三角函數的證明

LHS
= sin(750°+x) tan(765°+x) tan(225°-x) sec(x-60°)
= sin(720°+30°+x) tan(720°+45°+x) tan(225°-x) 1/cos(x-60°)
= sin(30°+x) tan(45°+x) tan(225°-x) 1/cos(x-60°)【720°=轉左兩個圈，即係同0°一樣】
= sin(30°+x) tan(45°+x) tan(180°+45°-x) 1/cos(x-60°)
= sin(30°+x) tan(45°+x) tan(45°-x) 1/cos(x-60°)【tan(180°+θ) = tanθ】
= cos(90° - (30°+x)) 1/tan(90° - (45°+x)) * tan(45°-x) 1/cos(x-60°)【tan(90°-θ)=1/tanθ 及 cos(90°-θ)=1/sinθ】
= cos(60°-x) 1/tan(45°-x) * tan(45°-x) 1/cos(x-60°)
= 1【分子分母約簡】
= RHS


希望幫倒你！^^

2006-12-10 22:52:17 補充：
As far as I know, I didn't see any mistake. Please mail me for the mistake I made. Thank you.

LHS = sin(750+x)tan(765+x)tan(225-x)sec(x-60)
= sin(30+x)tan(45+x)tan[-(135+x)] / cos(x-60)
= cos[90-(30+x)tan(45+x)tan[-(135+x)] / cos(60-x) <-- 考慮 cos(x-60) = cos[360-(60-x)]
= tan(45+x)[-tan(135+x)]<-- 考慮tan(-x) = -tan(x)
= tan(45+x)tan[180-(135+x)] <-- 考慮tan(180-x) = -tanx
= tan(45+x)tan(45-x)
= (tan45+tanx)(tan45-tanx) / (1-tan45tanx)(1+tan45tanx) <-- tan45 = 1
= (1+tanx)(1-tanx) / (1-tanx)(1+tanx)
= 1
=RHS
做完~