Question

Prove: arcsin[(x-1)/(x+1)] = 2arctan[sqrt(x)] - pi/2 Prove: arcsin[(x-1)/(x+1)] = 2arctan[sqrt(x)] - pi/2 @Mathematics

Answer 1

\[\sin^{-1}[ (x-1)\div(x+1)] = 2\tan^{-1} (\sqrt{x})-\pi \div 2\]

Answer 2

There are a few ways to skin this cat. My favorite today is this:
Define a function
\[ f(x) = \arcsin\left( \frac{x-1}{x+1} \right) - 2 \arctan\sqrt{x} \]
Show that \( f'(x) = 0 \) and thus \( f(x) = constant \).
Then show that constant is \(\pi/2\) (e.g., substituting \( x = 0 \) ).

Answer 3

correction: constant = \( -\pi/2 \).

Answer 4

what happens if we take sin( ) of both sides


\[\sin^{-1}[ \frac{x-1}{x+1}] = 2\tan^{-1} (\sqrt{x})-\frac{\pi}{2}\]


\[\sin(sin^{-1}[ \frac{x-1}{x+1}]) = sin(2\tan^{-1} (\sqrt{x})-\frac{\pi}{2})\]


\[\frac{x-1}{x+1}=\sin(2 \tan^{-1}(\sqrt{x})-\frac{\pi}{2})\]


\[\frac{x-1}{x+1}=\sin(2 \tan^{-1}(\sqrt{x})) \cos(\frac{\pi}{2})-\sin(\frac{\pi}{2})\cos(2 \tan^{-1}(\sqrt{x}))\]

\[\frac{x-1}{x+1}=0-\cos(2 \tan^{-1}(\sqrt{x}))\]


\[\frac{x-1}{x+1}=-\cos(2 \tan^{-1}(\sqrt{x}))\]

assume we have right triangle

now let \[u=\tan^{-1}(\sqrt{x}) => \tan(u)=\sqrt{x}=\frac{\sqrt{x}}{1} (=\frac{opp}{adj})\]

\[hyp=\sqrt{x+1}\]


so we have cos(2u)

\[\cos(2u)=\cos^2(u)-\sin^2(u)=(\frac{1}{\sqrt{x+1}})^2-(\frac{\sqrt{x}}{\sqrt{x+1}})^2\]
\[=\frac{1}{x+1}-\frac{x}{x+1}=\frac{1-x}{x+1}\]
so we have
\[\frac{x-1}{x+1}=-\frac{1-x}{x+1}\]
which is true

Answer 5

rofl no worries man tyvm :)

Answer 6

woman*

Answer 7

or lady

Answer 8

oh my mistake

Answer 9

or goddess will do

Answer 10

lmao

Answer 11

james's cal way sounds great too
but i like trig
its so neat-o

Answer 12

kk ty

Answer 13

i took the derivative like james said and i got to\[((x+1)^{2}-2\sqrt{x}) / (2x+2\sqrt{x})\]
and i need to show that it is a constant

Answer 14

u there goddess?

Answer 15

sorry the bottom was \[2x \sqrt{x} + 2\sqrt{x}\]

Answer 16

That's not quite right.

The derivative of the arcsin term is
\[ \frac{2}{(1+x)^2} \frac{1}{\sqrt{1 - \left(\frac{x-1}{x+1}\right)^2}} \]
\[= \frac{1}{1+x}\frac{2}{\sqrt{(x+1)^2 - (x-1)^2 }} \]
\[ = \frac{1}{1+x}\frac{2}{\sqrt{4x}} \]
\[ = \frac{1}{\sqrt{x}(1+x)} \]

Now I'll let you figure out the derivative of the arctan term again carefully and add it to this one.

Answer 17

where did the 2/(x+1)^2 come from?

Answer 18

that's the derivative of (x-1)/(x+1)

Answer 19

oh ok thx

Answer 20

let me know if you got it or not?
i have part of the solution using jame's way
you can compare if you want

Answer 21

i guess i will go in post what i have

\[\text{ let } u=\sin^{-1}(\frac{x-1}{x+1}) => \sin(u)=\frac{x-1}{x+1} (=\frac{opp}{hyp})\]
=>\[adj=\sqrt{(x+1)^2-(x-1)^2}=\sqrt{x^2+2x+1-(x^2-2x+1)}=\sqrt{4x}=2 \sqrt{x}\]
so anyways we are trying to find u'
\[u' \cos(u)=\frac{1(x+1)-1(x-1)}{(x+1)^2}=\frac{x+1-x+1}{(x+1)^2}=\frac{2}{(x+1)^2}\]
\[u'=\frac{2}{\cos(u) \cdot (x+1)^2}=\frac{2}{\frac{2 \sqrt{x}}{(x+1)}(x+1)^2}\]
\[=\frac{x+1}{ \sqrt{x} (x+1)^2}\]
ok now moving on to \[\text{ let } m=\tan^{-1}(\sqrt{x}) => \tan(m)=\sqrt{x}=\frac{\sqrt{x}}{1} (=\frac{opp}{adj})\]
=>\[hyp=\sqrt{x+1}\]
\[\sec^2(m) m'=\frac{1}{2 \sqrt{x}}\]
\[m'=\frac{1}{2 \sqrt{x} \sec^2(m)} =\frac{1}{2 \sqrt{x} (\sqrt{x+1})^2}=\frac{1}{2 \sqrt{x}(x+1)}\]
but we have 2m'
\[2m'=\frac{1}{\sqrt{x}(x+1)}\]
so
f'(x)=\[\frac{x+1}{\sqrt{x}(x+1)^2}-\frac{1}{\sqrt{x} (x+1)}=\frac{x+1-(x+1)}{(x+1)^2 \sqrt{x}}=0\]

Answer 22

ok looks good thank you both

Answer 23

so for the arctan part i got \[1/(2\sqrt{x}(x+1))\]
then you whould subtract them?
\[1/(\sqrt{x}(x+1)) - 1/(2\sqrt{x}(x+1))\]

Answer 24

2arctan part

Answer 25

Look again at the function f. The coefficient of the arctan term is -2.

Answer 26

yea what james said

Answer 27

Hence
\[ f'(x) = \frac{1}{\sqrt{x}(1+x)} - \frac{2}{2\sqrt{x}(1+x)} = 0\]

Answer 28

ok now i get it :)