Question

If $A+B=\frac{\pi }{3}$ and $\cos A+\cos B=1$ then $\cos (A-B)$ is

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

Answer 1

The correct option is A $-\frac{1}{3}$

$\cos A+\cos B=1\Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=1$

$\Rightarrow 2\cos \frac{\pi }{6}\cos \left(\frac{A-B}{2}\right)=1\Rightarrow \cos \left(\frac{A-B}{2}\right)=\frac{1}{\sqrt{3}}$

$\Rightarrow {\cos }^2\left(\frac{A-B}{2}\right)=\frac{1}{3}$

Now $\cos (A-B)=2{\cos }^2\left(\frac{A-B}{2}\right)-1=\frac{2}{3}-1=-\frac{1}{3}$