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

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TopNewestHi 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 ?

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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 \(:)\)

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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 .

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

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

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

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

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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

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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. \(:)\)

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Thank you .

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Can I say this ?

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

Thus \(\dfrac{\pi}{2}\)

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

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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

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A request can you post a wiki on double summation.

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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.

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