The value of cos left(sin ^{-1} frac{pi}{3}+c | vedclass

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

Question

(D) We know that the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $x \in [-1, 1]$.Here,the argument is $\frac{\pi}{3}$.Since $\pi \approx 3.14159$,w

Vedclass · Accepted Answer

Does not exist

Vedclass · Answer

(D) We know that the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $x \in [-1, 1]$.Here,the argument is $\frac{\pi}{3}$.Since $\pi \approx 3.14159$,we have $\frac{\pi}{3} \approx 1.047$.Because $1.047 > 1$,the value $\frac{\pi}{3}$ lies outside the domain $[-1, 1]$.Therefore,$\sin ^{-1} \left(\frac{\pi}{3}\right)$ and $\cos ^{-1} \left(\frac{\pi}{3}\right)$ are not defined in the set of real numbers.Thus,the expression $\cos \left(\sin ^{-1} \frac{\pi}{3}+\cos ^{-1} \frac{\pi}{3}\right)$ does not exist.

The value of $\cos \left(\sin ^{-1} \frac{\pi}{3}+\cos ^{-1} \frac{\pi}{3}\right)$ is

Similar Questions

If $\cos ^{-1} x = y$,then . . . . . . .

The domain of the function $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is $(-\infty, -a] \cup [a, \infty)$. Then $a$ is equal to

Let $f:(-1, 1) \to B$ be a function defined by $f(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$. Then $f$ is both one-one and onto when $B$ is the interval:

If $y = \operatorname{cosec}^{-1}(x)$ and $\frac{dy}{dx} = \frac{-1}{|x| \sqrt{x^2-1}}$,then

The range of the real valued function $f(x) = \operatorname{Cos}^{-1}\left(\frac{3}{\sqrt{9x^2 - 12x + 22}}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module