Prove using [[Angle Sum-or-Difference Identities]] ## Double Angle Formulas $\begin{aligned} \sin (2 \theta) &=2 \sin \theta \cos \theta \\ \cos (2 \theta) &=\cos ^{2} \theta-\sin ^{2} \theta &&=2 \cos ^{2} \theta-1 &&=1-2 \sin ^{2} \theta \\ \tan (2 \theta) &=\frac{2 \tan \theta}{1-\tan ^{2} \theta}&&=\frac{2}{\cot\theta-\tan\theta}&&=\frac{2\cot\theta}{\cot^2\theta-1}\\ \csc(2\theta)&=2\csc\theta\sec\theta\\ \sec(2\theta)&=\frac{1}{\cos ^{2} \theta-\sin ^{2} \theta}&&=\frac{\sec^2\theta\csc^2\theta}{\csc^2\theta-\sec^2\theta} \end{aligned}$ ## Half Angle Formulas The following are a corollary from $\cos(2\theta)$, but the since there are two possible angles, the quadrant of the half-angle should be known to narrow down the result to a single answer. $\sin \frac{\theta}{2}=\pm \sqrt{\frac{1-\cos \theta}{2}}$ $\cos \frac{\theta}{2}=\pm \sqrt{\frac{1+\cos \theta}{2}}$ $\tan \frac{\theta}{2}=\pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$ $\cos2\theta$ leads to [[Power Reduction Identities]]