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

Mathematics
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hint : max value of sin^-1 x ?
ye sawal kaha se liya hai
arre,,arent you getting the catch ? that was a great thing acc to me,,whats the max value of xin^-1 x? o.O

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Other answers:

great hint* :P
u mean max value of sinx-1 or sinx
sin^-1
inverse
pi/2 + pi/2 +pi/2 do u mean that
yep..the only possible ans..
hmm i don't think so
really,,what makes you argue ?
try that formula sin-1x + sin-1y=sin-1(x(1-y^2)^1/2+y(1-x^2)^1/2)
bring sins into one form like sin-1(.....)=3pi/2
ok u are may be right
i was just trying to confirm it
ofcorse am right,, suppose i ask you sinx +siny + sinz =3, then ofcorse its possible when all are =1
same with this case.. hence x=y=z =1
@shubhamsrg Why Did u Take sin as Maximum ? Sorry My Connection was Out Of Order :)
@shubhamsrg provided you with the answer
i go with @shubhamsrg\[\sin ^{-1}x+\sin ^{-1} y+\sin ^{-1}z \le \frac{ 3\pi }{ 2 }\]equality occurs when\[\sin ^{-1}x=\sin ^{-1} y=\sin ^{-1}z = \frac{\pi }{ 2 }\]
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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