Question

Verify the following equalities.
$\cos {90}^{\circ }=1-2{\sin }^2{45}^{\circ }=2{\cos }^2{45}^{\circ }-1$

Answer 1

$\cos 90°=0,1-2{\sin }^{2}45°=1-2{\left(\frac{1}{\sqrt{2}}\right)}^2=1-1=0$

$\Rightarrow 2{\cos }^{2}45°-1=2\times {\left(\frac{1}{\sqrt{2}}\right)}^2-1=1-1=0$

$\therefore \cos 90°=1-2{\sin }^{2}45°=2{\cos }^{2}45°$

Hence, the answer is verified.