Question

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

• 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)

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

We know that: $\cos A+\cos B=1$
Also, $\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$
$⟹2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=1$ and $A+B=\frac{\pi }{3}$
$⟹2\cos \left(\frac{A-B}{2}\right)=\frac{1}{\sqrt{3}}$
$⟹\cos (A-B)=2{\cos }^2\left(\frac{A-B}{2}\right)-1=\frac{2}{3}-1=-\frac{1}{3}$

Also,
$\cos A\cos B=\frac{\cos (A+B)+\cos (A-B)}{2}=\frac{1}{12}$ and $\cos A+\cos B=1$
$⟹12{\cos }^{2}A-12\cos A+1=0$
$⟹\cos A=\frac{12\pm \sqrt{96}}{24}$
So this the solution for $\cos A$ and $\cos B$. Now we can easily find $|\cos A-\cos B|$ by taking the difference of the two roots,
$|\cos A-\cos B|=2\times \frac{\sqrt{96}}{24}=\sqrt{\frac{2}{3}}$