MAths 問題 想左好內 救人解答@@

( 1 ) From the given conditions,
sin2x = 9 cos2 y
cos2 x = 1 – 9 cos2 y
cos2 x = 1 – 9 ( 1 – sin2 y )
cos2 x = 1 – 9 + 9 sin2y
cos2 x = 9 sin2y – 8
( 1 / 9 )( sin2 z ) = 9 ( 9 cos2 z ) – 8
sin2 z = 729 cos2 z – 72
1 – cos2 z = 729 cos2 z – 72
73 = 730 cos2 z
cos2 z = 1 / 10
sin2 z = 9 / 10
Then , you can find that sin2 x = 9 / 10 and sin2 y = 9 / 10
So,
sin2 x + sin2 y + sin2 z
= 9 / 10 + 9 / 10 + 9 / 10
= 2.7

( 2 ) cscx = 4m2n2 / ( m2 + n2 )2
Then sin x = ( m2 + n2 )2 / 4m2n2
-1 ≦ sin x ≦ 1
Consider sin x ≧ - 1
( m2 + n2 )2 / 4m2n2 ≧ - 1
( m2 + n2 )2 + 4m2n2 ≧ 0
So all real values of m and n can satisfy the condition ( m2 + n2 )2 + 4m2n2 ≧ 0
Then consider sin x ≦ 1
( m2 + n2 )2 / 4m2n2 ≦ 1
( m2 + n2 )2 – 4m2n2 ≦ 0
( m2 + n2 + 2mn )( m2 + n2 – 2mn ) ≦ 0
( m + n ) 2 ( m – n ) 2 ≦ 0
But as you know, ( m + n ) 2 ( m – n ) 2 is ≧ 0, so
( m + n ) 2 ( m – n ) 2 = 0
Hence combining the solutions, m = -n or m = n
For m = - n,
( 3m + 2n )/( 2m + n )
= ( - 3n + 2n ) / ( - 2n + n )
= 1
For m = n,
( 3m + 2n ) / ( 2m + n )
= ( 3n + 2n ) / ( 2n + n )
= 5 / 3

The first one is completely trivial

sin^2x+sin^2y+sin^2z
= (sin x)^2 + (sin y)^2 + (sin z)^2
= (3 cos y)^2 + (3 cos z)^2 + (3 cos x)^2
= 9 (cos y)^2 + 9 (cos z)^2 + 9 (cos x)^2
= 9 [(cos y)^2 + (cos z)^2 + (cos x)^2]
= 9 [ 1 - (sin y)^2 + 1 - (sin z)^2 + 1 - (sin x)^z ]
= 9 [ 3 - ((sin x)^2 + (sin y)^2 + (sin z)^2) ]
= 27 - 9 [(sin x)^2 + (sin y)^2 + (sin z)^2]
10 [(sin x)^2 + (sin y)^2 + (sin z)^2] = 27 [Here is the trick.]
(sin x)^2 + (sin y)^2 + (sin z)^2 = 27/10 = 2.7