## Trigonometry (11th Edition) Clone

$$2\sin22.5^\circ\cos22.5^\circ=\frac{\sqrt2}{2}$$ We match 3 with B.
$$X=2\sin22.5^\circ\cos22.5^\circ$$ Recall the double-angle identities: $$2\sin A\cos A=\sin2A$$ $X$ can be seen here as $2\sin A\cos A$ with $A=22.5^\circ$. Therefore, $$X=2\sin22.5^\circ\cos22.5^\circ=\sin(2\times22.5^\circ)$$ $$X=\sin45^\circ$$ $$X=\frac{\sqrt2}{2}$$ So, $$2\sin22.5^\circ\cos22.5^\circ=\frac{\sqrt2}{2}$$ 3 would come with B.