqrt2\sin2\alpha$$ I have previously found that $\tan\alpha=\dfrac{\sqrt2}{2}$. Using $\sin^2\alpha+\cos^2\alpha=1$, we can actually find the values of $\sin\alpha$ and $\cos\alpha$ and then we have $$\sin2\alpha=2\sin\alpha\cos\alpha$$ Is this necessary, though? The authors have written that $$\sin2\alpha=\dfrac{\tan\alpha}{1+\tan^2\alpha}$$ How can I derive that?

Answer 1

The real identity is

$$\sin 2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha}$$

which you get by noticing that $$\begin{align} \sin 2\alpha &= 2 \sin \alpha \cos \alpha \\ &=2\tan \alpha\cos^2\alpha \\ &=\frac{2\tan\alpha}{\sec^2\alpha} \\ &=\frac{2\tan\alpha}{1+\tan^2\alpha} \end{align}$$