Question

Prove$\sin 3A=3\sin A-4{\sin }^{3}A$

Answer 1

Prove that :$\sin 3A=3\sin A-4{\sin }^{3}A$.

Solve the L.H.S part:

$$\begin{gathered} \sin 3A=\sin (2A+A) \\ \Rightarrow =\sin 2A\cos A+\cos 2A\sin A\because \mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{A}\mathbf{+}\mathit{B}\mathbf{)}=\sin A\cos B+\cos A\sin B \\ \Rightarrow =2\sin A\cos A\cos A+(1-2si{n}^{2}A\mathbf{})\sin A\because \mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{A}=2\sin A\cos Aand\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{A}\mathbf{}\mathbf{=}1-2si{n}^{2}A\mathbf{} \\ \Rightarrow =2\sin A{\cos }^{2}A+\sin A-2si{n}^{3}A \\ \Rightarrow =\sin A(2{\cos }^{2}A+1)-2si{n}^{3}A \\ \Rightarrow =\sin A(2(\mathbf{}1-\mathbf{}{\sin }^{2}A)+1)-2si{n}^{3}A\because {\sin }^{2}A+{\cos }^{2}A=1\mathit{s}\mathit{o}\mathbf{},\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}\mathbf{}\mathbf{}}\mathit{A}\mathbf{=}\mathbf{}1-\mathbf{}{\sin }^{2}A \\ \Rightarrow =\sin A(3-\mathbf{}2{\sin }^{2}A)-2{\sin }^{3}A \\ \Rightarrow =3\sin A-\mathbf{}2{\sin }^{3}A-2{\sin }^{3}A \\ \Rightarrow =3sinA-4si{n}^{3}A \end{gathered}$$

We can see that L.H.S = R.H.S

Hence, proved.