Double Angle Formulas

In the Angle Addition Identities post it was shown that $\displaystyle \sin(x+y) = \sin x \cos y + \sin y \cos x$ and $\displaystyle \cos(x+y) = \cos x \cos y – \sin x \sin y$. These identities are valid if $\sin \theta \equiv \frac{opp}{hyp}$ and $\cos \theta \equiv \frac{adj}{hyp}$, which implies that $\theta \in [0,\frac{\pi}{2})$ in order for $\theta$ to be an acute angle of a right triangle. These constraints are imposed because the two Angle Addition Identities were derived using these constraints. The next step, following the steps in reference [1], is setting $y$ equal to $x$. Doing this yields $\displaystyle \sin(2x) = \sin x \cos x + \sin x \cos x = 2 \sin x \cos x$. Hence $\boxed{ \displaystyle \sin(2x) = 2 \sin x \cos x }$. Similarly, following the steps in reference [1], substituting $x$ for $y$ in $\displaystyle \cos(x+y) = \cos x \cos y – \sin x \sin y$ yields $\displaystyle \cos(2x) = \cos x \cos x – \sin x \sin x = \cos ^2 x – \sin ^2 x$. Hence $\boxed{ \displaystyle \cos(2x) = \cos ^2 x – \sin ^2 x }$. Using this Identity, the previous equation becomes $\displaystyle \cos(2x) = (1 – \sin^2 x) – \sin^2 x = 1 – 2 \sin^2 x$. Hence $\boxed{ \displaystyle \cos(2x) = 1 – 2 \sin^2 x }$. Solving for $\sin x$ and taking the positive root to ensure that $\sin x$, a ratio of lengths, is not negative yields $\displaystyle \sin x = \sqrt{ \frac{1-\cos(2x)}{2} }$. Finally, define $x \equiv \frac{\alpha}{2}$ and substitute to get the half-angle identity [2], $\displaystyle \boxed{ \sin \frac{\alpha}{2} = \sqrt{ \frac{1-\cos \alpha }{2}} }$. Therefore, $\displaystyle 2\sin^2 \frac{\alpha}{2} = 1-\cos \alpha$.   […]