Question

The value of $\cos \left({36}^{\circ }-A\right)\cos \left({36}^{\circ }+A\right)+\cos \left({54}^{\circ }+A\right)\cos \left({54}^{\circ }-A\right)$ is

• A. ${\sin }^{2}A$
• B. $\cos 2A$
• C. $\cos 3A$
• D. $\sin 3A$

Answer 1

The correct option is B $\cos 2A$

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

here, $A={36}^o,B=A$

$\Rightarrow \cos ({36}^o-A)\cos ({36}^o+A)+\cos ({54}^o+A)\cos ({54}^o-A)$

$\Rightarrow {\cos }^2{36}^o-{\sin }^{2}A+{\cos }^2{54}^o-{\sin }^{2}A$

$\Rightarrow -2{\sin }^{2}A+{\cos }^2{36}^o+\left(\cos \right(90-36){)}^2$

$\Rightarrow -2{\sin }^{2}A+{\cos }^2{36}^o+{\sin }^2{36}^o$

$co{s}^2\theta +{\sin }^2\theta =1$

$\Rightarrow 1-2{\sin }^{2}A$

$\Rightarrow \cos 2A$