Câu `2:`

Vì `0 < a,b < pi/2`

`to sin a , cos a , sin b , cos b > 0`

Ta có : `sin^2 a + cos^2 a = 1`

`to sin a = sqrt{1 - cos^2 a} = sqrt{1 - (3/5)^2} = 4/5`

`sin^2 b + cos^2 b = 1`

`to cos b = sqrt{1 - sin^2 b} = sqrt{1 - (5/13)^2} = 12/13`

`a) sin(a + b) = sin a . cos b + cos a . sin b = 4/5 . 12/13 + 3/5 . 5/13 = 63/65`

`cos(a - b) = cos a . cos b + sin a . sin b = 3/5 . 12/13 + 4/5 . 5/13 = 56/65`

`b)` Ta có :

`sin^4 x + cos^4 x = sin^4 x + cos^4 x + 2sin^2 x . cos^2 x - 2sin^2 x . cos^2 x`

` = (sin^2 x)^2 + (cos^2 x)^2 + 2sin^2 x . cos^2 x - 1/2 . 4sin^2 x . cos^2 x`

` = (sin^2 x + cos^2 x)^2 - 1/2 . (2sin x . cos x)^2`

Ta có : `sin^2 x + cos^2 x = 1 ; sin 2x = 2sin x . cos x`

` = 1^2 - 1/2 . (sin 2x)^2`

` = 1 - 1/2 sin^2 2x`

`to` Hoàn tất chứng minh