Question

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

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

Answer 1

Answer: (B), (C)

The correct options are
B $|\cos A-\cos B|=\sqrt{\frac{2}{3}}$
C $\cos (A-B)=-\frac{1}{3}$

$\cos A+\cos B=1$

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

Since $A+B=\frac{\pi }{3}\Rightarrow \frac{A+B}{2}=\frac{\pi }{6}$
Hence $\cos \left(\frac{A+B}{3}\right)=\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$

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

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

Squaring both sides, we get

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

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

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

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

$=2\times \frac{1}{2}\sqrt{1-\frac{1}{3}}$

$=\sqrt{\frac{2}{3}}$ (on simplification)