Question

Evaluate:
${\sin }^{2}37{\frac{1}{2}}^0-{\cos }^{2}67{\frac{1}{2}}^0$

Answer 1

${\sin }^{2}x-{\cos }^{2}B=-\cos (A+B)\cos (A-B)$

So,

${\sin }^{2}x{\frac{1}{2}}^{\circ }-{\cos }^{2}67{\frac{1}{2}}^{\circ }$

$=-\cos (37{\frac{1}{2}}^{\circ }+67{\frac{1}{2}}^{\circ })\cos (37{\frac{1}{2}}^{\circ }-67{\frac{1}{2}}^{\circ })$

$=-\cos \left({105}^{\circ }\right)\cos (-{30}^{\circ })$

$\because \cos (-x)=\cos x$

$=-\cos \left({105}^{\circ }\right)\cos \left({30}^{\circ }\right)$

$=-\cos (90+{15}^{\circ })\cos \left({30}^{\circ }\right)$

$=+\sin \left({15}^{\circ }\right)\cos \left({30}^{\circ }\right)$

$\sin {15}^{\circ }=\frac{\sqrt{6}-\sqrt{2}}{4}$ , $\cos \left({30}^{\circ }\right)=\frac{\sqrt{3}}{2}$

${\sin }^{2}37{\frac{1}{2}}^{\circ }-{\cos }^{2}67{\frac{1}{2}}^{\circ }$ $=\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)$ $\left(\frac{\sqrt{3}}{2}\right)$

$=\frac{\sqrt{18}-\sqrt{6}}{8}$