Question

If $A+B=\frac{\pi }{3}$ and $\cos A+\cos B=1$, then which of the following are true:

• A. $\cos (A-B)=\frac{1}{3}$
• B. $\cos (A-B)=-\frac{1}{3}$
• C. $|\cos A-\cos B|=\sqrt{2/3}$
• D. $|\cos A-\cos B|=\frac{1}{\sqrt{3}}$

Answer 1

Answer: (B), (C)

$\cos (A-B)=-\frac{1}{3}$
$|\cos A-\cos B|=\sqrt{2/3}$

Ans. $\left(b\right)$, $\left(c\right)$

From the given relation, we have

$2\cos \frac{A+B}{2}\cos \frac{A-B}{2}=1$

Or $2\cos {30}^o\cos \frac{A-B}{2}=1$

$\therefore \cos \frac{A-B}{2}=\frac{1}{\sqrt{3}}$

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

Again $|\cos A-\cos B|=2\sin \frac{A+B}{2}\sin \frac{B-A}{2}$

$=2\sin {30}^o\sqrt{1-{\cos }^2\frac{A-B}{2}}=1\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}$