sin (2{sin ^{ - 1}}0.8) = | vedclass

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

Question

(A) Let $	heta = \sin^{-1}(0.8)$. Then $\sin 	heta = 0.8 = \frac{4}{5}$.We need to find $\sin(2	heta)$.Using the identity $\sin(2	heta) = 2 \sin \

Vedclass · Accepted Answer

$0.96$

Vedclass · Answer

(A) Let $	heta = \sin^{-1}(0.8)$. Then $\sin 	heta = 0.8 = \frac{4}{5}$.We need to find $\sin(2	heta)$.Using the identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$.Since $\sin 	heta = \frac{4}{5}$,we have $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.Therefore,$\sin(2	heta) = 2 	imes \frac{4}{5} 	imes \frac{3}{5} = \frac{24}{25} = 0.96$.

$\sin (2{\sin ^{ - 1}}0.8) = $

Similar Questions

If $4 \sin ^{-1} x + 6 \cos ^{-1} x = 3 \pi$,where $-1 \leq x \leq 1$,then $x =$

Find the value of $\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$

$\cot \left[ \cos^{-1} \left( \frac{7}{25} \right) \right] = $

Let the function $g: (-\infty, \infty) \rightarrow \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ be given by $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$. Then,$g$ is

If $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)$,then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module