The graph of the function f(x) = cos x cos(x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $f(x) = \cos x \cos(x + 2) - \cos^2(x + 1)$.Using the formula $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$,we have:$f(x) = \frac{1}{2} [\cos(

Vedclass · Accepted Answer

a straight line passing through the point $(\frac{\pi}{2}, -\sin^2 1)$ and parallel to the $x$-axis

Vedclass · Answer

(D) Let $f(x) = \cos x \cos(x + 2) - \cos^2(x + 1)$.Using the formula $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$,we have:$f(x) = \frac{1}{2} [\cos(2x + 2) + \cos(-2)] - \cos^2(x + 1)$.Recall that $2 \cos^2 	heta = 1 + \cos(2	heta)$,so $\cos^2(x + 1) = \frac{1 + \cos(2x + 2)}{2}$.Substituting this into the expression for $f(x)$:$f(x) = \frac{1}{2} \cos(2x + 2) + \frac{1}{2} \cos 2 - \frac{1}{2} - \frac{1}{2} \cos(2x + 2)$.$f(x) = \frac{1}{2} \cos 2 - \frac{1}{2} = -\frac{1}{2} (1 - \cos 2)$.Using $1 - \cos 2	heta = 2 \sin^2 	heta$,we get $1 - \cos 2 = 2 \sin^2 1$.Thus,$f(x) = -\frac{1}{2} (2 \sin^2 1) = -\sin^2 1$.Since $f(x) = -\sin^2 1$ is a constant,the graph is a horizontal straight line parallel to the $x$-axis passing through the point $(0, -\sin^2 1)$ and $(\frac{\pi}{2}, -\sin^2 1)$.

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

Similar Questions

$4 \cos \frac{7 \theta}{2} \cos \frac{3 \theta}{2} \sin 5 \theta = $

$3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin ^{6} x+\cos ^{6} x)$ is equal to

The value of $\cos^{3}\left(\frac{\pi}{8}\right) \cdot \cos\left(\frac{3\pi}{8}\right) + \sin^{3}\left(\frac{\pi}{8}\right) \cdot \sin\left(\frac{3\pi}{8}\right)$ is:

The value of $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{2 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{4 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{6 \pi}{8}+\cos ^4 \frac{7 \pi}{8}+\cos ^4 \frac{8 \pi}{8}$ is equal to:

If $a = \frac{x}{y-z}$,$b = \frac{y}{z-x}$,and $c = \frac{z}{x-y}$,where $x, y$,and $z$ are distinct such that $x-y, y-z, z-x \neq 0$,then what is the value of $ab + bc + ca + abc$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module