Astonishing Property of Inverse Trigonometric Functions

I'm not sure if someone has noticed it before or not.

But anyways, I found this result today,

$\Re{(\mbox{arcsin } x)} = \left\{ \begin{array}{rl} \frac{\pi}{2} & \forall\ x \geq 1 \\ -\frac{\pi}{2} & \forall\ x \leq -1 \end{array} \right.$

$\Re{(\mbox{arccos } x)} = \left\{ \begin{array}{rl} 0 & \forall\ x \geq 1 \\ \pi & \forall\ x \leq -1 \end{array} \right.$

As @megh choksi suggested in my previous note, I'll not be sharing the proof right now. Instead, I'll wait for a few days for others to try.

Hint - You may read this wiki - Generalizing The Circular Functions

#Geometry #Trigonometry #InverseTrigonometricFunctions #ComplexAnalysis #Kishlaya

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Hi Kishlaya , you considering the aspect of doing Research in Maths ? You'd do well . Given the standard of your questions, the ideas and solving techniques that you come up with I think I know who's gonna come out to be the AIR 1 in JEE Advance in a few months . Don't you think so too ?

A Former Brilliant Member - 6 years, 4 months ago

Even I think, I know who's gonna be JEE AIR 1.

Ofcourse, it's you and who else $:)$

Kishlaya Jaiswal - 6 years, 4 months ago

Dude there is a lot of competition and I can cite examples from Brilliant also with you , Pranjal , Ronak, Megh , Deepanshu topping the list ? Plus Physics and Inorganic Chemistry are Barriers that I have to overcome .

A Former Brilliant Member - 6 years, 4 months ago

@A Former Brilliant Member – Same is the story with me. I find Chemistry to be the biggest hurdle that I have to cross anyhow, to get a good rank in JEE.

Kishlaya Jaiswal - 6 years, 4 months ago

@Kishlaya Jaiswal – Well, I wish you the Best of Luck and hope that we both get into the same IIT and a good one at that !!

A Former Brilliant Member - 6 years, 4 months ago

Thank you .

U Z - 6 years, 4 months ago

@Jake Lai, @Shashwat Shukla , @Azhaghu Roopesh M , @megh choksi , @Pranjal Jain , @Calvin Lin , @Michael Mendrin, @Daniel Liu, @Jon Haussmann @Krishna Sharma

Kishlaya Jaiswal - 6 years, 4 months ago

Yeah , I'm on it . And check out my latest problem Kishlaya , it might be of significant interest to you !!!

Check it out !!!

@Kishlaya Jaiswal , @megh choksi

A Former Brilliant Member - 6 years, 4 months ago

Wow, that's an amazing problem and even the title being more amazing. And now, you're completely flattering me. Honestly speaking, I don't deserve that much brag.

You all (each and every Brilliantian) has been my inspiration. $:)$

Kishlaya Jaiswal - 6 years, 4 months ago

@Kishlaya Jaiswal – Hey, I have a series of questions planned in which I'll be using concepts I have learned from you . So where do you want me to provide the links to those questions ? Seeing currently you won't get them as soon as I post them .

A Former Brilliant Member - 6 years, 4 months ago

I'll just prove for $\Re(\arcsin(x)) = \frac{\pi}{2} \forall x \geq 1$ as a HINT for others to prove the rest $\ddot\smile$

We all know that $\arcsin (n) = -i\cdot ln(i\cdot n \pm \sqrt{1-n^{2}}) \\= -i\cdot ln(i\cdot n \pm i\sqrt{m} \\= -i\cdot ln(i(n \pm \sqrt{m}))$

Now we use $e^{ix} = cos(x) + i\cdot sin(x) \\ \quad \text{and} \quad \\= ln(i\cdot x) = ln(x\cdot e^{i\cdot \frac{\pi}{2}}) \\= -i\cdot ( ln(x) + i\cdot \frac{\pi}{2}) \\= \frac{\pi}{2} - ln(x)$

So we get $\\ \rightarrow \frac{\pi}{2} - i\cdot ln(n \pm \sqrt{m})$

Therefore $\Re(\arcsin(x)) = \frac{\pi}{2}$

A Former Brilliant Member - 6 years, 4 months ago

@Jake Lai Is it possible to make the fonts look big ?

A Former Brilliant Member - 6 years, 4 months ago

Perfect!!! And similarly, we can work out the other relations too.

Kishlaya Jaiswal - 6 years, 4 months ago

Can I say this ?

$\Re{(\mbox{arcsin } x)} = \Re(\dfrac{\pi}{2} - cos^{-1} x)$

Thus $\dfrac{\pi}{2}$

U Z - 6 years, 4 months ago

Yep, but not exactly because there will be a contradiction.

Because $\Re\left(\frac{\pi}{2} - \cos^{-1}x \right) = \frac{\pi}{2} - \Re\left(\cos^{-1}x \right)$

And then you need to make assumption that $\Re\left(\cos^{-1}x \right) = 0$

And then to prove the above assumption, you will again need $\Re(\cos^{-1}x) = \Re\left(\frac{\pi}{2} - \sin^{-1}x \right)$

Kishlaya Jaiswal - 6 years, 4 months ago

I mean for x>0 , arc(cosx) will always be imaginary , so real part of will be pi/2.

For $cos^{-1} x$ , the same way you proved,

$cos^{-1} x = ilog(x + i\sqrt{1 - x^2})$

For imaginary as well as real , it gives

$cos^{-1} x + sin^{-1} x = \dfrac{\pi}{2}$

Please correct me if I am wrong

U Z - 6 years, 4 months ago

@U Z – Hmm, that seems perfectly reasonable.

Kishlaya Jaiswal - 6 years, 4 months ago

@Kishlaya Jaiswal – Sorry for the suggestion , now as you wish please you post a wiki on it , will enjoy learning from you.

A request can you post a wiki on double summation.

U Z - 6 years, 4 months ago

@U Z – Actually, I'll add that to the existing wiki itself.

And yes, I'll try my best to post a wiki on double summation as soon as I get time because I've my board examinations going on so it's a little difficult. Ok, so I'll write little by little each day and once the article gets completed, I'll be posting it.

Kishlaya Jaiswal - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$