Question

A solution $(x,y)$ of the system of equations $x-y=\frac{1}{3}$ and ${\cos }^2\pi x-{\sin }^2\pi y=\frac{1}{2}$ is given by

• A. $(\frac{11}{6},\frac{13}{6})$
• B. $(\frac{5}{4},\frac{4}{3})$
• C. $(\frac{13}{6},\frac{11}{6})$
• D. $(\frac{5}{12},\frac{1}{12})$

Answer 1

The correct option is C $(\frac{13}{6},\frac{11}{6})$

${\cos }^2\pi x-{\sin }^2\pi y=\frac{1}{2}$
$\Rightarrow \cos \left(\pi \right(x+y\left)\right)\cos \left(\pi \right(x-y\left)\right)=\frac{1}{2}$
$[\because \cos (A+B\left)\cos \right(A-B)={\cos }^{2}A-{\sin }^{2}B]$
$\Rightarrow \cos \left(\pi \right(x+y\left)\right)\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}$
$\Rightarrow \cos \left(\pi (x+y)\right)=1$
$\Rightarrow \pi (x+y)=2n\pi$, where $n\in Z$
$\Rightarrow x+y=2n$ and $x-y=\frac{1}{3}$
$\Rightarrow x=n+\frac{1}{6}$ and $y=n-\frac{1}{6}$
$(x,y)=(n+\frac{1}{6},n-\frac{1}{6})$ which is satisfied by $(\frac{13}{6},\frac{11}{6})$ for $n=2$.