The maximum value of sin (cos (tan x)) is | vedclass

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

Question

(B) Let $f(x) = \sin (\cos (	an x))$.We know that the range of the tangent function $	an x$ is $(-\infty, \infty)$.Therefore,the range of $\cos (	a

Vedclass · Accepted Answer

$\sin 1$

Vedclass · Answer

(B) Let $f(x) = \sin (\cos (	an x))$.We know that the range of the tangent function $	an x$ is $(-\infty, \infty)$.Therefore,the range of $\cos (	an x)$ is $[-1, 1]$.Let $u = \cos (	an x)$,so $u \in [-1, 1]$.We want to find the maximum value of $g(u) = \sin u$ for $u \in [-1, 1]$.Since the sine function is strictly increasing on the interval $[-1, 1]$,the maximum value occurs at the largest value of $u$,which is $u = 1$.Thus,the maximum value is $\sin(1)$.

The maximum value of $\sin (\cos (\tan x))$ is

Similar Questions

Secondary productivity is the rate of formation of new organic matter by:

The utility of a microorganism like $Trichoderma$ $harzianum$ has been proven in which of the following?

If $f: R \rightarrow C$ is defined by $f(x)=e^{2 i x}$ for $x \in R$,then $f$ is (where $C$ denotes the set of all complex numbers)

If $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$ for $x \in R$,then $f(2002) = $

If $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$,then $a$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module