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Yahoo!

  • 3 years ago

Find x^2 + y^2+ z^2 - 4xyz if \[\sin ^{-1}x+\sin ^{-1} y+\sin ^{-1}z = \frac{ 3\Pi }{ 2 }\]?

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  1. shubhamsrg
    • 3 years ago
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    hint : max value of sin^-1 x ?

  2. Fazeelayaz
    • 3 years ago
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    ye sawal kaha se liya hai

  3. shubhamsrg
    • 3 years ago
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    arre,,arent you getting the catch ? that was a great thing acc to me,,whats the max value of xin^-1 x? o.O

  4. shubhamsrg
    • 3 years ago
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    great hint* :P

  5. Fazeelayaz
    • 3 years ago
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    u mean max value of sinx-1 or sinx

  6. shubhamsrg
    • 3 years ago
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    sin^-1

  7. shubhamsrg
    • 3 years ago
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    inverse

  8. Fazeelayaz
    • 3 years ago
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    pi/2 + pi/2 +pi/2 do u mean that

  9. shubhamsrg
    • 3 years ago
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    yep..the only possible ans..

  10. Fazeelayaz
    • 3 years ago
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    hmm i don't think so

  11. shubhamsrg
    • 3 years ago
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    really,,what makes you argue ?

  12. Fazeelayaz
    • 3 years ago
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    try that formula sin-1x + sin-1y=sin-1(x(1-y^2)^1/2+y(1-x^2)^1/2)

  13. Fazeelayaz
    • 3 years ago
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    bring sins into one form like sin-1(.....)=3pi/2

  14. Fazeelayaz
    • 3 years ago
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    ok u are may be right

  15. Fazeelayaz
    • 3 years ago
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    i was just trying to confirm it

  16. shubhamsrg
    • 3 years ago
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    ofcorse am right,, suppose i ask you sinx +siny + sinz =3, then ofcorse its possible when all are =1

  17. shubhamsrg
    • 3 years ago
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    same with this case.. hence x=y=z =1

  18. Yahoo!
    • 3 years ago
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    @shubhamsrg Why Did u Take sin as Maximum ? Sorry My Connection was Out Of Order :)

  19. Yahoo!
    • 3 years ago
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    @satellite73 @RadEn

  20. anonymous
    • 3 years ago
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    @shubhamsrg provided you with the answer

  21. mukushla
    • 3 years ago
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    i go with @shubhamsrg\[\sin ^{-1}x+\sin ^{-1} y+\sin ^{-1}z \le \frac{ 3\pi }{ 2 }\]equality occurs when\[\sin ^{-1}x=\sin ^{-1} y=\sin ^{-1}z = \frac{\pi }{ 2 }\]

  22. anonymous
    • 3 years ago
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    you have \[\sin^{-1}(x)+\sin^{-1}(y)+\sin^{-1}(z)=\frac{3\pi}{2}\] but the the very largest \(\sin^{-1}(x)\) can be is \(\frac{\pi}{2}\) so they must each be \(\frac{\pi}{2}\), otherwise they cannot add to \(\frac{3\pi}{2}\)

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