Question

Prove that: $\sin 5\mathrm{\theta }=5\sin \mathrm{\theta }-20{\sin }^3\mathrm{\theta }+16{\sin }^5\mathrm{\theta }$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sin 5\mathrm{\theta } \\ =\sin \left(3\mathrm{\theta }+2\mathrm{\theta }\right) \\ =\sin 3\mathrm{\theta }\times \cos 2\mathrm{\theta }+\cos 3\mathrm{\theta }\times \sin 2\mathrm{\theta } \\ =\left(3\mathrm{sin\theta }-4{\sin }^3\mathrm{\theta }\right)\left(1-2{\sin }^2\mathrm{\theta }\right)+\left(4{\cos }^3\mathrm{\theta }-3\mathrm{cos\theta }\right)\times 2\mathrm{sin\theta cos\theta } \\ =3\mathrm{sin\theta }-6{\sin }^3\mathrm{\theta }-4{\sin }^3\mathrm{\theta }+8{\sin }^5\mathrm{\theta }+\left(8{\cos }^4\mathrm{\theta }-6{\cos }^2\mathrm{\theta }\right)\mathrm{sin\theta } \\ =3\mathrm{sin\theta }-10{\sin }^3\mathrm{\theta }+8{\sin }^5\mathrm{\theta }+\left\{8\mathrm{sin\theta }{\left(1-{\sin }^2\mathrm{\theta }\right)}^2-6\mathrm{sin\theta }\left(1-{\sin }^2\mathrm{\theta }\right)\right\} \\ =3\mathrm{sin\theta }-10{\sin }^3\mathrm{\theta }+8{\sin }^5\mathrm{\theta }+\left\{8\mathrm{sin\theta }\left(1-2{\sin }^2\mathrm{\theta }+{\sin }^4\mathrm{\theta }\right)-6\mathrm{sin\theta }+6{\sin }^3\mathrm{\theta }\right\} \\ =3\mathrm{sin\theta }-10{\sin }^3\mathrm{\theta }+8{\sin }^5\mathrm{\theta }+8\mathrm{sin\theta }-16{\sin }^3\mathrm{\theta }+8{\sin }^5\mathrm{\theta }-6\mathrm{sin\theta }+6{\sin }^3\mathrm{\theta } \\ =5\mathrm{sin\theta }-20{\sin }^3\mathrm{\theta }+16{\sin }^5\mathrm{\theta } \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$