Question

If $x-y=\frac{1}{3}$ and ${\cos }^2\left(\pi x\right)-{\sin }^2\left(\pi y\right)=\frac{1}{2}$, then $(x,y)$ can be

• A. $(\frac{2}{3},\frac{1}{3})$
• B. $(\frac{13}{6},\frac{11}{6})$
• C. $(\frac{7}{6},\frac{5}{6})$
• D. $(\frac{19}{6},\frac{17}{6})$

Answer 1

Answer: (B), (C), (D)

$(\frac{19}{6},\frac{17}{6})$

$x-y=\frac{1}{3}$
${\cos }^2\left(\pi x\right)-{\sin }^2\left(\pi y\right)=\frac{1}{2}$
$\Rightarrow \cos \pi (x+y)\cdot \cos \pi (x-y)=\frac{1}{2}$
$\Rightarrow \cos \pi (x+y)\cdot \cos \frac{\pi }{3}=\frac{1}{2}$
$\Rightarrow \cos \pi (x+y)=1$
​​​​​​​​​​​​​​​​​​​​​$\Rightarrow \pi (x+y)=2n\pi$
​​​​​​​​​​​​​​​​​​​​​$\Rightarrow (x+y)=2n,n\in Z$

Hence, from the given options the value of $(x,y)$ are
$(\frac{13}{6},\frac{11}{6}),(\frac{7}{6},\frac{5}{6}),(\frac{19}{6},\frac{17}{6})$