Question

Find the value of ${\cos }^2{48}^{\circ }-{\sin }^2{12}^{\circ }$.

• A. $\frac{\sqrt{5}-1}{4}$
• B. $\frac{\sqrt{5}+1}{8}$
• C. $\frac{\sqrt{3}-1}{4}$
• D. $\frac{\sqrt{5}+1}{2\sqrt{2}}$

Answer 1

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

As we know that

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

So ${\cos }^2{48}^{\circ }-{\sin }^2{12}^{\circ }=\cos ({48}^{\circ }+{12}^{\circ })\cos ({48}^{\circ }-{12}^{\circ })=\cos {60}^{\circ }\cos {36}^{\circ }=\frac{1}{2}\times \frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}+1}{8}$