Question

A solution $$\left(x,y\right)$$ of the system of equations $$x-y={\displaystyle \frac{1}{3}}$$ and $${\cos }^2\left(\pi x\right)-{\sin }^2\left(\pi y\right)={\displaystyle \frac{1}{2}}$$ is given by
A) $${\displaystyle \frac{2}{3}},{\displaystyle \frac{1}{3}}$$
B) $${\displaystyle \frac{7}{6}},{\displaystyle \frac{1}{6}}$$
C) $${\displaystyle \frac{13}{6}},{\displaystyle \frac{11}{6}}$$
D) $${\displaystyle \frac{1}{6}},{\displaystyle \frac{5}{6}}$$

Answer 1

Hint:
To solve this we can use the formula $${\cos }^{2}A-{\sin }^{2}B=\cos \left(A+B\right).\cos \left(A-B\right)$$. Here $$A=\pi x$$ and $$B=\pi y$$. So $$A-B$$ will be $$\pi \left(x-y\right)$$ substitute the value of $$x-y={\displaystyle \frac{1}{3}}$$ in this. In this way you will obtain the value of $$\cos \left(A+B\right)$$. Use $$\cos \left(A+B\right)$$ function to arrive at the correct answer among the given options.

Complete step by step solution:
Given: $$x-y={\displaystyle \frac{1}{3}}$$,
$${\cos }^2\left(\pi x\right)-{\sin }^2\left(\pi y\right)={\displaystyle \frac{1}{2}}$$
We know that $${\cos }^{2}A-{\sin }^{2}B=\cos \left(A+B\right).\cos \left(A-B\right)$$ assuming $$A=\pi x$$ and $$B=\pi y$$ we get
$$\begin{gathered} {\cos }^2\left(\pi x\right)-{\sin }^2\left(\pi y\right)=\cos \left(\pi x+\pi y\right)\cos \left(\pi x-\pi y\right) \\ \Rightarrow {\displaystyle \frac{1}{2}}=\cos \left[\pi \left(x+y\right)\right]\cos \left[\pi \left(x-y\right)\right] \\ \Rightarrow {\displaystyle \frac{1}{2}}=\cos \left[\pi \left(x+y\right)\right]\cos \left[\pi \left({\displaystyle \frac{1}{3}}\right)\right] \\ \Rightarrow \cos \left[\pi \left(x+y\right)\right]={\displaystyle \frac{1}{{\displaystyle \frac{2}{\cos {\displaystyle \frac{\pi }{3}}}}}}={\displaystyle \frac{1}{{\displaystyle \frac{{\displaystyle \frac{2}{1}}}{2}}}}=1 \\ \Rightarrow \cos \left[\pi \left(x+y\right)\right]=1 \end{gathered}$$
Let $$x+y=n$$
Then $$\cos n\pi =1$$
$$x-y={\displaystyle \frac{1}{3}}$$ (where $$n$$ is an integer)
$$\begin{gathered} \Rightarrow 2x=2n+{\displaystyle \frac{1}{3}} \\ \Rightarrow x=\left(n+{\displaystyle \frac{1}{6}}\right) \\ \Rightarrow y=2n-x=2n-n-{\displaystyle \frac{1}{6}} \\ \Rightarrow y=n-{\displaystyle \frac{1}{6}} \end{gathered}$$
Putting
$$\begin{gathered} \left(x,y\right)=\left(x,y\right)\equiv \left({\displaystyle \frac{1}{6}},-{\displaystyle \frac{1}{6}}\right) \\ n=1 \\ \Rightarrow \left(x,y\right)=\left({\displaystyle \frac{4}{6}},{\displaystyle \frac{5}{6}}\right) \\ n=2 \end{gathered}$$
$$\Rightarrow \left(x,y\right)\equiv \left({\displaystyle \frac{13}{6}},{\displaystyle \frac{11}{6}}\right)$$ (Only this option matches)

The correct answer is option (c) $$\left({\displaystyle \frac{13}{6}},{\displaystyle \frac{11}{6}}\right)$$

Note:
In multiple questions like these it might not be possible to obtain the correct solution only from the question. You may require to analyze the options to arrive at the correct answer. Similar questions might be asked that requires the knowledge of other trigonometric equations.