Question

$({\cos }^2{66}^{\circ }-{\sin }^2{6}^{\circ })({\cos }^2{48}^{\circ }-{\sin }^2{12}^{\circ })$ equals -

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{5}{16}$
• D. $\frac{3}{16}$

Answer 1

The correct option is A $\frac{1}{16}$

As we know that

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

${\cos }^2{66}^{\circ }-{\sin }^2{6}^{\circ }=\cos ({66}^{\circ }+6^{\circ })\cos ({66}^{\circ }-6^{\circ })=\cos {72}^{\circ }\cos {60}^{\circ }=\frac{1}{2}\sin {18}^{\circ }$

${\cos }^2{48}^{\circ }-{\sin }^2{12}^{\circ }=\cos ({48}^{\circ }+{12}^{\circ })\cos ({48}^{\circ }-{12}^{\circ })=\cos {60}^{\circ }\cos {36}^{\circ }=\frac{1}{2}\cos {36}^{\circ }$

$({\cos }^2{66}^{\circ }-{\sin }^2{6}^{\circ })({\cos }^2{48}^{\circ }-{\sin }^2{12}^{\circ })=\frac{1}{4}\sin {18}^{\circ }\cos {36}^{\circ }=\frac{1}{4}\times \frac{(\sqrt{5}+1)(\sqrt{5}-1)}{16}=\frac{1}{16}$