Question

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

Answer 1

hint :

max value of sin^-1 x ?

Answer 2

ye sawal kaha se liya hai

Answer 3

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

Answer 4

great hint* :P

Answer 5

u mean max value of sinx-1 or sinx

Answer 6

sin^-1

Answer 7

inverse

Answer 8

pi/2 + pi/2 +pi/2 do u mean that

Answer 9

yep..the only possible ans..

Answer 10

hmm i don't think so

Answer 11

really,,what makes you argue ?

Answer 12

try that formula sin-1x + sin-1y=sin-1(x(1-y^2)^1/2+y(1-x^2)^1/2)

Answer 13

bring sins into one form like sin-1(.....)=3pi/2

Answer 14

ok u are may be right

Answer 15

i was just trying to confirm it

Answer 16

ofcorse am right,,
suppose i ask you sinx +siny + sinz =3, then ofcorse its possible when all are =1

Answer 17

same with this case..
hence x=y=z =1

Answer 18

@shubhamsrg Why Did u Take sin as Maximum ? Sorry My Connection was Out Of Order :)

Answer 19

@satellite73 @RadEn

Answer 20

@shubhamsrg provided you with the answer

Answer 21

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

Answer 22

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 \(\frac{3\pi}{2}\)