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,

(arcsin x)={π2 x1π2 x1 \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.

(arccos x)={0 x1π x1 \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
4 years, 9 months ago

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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 - 4 years, 9 months ago

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

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.

.

.

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

Kishlaya Jaiswal - 4 years, 9 months ago

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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 - 4 years, 9 months ago

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@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 - 4 years, 9 months ago

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@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 - 4 years, 9 months ago

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Thank you .

U Z - 4 years, 9 months ago

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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 - 4 years, 9 months ago

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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 - 4 years, 9 months ago

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@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 - 4 years, 9 months ago

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I'll just prove for (arcsin(x))=π2x1\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)=iln(in±1n2)=iln(in±im=iln(i(n±m)) \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 eix=cos(x)+isin(x)and=ln(ix)=ln(xeiπ2)=i(ln(x)+iπ2)=π2ln(x) 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 π2iln(n±m) \\ \rightarrow \frac{\pi}{2} - i\cdot ln(n \pm \sqrt{m})

Therefore (arcsin(x))=π2\Re(\arcsin(x)) = \frac{\pi}{2}

A Former Brilliant Member - 4 years, 9 months ago

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

A Former Brilliant Member - 4 years, 9 months ago

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

Kishlaya Jaiswal - 4 years, 9 months ago

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Can I say this ?

(arcsin x)=(π2cos1x) \Re{(\mbox{arcsin } x)} = \Re(\dfrac{\pi}{2} - cos^{-1} x)

Thus π2\dfrac{\pi}{2}

U Z - 4 years, 9 months ago

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

Because (π2cos1x)=π2(cos1x)\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 (cos1x)=0\Re\left(\cos^{-1}x \right) = 0

And then to prove the above assumption, you will again need (cos1x)=(π2sin1x)\Re(\cos^{-1}x) = \Re\left(\frac{\pi}{2} - \sin^{-1}x \right)

Kishlaya Jaiswal - 4 years, 9 months ago

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

For cos1xcos^{-1} x , the same way you proved,

cos1x=ilog(x+i1x2) cos^{-1} x = ilog(x + i\sqrt{1 - x^2})

For imaginary as well as real , it gives

cos1x+sin1x=π2 cos^{-1} x + sin^{-1} x = \dfrac{\pi}{2}

Please correct me if I am wrong

U Z - 4 years, 9 months ago

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@U Z Hmm, that seems perfectly reasonable.

Kishlaya Jaiswal - 4 years, 9 months ago

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@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 - 4 years, 9 months ago

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@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 - 4 years, 9 months ago

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