Question

$\sin 5A=5{\cos }^{4}A\sin A-10{\sin }^{3}A{\cos }^{2}A+{\sin }^{5}A$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sin 5\mathrm{A} \\ =\sin \left(3\mathrm{A}+2\mathrm{A}\right) \\ =\sin 3\mathrm{A}\times \cos 2\mathrm{A}+\cos 3\mathrm{A}\times \sin 2\mathrm{A} \\ =\left(3\mathrm{sinA}-4{\sin }^3\mathrm{A}\right)\left(2{\cos }^2\mathrm{A}-1\right)+\left(4{\cos }^3\mathrm{A}-3\mathrm{cosA}\right)\times 2\mathrm{sinAcosA} \\ =-3\mathrm{sinA}+4{\sin }^3\mathrm{A}+6{\mathrm{sinAcos}}^2\mathrm{A}-8{\sin }^3{\mathrm{Acos}}^2\mathrm{A}+8{\mathrm{sinAcos}}^4\mathrm{A}-6{\mathrm{sinAcos}}^2\mathrm{A} \\ =8{\mathrm{sinAcos}}^4\mathrm{A}-8{\sin }^3{\mathrm{Acos}}^2\mathrm{A}-3\mathrm{sinA}+4{\sin }^3\mathrm{A} \\ =5{\mathrm{sinAcos}}^4\mathrm{A}-10{\sin }^3{\mathrm{Acos}}^2\mathrm{A}-3\mathrm{sinA}+3{\mathrm{sinAcos}}^4\mathrm{A}+4{\sin }^3\mathrm{A}+2{\sin }^3{\mathrm{Acos}}^2\mathrm{A} \\ =5{\mathrm{sinAcos}}^4\mathrm{A}-10{\sin }^3{\mathrm{Acos}}^2\mathrm{A}-3\mathrm{sinA}\left(1-{\cos }^4\mathrm{A}\right)+2{\sin }^3\mathrm{A}\left(2+{\cos }^2\mathrm{A}\right) \end{gathered}$$
$$\begin{gathered} =5{\mathrm{sinAcos}}^4\mathrm{A}-10{\sin }^3{\mathrm{Acos}}^2\mathrm{A}-3\mathrm{sinA}\left(1-{\cos }^2\mathrm{A}\right)\left(1+{\cos }^2\mathrm{A}\right)+2{\sin }^3\mathrm{A}\left(2+{\cos }^2\mathrm{A}\right) \\ =5{\mathrm{sinAcos}}^4\mathrm{A}-10{\sin }^3{\mathrm{Acos}}^2\mathrm{A}-3{\sin }^3\mathrm{A}\left(1+{\cos }^2\mathrm{A}\right)+2{\sin }^3\mathrm{A}\left(2+{\cos }^2\mathrm{A}\right) \\ =5{\mathrm{sinAcos}}^4\mathrm{A}-10{\sin }^3{\mathrm{Acos}}^2\mathrm{A}-{\sin }^3\mathrm{A}\left[3\left(1+{\cos }^2\mathrm{A}\right)-2\left(2+{\cos }^2\mathrm{A}\right)\right] \\ =5{\mathrm{sinAcos}}^4\mathrm{A}-10{\sin }^3{\mathrm{Acos}}^2\mathrm{A}-{\sin }^3\mathrm{A}\left[3+3{\cos }^2\mathrm{A}-4-2{\cos }^2\mathrm{A}\right] \\ =5{\mathrm{sinAcos}}^4\mathrm{A}-10{\sin }^3{\mathrm{Acos}}^2\mathrm{A}-{\sin }^3\mathrm{A}\left[{\cos }^2\mathrm{A}-1\right] \\ =5{\mathrm{sinAcos}}^4\mathrm{A}-10{\sin }^3{\mathrm{Acos}}^2\mathrm{A}-{\sin }^3\mathrm{A}\times \left(-{\sin }^2\mathrm{A}\right) \\ =5{\mathrm{sinAcos}}^4\mathrm{A}-10{\sin }^3{\mathrm{Acos}}^2\mathrm{A}+{\sin }^5\mathrm{A} \\ =5{\cos }^4\mathrm{A}\mathrm{sinA}-10{\cos }^2\mathrm{A}{\sin }^3\mathrm{A}+{\sin }^5\mathrm{A} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$