Waste less time on Facebook — follow Brilliant.
×

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

Note by Kishlaya Jaiswal
2 years, 5 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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 ? Azhaghu Roopesh M · 2 years, 5 months ago

Log in to reply

@Azhaghu Roopesh M Even I think, I know who's gonna be JEE AIR 1.

.

.

.

.

.

Ofcourse, it's you and who else \(:)\) Kishlaya Jaiswal · 2 years, 5 months ago

Log in to reply

@Kishlaya Jaiswal 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 . Azhaghu Roopesh M · 2 years, 5 months ago

Log in to reply

@Azhaghu Roopesh M 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 · 2 years, 5 months ago

Log in to reply

@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 !! Azhaghu Roopesh M · 2 years, 5 months ago

Log in to reply

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 \(\Re(\arcsin(x)) = \frac{\pi}{2}\) Azhaghu Roopesh M · 2 years, 5 months ago

Log in to reply

@Azhaghu Roopesh M Perfect!!! And similarly, we can work out the other relations too. Kishlaya Jaiswal · 2 years, 5 months ago

Log in to reply

@Azhaghu Roopesh M @Jake Lai Is it possible to make the fonts look big ? Azhaghu Roopesh M · 2 years, 5 months ago

Log in to reply

Log in to reply

@Kishlaya Jaiswal 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 Azhaghu Roopesh M · 2 years, 5 months ago

Log in to reply

@Azhaghu Roopesh M 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 · 2 years, 5 months ago

Log in to reply

@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 . Azhaghu Roopesh M · 2 years, 5 months ago

Log in to reply

Thank you . Megh Choksi · 2 years, 5 months ago

Log in to reply

Can I say this ?

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

Thus \(\dfrac{\pi}{2}\) Megh Choksi · 2 years, 5 months ago

Log in to reply

@Megh Choksi 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 · 2 years, 5 months ago

Log in to reply

@Kishlaya Jaiswal 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 Megh Choksi · 2 years, 5 months ago

Log in to reply

@Megh Choksi Hmm, that seems perfectly reasonable. Kishlaya Jaiswal · 2 years, 5 months ago

Log in to reply

@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. Megh Choksi · 2 years, 5 months ago

Log in to reply

@Megh Choksi 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 · 2 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...