The graph of the function $\cos x \cos (x+2)-\cos ^2(x+1)$ is

A straight line passing through $\left(0,-\sin ^2 \theta\right)$ with slope 2

A straight line passing through $(0,0)$

A parabola with vertex $\left(1,-\sin ^2 1\right)$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Medium

1 of 1

Analyzing +4 | -1

The graph of the function

\cos x \cos (x+2)-\cos ^2(x+1)

View answer and explanation

Correct option: D

A straight line passing through the point

\left(\frac{\pi}{2},-\sin ^2 1\right)

and parallel to the

X

-axis

Option D (A straight line passing through the point

\left(\frac{\pi}{2},-\sin ^2 1\right)

and parallel to the

X

-axis) is correct because the given expression simplifies to a constant value,

-\sin^2 1

, which represents a horizontal line.

Practice All Chapter MCQs →

\sin \,\,x\,\,\;cos\,\,3x = \sin \,\,3x\,\,\cos \,\,5x

\left[ {0,\;\,\frac{\pi }{2}} \right]

\left( {x,\,y} \right)

x,\,\,y \in \left[ {0,10} \right]

\left( {\sqrt {{{\sin }^2} - \sin \,\,x + \frac{1}{2}} } \right){2^{{{\sec }^2}y}}\,\, \cdot \,\, \le 1

\mathrm{X}

\theta \rightarrow \frac{\pi}{2}

\sin x+\cos x=\frac{\sqrt{7}}{2}

x \in\left[0, \frac{\pi}{4}\right]

\tan \frac{x}{2}

\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}

0 < x < \pi

\cos x+\sin x=\frac{1}{2}

\tan x