Question

What is the formula of $\sin 4A$

Answer 1

$$\begin{gathered} \sin 4A=\sin 2\left(2A\right) \\ =2\sin 2A\cos 2A \\ =2\left(2\sin A\cos A\right)\left({\cos }^{2}A-{\sin }^{2}A\right) \\ =4\sin A\cos A\left({\cos }^{2}A\right)-4\sin A\cos A\left({\sin }^{2}A\right) \\ =4\sin A{\cos }^{3}A-4{\sin }^{3}A\cos A \end{gathered}$$

Hence, $\sin 4A=4\sin A{\cos }^{3}A-4{\sin }^{3}A\cos A$