Question

Prove that ${\cos }^2{45}^o-{\sin }^2{15}^o=\frac{\sqrt{3}}{4}$

Answer 1

L.H.S$={\cos }^2{45}^{\circ }-{\sin }^2{15}^{\circ }$

$={\cos }^2{45}^{\circ }-{\left(\sin {15}^{\circ }\right)}^2$

$={\left(\frac{1}{\sqrt{2}}\right)}^2-{\left(\sin ({45}^{\circ }-{30}^{\circ })\right)}^2$

$=\frac{1}{2}-{(\sin {45}^{\circ }\cos {30}^{\circ }-\cos {45}^{\circ }\sin {30}^{\circ })}^2$

$=\frac{1}{2}-{(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\frac{1}{\sqrt{2}})}^2$

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

$=\frac{1}{2}-\left(\frac{3+1-2\sqrt{3}}{8}\right)$

$=\frac{4-4+2\sqrt{3}}{8}$

$=\frac{\sqrt{3}}{4}=$R.H.S