Hello !

Today was my Pre-RMO . I was stuck in the following question please help .

**If**

\[ \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \dfrac{3 \pi}{2} \]

**Then find the value of :**

\[ x^9 + y^9 + z^9 + \dfrac{1}{x^9y^9z^9} \]

Well , I had got the following result from this : ( Might be wrong )

\[ x + y + z = -1 \]

but I couldn't manage to do it any further . Please help .

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## Comments

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TopNewestThe range of \(\sin^{-1}\) is \([-\tfrac12\pi,\tfrac12\pi]\). Since \(\sin^{-1}x+\sin^{-1}y+\sin^{-1}z = \tfrac32\pi\), we know the values of \(\sin^{-1}x,\sin^{-1}y,\sin^{-1}z\) straight away (what three numbers between \(-a\) and \(a\) add to \(3a\)?). Thus we know \(x,y,z\).

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Thank You !

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Very simple one! As, \(\sin^{-1} x \to [\frac{- \pi}{2}, \frac{\pi}{2}]\)

So, it leads us to the conclusion that \(x=1, y=1, z=1\)

Then, required value becomes \(\boxed{4}\)

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i thought same . quite easy.

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max value of the following equation is 3pi/2 and hence each angle should be pi/2. if angle is pi/2 then x=y=z=1 then 1 + 1 + 1 + 1 = 4. That's it

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Priansh, can I get the copy of your PRMO paper??

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Well they didn't allow us to take it back home :/

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What is Pre-RMO? Is the full paper available somewhere?

Thanks!

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Promo paper

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In Mumbai region there is Pre-RMO, as the no. of students taking part was much more. While the other regions in the country have RMO directly (including other-than-Mumbai Maharashtra part. We are going to give it on 1st December. Let's discuss some good problems thereafter...!!!

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Well this time , the government has allowed every state to set their own papers for pre-RMO . That is the qualification stage for RMO level 1 .

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In Telangana region,there is no pre-RMO....lucky for me I can directly write RMO paper without qualifying.... BTW....The RMO exam is not so easy as pre-RMO....I was dying to solve some of them...

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