Question

Inverse Trigonometry

$\bf Verify \: \red{ LHS }= \red{ RHS }$

$\sin^{-1} (2x\sqrt{1-x^2}) = \cos^{-1} x$

Answer should be in proper way!​

Answer 1

Correct Question:

To prove L.H.S = R.H.S

$\longmapsto \sin^{-1} (2x\sqrt{1-x^2}) = 2\cos^{-1} x$

Solution:

$\bf Let \: x = \cos\theta$

$\dag\large\underbrace{\rm \red{L.H.S}}:$

$\longmapsto \sin^{-1} (2x\sqrt{1-x^2})$

Putting x = cos θ

$\dashrightarrow \sin^{-1} (2\cos\theta\sqrt{1\cos^2\theta})\\\\\dashrightarrow\sin^{-1} (2\cos\theta\sin\theta$

$\dashrightarrow \sin^{-1} (\sin2\theta) \\\\ \dashrightarrow 2\theta$

$\dag\large\underbrace{\rm \red{R.H.S}}:$

$\longmapsto2\cos^{-1} x$

$\dashrightarrow 2\cos^{-1} (\cos\theta) \\\dashrightarrow 2(\theta)\\\dashrightarrow 2\theta$

L.H.S = R.H.S

$\quad\large \underline{\mathfrak{\green{Hence\:Proved..{\displaystyle !\,}}}}$

• Note : There would be $2\cos^{-1} x$ on right hand side. If there is $\cos^{-1} x$ then it cannot be equal .