超簡單三角函數(急)!!!要求詳解

設
x = cos^-1 (- 4/5) , (0 ≤ x ≤ π) ,
y = tan^-1 (- 1) , (-π/2 ≤ y ≤ π/2)則
cos x = - 4/5 , 故 sin x = √(1 - (-4/5)²) = 3/5 , 得 tan x = (3/5) / (-4/5) = - 3/4
tan y = - 1 原式 = tan(x + y) = (tan x + tan y) / (1 - tan x tany)= (- 3/4 - 1) / (1 - (-3/4)(- 1))= - 7

2011-05-01 17:41:46 補充：
y = tan^-1 (- 1) , (-π/2 < y < π/2)