Question

The value of ${\cos }^2{45}^o-{\sin }^2{15}^o$ is

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

Answer 1

Answer: (D)

$\frac{\sqrt{3}}{4}$

To find ${\cos }^2{45}^o-{\sin }^2{15}^o$

We know that, $\cos 2\theta =2{\cos }^2\theta -1⟹{\cos }^2\theta =\frac{1+\cos 2\theta }{2}$

And $\cos 2\theta =1-2{\sin }^2\theta ⟹{\sin }^2\theta =\frac{1-\cos 2\theta }{2}$

$\therefore {\cos }^2{45}^o=\frac{1+\cos (2\times 45)}{2}=\frac{1+\cos {90}^o}{2}=\frac{1+0}{2}=\frac{1}{2}$

Also, $\therefore {\sin }^2{15}^o=\frac{1-\cos (2\times 15)}{2}=\frac{1-\cos {30}^o}{2}=\frac{1-\frac{\sqrt{3}}{2}}{2}=\frac{1}{2}-\frac{\sqrt{3}}{4}$

Then we get,

${\cos }^2{45}^o-{\sin }^2{15}^o$

$=\frac{1}{2}-\frac{1}{2}+\frac{\sqrt{3}}{4}$

$=\frac{\sqrt{3}}{4}$

Hence, ${\cos }^2{45}^o-{\sin }^2{15}^o=\frac{\sqrt{3}}{4}$