Question

Evaluate:

${\cos }^2{72}^{\circ }-{\sin }^2{54}^{\circ }$

• A. $\frac{\sqrt{5}}{2}$
• B. $-\frac{\sqrt{3}}{4}$
• C. $-\frac{\sqrt{5}}{4}$
• D. $\frac{\sqrt{5}}{8}$

Answer 1

The correct option is C $-\frac{\sqrt{5}}{4}$

We know that;

$\cos {72}^{\circ }=\sin ({90}^{\circ }-{72}^{\circ })=\sin {18}^{\circ }=\frac{-1+\sqrt{5}}{4}$

$\sin {54}^{\circ }=\cos ({90}^{\circ }-{54}^{\circ })=\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}$

Therefore,

$\Rightarrow {\cos }^2{72}^{\circ }-{\sin }^2{54}^{\circ }$

$\Rightarrow {\sin }^2{18}^{\circ }-{\cos }^2{36}^{\circ }$

$\Rightarrow {\left(\frac{-1+\sqrt{5}}{4}\right)}^2-{\left(\frac{\sqrt{5}+1}{4}\right)}^2$

$\Rightarrow \left(\frac{1+5-2\sqrt{5}}{16}\right)-\left(\frac{5+1+2\sqrt{5}}{16}\right)$

$\Rightarrow \left(\frac{6-2\sqrt{5}-6-2\sqrt{5}}{16}\right)$

$\Rightarrow (-\frac{4\sqrt{5}}{16})$

$\Rightarrow (-\frac{\sqrt{5}}{4})$

Hence, the value of ${\cos }^2{72}^{\circ }-{\sin }^2{54}^{\circ }$ is $-\frac{\sqrt{5}}{4}$.