Question

The value of ${\cos }^2{48}^{\circ }-{\sin }^2{12}^{\circ }$ is

• A. $\frac{\sqrt{5}+1}{8}$
• B. $\frac{\sqrt{5}-1}{8}$
• C. $\frac{\sqrt{5}+1}{16}$
• D. None of these

Answer 1

The correct option is A $\frac{\sqrt{5}+1}{8}$

The value of ${\cos }^2{48}^{\circ }-{\sin }^2{12}^{\circ }$ is

$=\frac{1}{2}(2{\cos }^2{48}^{\circ }-2{\sin }^2{12}^{\circ })$

$=\frac{1}{2}\left[\right(1+\cos {96}^{\circ })-(1-\cos {24}^{\circ }\left)\right]$

$=\frac{1}{2}(\cos {96}^{\circ }+\cos {24}^{\circ })$

$=\frac{1}{2}\left[2\cos \left(\frac{{96}^{\circ }+{24}^{\circ }}{2}\right)\cos \left(\frac{{96}^{\circ }-{24}^{\circ }}{2}\right)\right]$

$=\cos {60}^{\circ }\cos {36}^{\circ }$

$=\frac{1}{2}\times \left(\frac{\sqrt{5}+1}{4}\right)$

$=\frac{\sqrt{5}+1}{8}$