Question:
If $\sin \alpha=\frac{1}{2}$, prove that $\left(3 \cos \alpha-4 \cos ^{3} \alpha\right)=0$
Solution:
LHS $=\left(3 \cos \alpha-4 \cos ^{3} \alpha\right)$
$=\cos \alpha\left(3-4 \cos ^{2} \alpha\right)$
$=\sqrt{1-\sin ^{2} \alpha}\left[3-4\left(1-\sin ^{2} \alpha\right)\right]$
$=\sqrt{1-\left(\frac{1}{2}\right)^{2}}\left[3-4\left(1-\left(\frac{1}{2}\right)^{2}\right)\right]$
$=\sqrt{\frac{1}{1}-\frac{1}{4}}\left[3-4\left(\frac{1}{1}-\frac{1}{4}\right)\right]$
$=\sqrt{\frac{3}{4}}\left[3-4\left(\frac{3}{4}\right)\right]$
$=\sqrt{\frac{3}{4}}[3-3]$
$=\sqrt{\frac{3}{4}}[0]$
$=0$
$=\mathrm{RHS}$
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