The value of cos left(2 cos ^{-1} x+sin ^{-1} | vedclass

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

Question

(C) Given expression: $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ We can rewrite this as: $\cos \left(\cos ^{-1} x + (\cos ^{-1} x+\sin ^{-1} x)\r

Vedclass · Accepted Answer

$-\frac{2 \sqrt{6}}{5}$

Vedclass · Answer

(C) Given expression: $\cos \left(2 \cos ^{-1} x+\sin ^{-1} xight)$ We can rewrite this as: $\cos \left(\cos ^{-1} x + (\cos ^{-1} x+\sin ^{-1} x)ight)$ Using the property $\cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}$,the expression becomes: $\cos \left(\cos ^{-1} x+\frac{\pi}{2}ight)$ Using the trigonometric identity $\cos \left(	heta+\frac{\pi}{2}ight) = -\sin 	heta$: $= -\sin \left(\cos ^{-1} xight)$ Since $\cos ^{-1} x = \sin ^{-1} \sqrt{1-x^2}$,we have: $= -\sin \left(\sin ^{-1} \sqrt{1-x^2}ight) = -\sqrt{1-x^2}$ Substituting $x=\frac{1}{5}$: $= -\sqrt{1-\left(\frac{1}{5}ight)^2} = -\sqrt{1-\frac{1}{25}} = -\sqrt{\frac{24}{25}} = -\frac{2 \sqrt{6}}{5}$

The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$,where $0 \leq \cos ^{-1} x \leq \pi$ and $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$,is

Similar Questions

If $\cos \left(\cos ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} x\right)=1$,then find the value of $x$.

If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$,then the value of $x$ is

If $(\sin ^{-1} x)^{2}-(\cos ^{-1} x)^{2}=a ; 0 < x < 1, a \neq 0$,then the value of $2 x^{2}-1$ is :

If $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$,then $x$ =

The value of $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$ (where,$x, y>0$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module