Question

If `(sin^-1x)^2 + (sin^-1y)^2+(sin^-1z)^2=3/4pi^2,`  find the value of x² + y² + z² 

Answer 1

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