Question

Verify the following equalities.
$\cos {60}^{\circ }=1-2{\sin }^2{30}^{\circ }=2{\cos }^2{30}^{\circ }-1$

Answer 1

$\cos 60°=\frac{1}{2}$

$\Rightarrow 1-2{\sin }^{2}30°=1-2{\left(\frac{1}{2}\right)}^2=1-\frac{1}{2}=\frac{1}{2}$

$\Rightarrow 2{\cos }^{2}30°-1=2{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{2}-1=\frac{1}{2}$

$\therefore \cos 60°=1-2{\sin }^{2}30°=2{\cos }^{2}30°-1$

Hence, the answer is verify.