Question

The graph of the function $\cos x.\cos \left( x+2 \right)-{{\cos }^{2}}\left( x+1 \right)$ is a
(A) straight line passing through the point $\left( 0,-{{\sin }^{2}}1 \right)$ with slope 2.
(B) straight line passing through the origin.
(C) parabola with vertex $\left( 1,-{{\sin }^{2}}1 \right)$
(D) straight line passing through the point $\left( \dfrac{\pi }{2},-{{\sin }^{2}}1 \right)$ and parallel to the x-axis.

Answer 1

We know the formula, $\cos \left( A-B \right)\cos \left( A+B \right)={{\cos }^{2}}A-{{\sin }^{2}}B$ . Replace A by (x+1) and B by 1, in this formula and get the value of $\cos \left( x \right)\cos \left( x+2 \right)$ . Now, put the value of $\cos \left( x \right)\cos \left( x+2 \right)$ in the equation $\cos x.\cos \left( x+2 \right)-{{\cos }^{2}}\left( x+1 \right)$ and the equation of the curve. Now, plot the graph and after plotting we get a line parallel to x-axis. When a line is parallel to the x-axis then the x-coordinate of points which is on the straight line can be any real number. Now, conclude the answer.

Complete step by step answer:
According to the question, we have the equation of the curve,
$y=\cos x.\cos \left( x+2 \right)-{{\cos }^{2}}\left( x+1 \right)$ ……………………….(1)
The given equation is not in a simplified way. Therefore, we need to simplify it. The, only we will be able to plot its graph.
Now, simplifying equation (1), we get
$y=\cos x.\cos \left( x+2 \right)-{{\cos }^{2}}\left( x+1 \right)$
$\Rightarrow y=\cos \left( x+1-1 \right).\cos \left( x+1+1 \right)-{{\cos }^{2}}\left( x+1 \right)$ …………………..(2)
We know the identity, $\cos \left( A-B \right)\cos \left( A+B \right)={{\cos }^{2}}A-{{\sin }^{2}}B$ ……………..(3)
Replacing A by (x+1) and B by 1, in equation (3), we get
$\cos \left( x+1-1 \right)\cos \left( x+1+1 \right)={{\cos }^{2}}\left( x+1 \right)-{{\sin }^{2}}1$
$\Rightarrow \cos \left( x \right)\cos \left( x+2 \right)={{\cos }^{2}}\left( x+1 \right)-{{\sin }^{2}}1$ ……………………….(4)
From equation (4), we have the value of $\cos \left( x \right)\cos \left( x+2 \right)$ .
Now, putting the value of $\cos \left( x \right)\cos \left( x+2 \right)$ in equation (2), we get

& y=\cos \left( x+1-1 \right).\cos \left( x+1+1 \right)-{{\cos }^{2}}\left( x+1 \right) \\\ & \Rightarrow y={{\cos }^{2}}\left( x+1 \right)-{{\sin }^{2}}1-{{\cos }^{2}}\left( x+1 \right) \\\ & \Rightarrow y=-{{\sin }^{2}}1 \\\ \end{aligned}$$ Now, we have got the equation of the curve which is, $$y=-{{\sin }^{2}}1$$ . We can see that the equation of the curve is of straight line. Now, plotting the graph of the curve, $$y=-{{\sin }^{2}}1$$ .  From the graph we can see that the slope of the straight line is zero and also parallel to x-axis. Any point on the straight line has y coordinate always equal to $$-{{\sin }^{2}}1$$ and the x coordinate of the point is any real number. That is, $$x\in R$$ and $$y=-{{\sin }^{2}}1$$ . From the graph, we can see that the straight line is passing through the point $$\left( \dfrac{\pi }{2},-{{\sin }^{2}}1 \right)$$ and parallel to the x-axis. **So, the correct answer is “Option D”.** **Note:** We can also solve this question by another method. Using the formula, $$\cos A.\cos B=\dfrac{\cos \left( A+B \right)+\cos \left( A-B \right)}{2}$$ for the simplification of the equation $$y=\cos x.\cos \left( x+2 \right)-{{\cos }^{2}}\left( x+1 \right)$$ . Now, simplifying we get $$\begin{aligned} & y=\cos x.\cos \left( x+2 \right)-{{\cos }^{2}}\left( x+1 \right) \\\ & \Rightarrow y=\dfrac{\cos \left( 2x+2 \right)+\cos 2}{2}-{{\cos }^{2}}\left( x+1 \right) \\\ \end{aligned}$$ $$\Rightarrow y=\dfrac{\cos 2\left( x+1 \right)+\cos 2\left( 1 \right)}{2}-{{\cos }^{2}}\left( x+1 \right)$$ ……………………….(1) We know the formula, $$\cos 2A=2{{\cos }^{2}}A-1$$ and $$\cos 2A=1-2{{\sin }^{2}}A$$ . Using this formula in equation (1), we get $$\begin{aligned} & \Rightarrow y=\dfrac{2{{\cos }^{2}}\left( x+1 \right)-1+1-2{{\sin }^{2}}1}{2}-{{\cos }^{2}}\left( x+1 \right) \\\ & \Rightarrow y=\dfrac{2\left( {{\cos }^{2}}\left( x+1 \right)-y={{\sin }^{2}}1 \right)}{2}-{{\cos }^{2}}\left( x+1 \right) \\\ & \Rightarrow y={{\cos }^{2}}\left( x+1 \right)-{{\sin }^{2}}1-{{\cos }^{2}}\left( x+1 \right) \\\ & \Rightarrow y=-{{\sin }^{2}}1 \\\ \end{aligned}$$ Now, we have got the equation of the curve which is, $$y=-{{\sin }^{2}}1$$ . We can see that the equation of the curve is of straight line. Now, plotting the graph of the curve, $$y=-{{\sin }^{2}}1$$ .  From the graph we can see that the slope of the straight line is zero and also parallel to x-axis. Any point on the straight line has y coordinate always equal to $$-{{\sin }^{2}}1$$ and the x coordinate of the point is any real number. That is, $$x\in R$$ and $$y=-{{\sin }^{2}}1$$ . From the graph, we can see that the straight line is passing through the point $$\left( \dfrac{\pi }{2},-{{\sin }^{2}}1 \right)$$ and parallel to the x-axis. Hence, option (D) is the correct one.