Question

2Sin^(-1) x = Sin^(-1)⁡ {2x*√(1-x^2)}⁡ it's okay. but i think it's proof is if we assume x= sin z?
2Sin^(-1) x = Sin^(-1)⁡ {2x*√(1-x^2)}⁡ it's okay. but i think
it's proof is
if we assume x= sin z

sin⁻¹(2sin z*sqrt(1-sin^2 z))
=sin⁻¹(2sin z cos z )
=sin⁻¹(sin 2z)
=2z
=2 sin⁻¹x
2cos^(-1) x = Sin^(-1)⁡ {2x*√(1-x^2)}⁡ it's also okay

it's my thinking
if we assume x= cos z

sin⁻¹(2cos z*sqrt(1-cos^2 z))
=sin⁻¹(2cos z sin z )
=sin⁻¹(sin 2z)
=2z
=2 cos⁻¹x

so now my question is are they both correct?
i want to differentiate this Sin^(-1)⁡ {2x*√(1-x^2)}
so if we use 2 sin^-1 x , then the answer will be 2/√(1-x^2)

and if we use 2 cos^-1 x , then the answer will be -2/√(1-x^2). there are two different answer after differentiate. please let me know.which one is correct or both correct??

Answer 1

Answer on Question #50654 – Math - Trigonometry

Problem.

2Sin^(-1) x = Sin^(-1) 📄 {2x*√(1-x^2)} 📄 it's okay. but i think it's proof is if we assume x = sin z?

2Sin^(-1) x = Sin^(-1) 📄 {2x*√(1-x^2)} 📄 it's okay. but i think

it's proof is

if we assume $x = \sin z$

$$\sin^{-1}(2 \sin z * \sqrt{1 - \sin^2 z})$$ $$= \sin^{-1}(2 \sin z \cos z)$$ $$= \sin^{-1}(\sin 2z)$$ $$= 2z$$ $$= 2 \sin^{-1}x$$ $$2 \cos^(-1) x = \sin^(-1) 📄 {2x*√(1-x^2)} 📄 it's also okay$$

it's my thinking

if we assume $x = \cos z$

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

so now my question is are they both correct?

i want to differentiate this Sin^(-1) 📄 {2x*√(1-x^2)}

so if we use $2 \sin^2 - 1x$, then the answer will be $2 / \sqrt{1 - x^2}$

and if we use $2 \cos^2 - 1x$, then the answer will be $-2 / \sqrt{1 - x^2}$. there are two different answers after differentiate. please let me know which one is correct or both correct?

Solution:

In your proof there is a mistake $\sin^{-1}(\sin 2z) = 2z$ only for $-\frac{\pi}{2} \leq 2z \leq \frac{\pi}{2}$.

Also the formula $2\sin^{-1}x = \sin^{-1}x\sqrt{1 - x^2}$ is correct only for $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ .

$y(x) = 2\sin^{-1}x$ black curve, $y(x) = \sin^{-1}x\sqrt{1 - x^2}$ red curve.

Also function $\sin^{-1}(2\sin z\sqrt{1 - \sin^2z})$ and $\sin^{-1}(2\cos z\sqrt{1 - \cos^2z})$ are different functions of argument $x$ , as in the first case $x = \sin z$ and in the second case $x = \cos z$ , so they could have different derivatives and both is true.

www.AssignmentExpert.com