Question

If $\sin \alpha =\frac{1}{2}$, prove that $\left(3\cos \alpha -4{\cos }^3\alpha \right)=0$.

Answer 1

$$\begin{gathered} \mathrm{LHS}=\left(3\cos \alpha -4{\cos }^3\alpha \right) \\ =\cos \alpha \left(3-4{\cos }^2\alpha \right) \\ =\sqrt{1-{\sin }^2\alpha }\left[3-4\left(1-{\sin }^2\alpha \right)\right] \\ =\sqrt{1-{\left(\frac{1}{2}\right)}^2}\left[3-4\left(1-{\left(\frac{1}{2}\right)}^2\right)\right] \\ =\sqrt{\frac{1}{1}-\frac{1}{4}}\left[3-4\left(\frac{1}{1}-\frac{1}{4}\right)\right] \\ =\sqrt{\frac{3}{4}}\left[3-4\left(\frac{3}{4}\right)\right] \\ =\sqrt{\frac{3}{4}}\left[3-3\right] \\ =\sqrt{\frac{3}{4}}\left[0\right] \\ =0 \\ =\mathrm{RHS} \end{gathered}$$