Question

Given that $\cos (A-B)=\cos A\cos B+\sin A\sin B$, find the value of $\cos {15}^o$.

• A. $\cos {15}^o=\frac{\sqrt{2}+1}{2\sqrt{2}}$
• B. $\cos {15}^o=\frac{\sqrt{3}+1}{2\sqrt{2}}$
• C. $\cos {15}^o=\frac{\sqrt{4}+1}{2\sqrt{2}}$
• D. $\cos {15}^o=\frac{\sqrt{6}+1}{2\sqrt{2}}$

Answer 1

The correct option is B $\cos {15}^o=\frac{\sqrt{3}+1}{2\sqrt{2}}$

To find the value of $\cos 15$, we have:

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

Putting $A=45$ and $B=30$

We get $\cos (45-30)=\cos 45\cos 30+\sin 45\sin 30$

$\cos 15=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}$

$\Rightarrow \cos 15=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$