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प्रश्न
If `(sin^-1x)^2 + (sin^-1y)^2+(sin^-1z)^2=3/4pi^2,` find the value of x2 + y2 + z2
उत्तर
We know that the maximum value of `sin^-1x. sin^-1y, sin^-1z is pi/2` and minimum value of `sin^-1x, sin^-1y, sin^-1z is pi/2`
Now,
For maximum value
LHS `=(sin^-1x)^2+(sin^-1y)^2+(sin^-1z)^2`
`=(pi/2)^2+(pi/2)^2+(pi/2)^2`
`=3/4pi^2=`RHS
and For minimum value
LHS `=(sin^-1x)^2+(sin^-1y)^2+(sin^-1z)^2`
`=(-pi/2)^2+(-pi/2)^2+(-pi/2)^2`
`=3/4pi^2` = RHS
Now, For maximum value
`sin^-1x=pi/2,sin^-1y=pi/2,sin^-1z=pi/2`
⇒ `x = sin pi/2,y=sin pi/2, z = sin pi/2`
⇒ x = 1, y = 1, z = 1
∴ x2 + y2 + z2 = 1 + 1 + 1 = 3
and for minimum value
`sin^-1x=-pi/2,sin^-1y=-pi/2,sin^-1z=-pi/2`
⇒ `x=sin(-pi/2),y=sin(-pi/2),z=sin(-pi/2)`
⇒ x = -1, y = -1, z = -1
∴ x2 + y2 + z2 = 1 + 1 + 1 = 3
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