Question

Solve ${\cos }^2{48}^0-{\sin }^2{12}^0=$

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

Answer 1

Answer: (B)

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

${\cos }^2{48}^o-{\sin }^2{12}^o$

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

$A={48}^o,B={12}^o$

$\because \cos {60}^o=\frac{1}{2}$, $cos\left({36}^o\right)=\frac{\sqrt{5}+1}{4}$

$\Rightarrow {\cos }^2{48}^o-{\sin }^2{12}^o=\cos (48+12)\cos (48-12)$

$=\cos \left({60}^o\right)\cos \left({36}^o\right)$

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

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