(2) 3條sin cos tan

7.
(sin(3π - θ)) / (tan(π/2 +θ)) ‧ (cos(4π -θ)) / (sin (-θ - 2π))
=(sinθ/-cotθ)‧ (-cosθ) / (sinθ)
=-sinθ
8
LHS
=sin^2(θ - 270度)　+ cos^2(90度 + θ ) + tan^2 (θ - 360度 )
=sin^2(270度-θ)　+ cos^2(90度 + θ ) + tan^2 (360度-θ )
=cos^2θ+sin^2θ+tan^2θ
=1+tan^2θ
= sec^2 θ
=RHS
9
(sec(-θ) + sin (-θ - 90度) ) / ((cosec(540度 - θ ) - cos(270度 + θ)
=(-sec θ+cos θ)/((cosec(180度 - θ ) - cos(270度 + θ)
=(-sec θ+cos θ)/(csc θ - sinθ)
=(cos^2 θ-1)(sinθ)/(1-sin^2θ)(cosθ)
=sin^3θ/cos^3θ
=tan^3θ



2007-01-28 19:45:57 補充：
因為任何一種三角函數加2nπ都不改其值如sin(2nπ+θ)=sinθ你明白此點就可以完全破解0西此類題目

2007-01-28 20:01:26 補充：
還有少少技巧﹐好似sin^2(270度-θ)你只要知道sin在這情況下轉cos便成﹐不用考慮正負﹐因為平方完一定變正所以我直接寫cos^2θ我也不是亂跳步的

7) (sin(3π - θ)) / (tan(π/2 +θ)) ‧ (cos(4π -θ)) / (sin (-θ - 2π))
= (sin(2π + (π - θ)))/-cot θ ‧ cos (2*2π - θ) /-(sin (θ + 2π))
=sinθ/(cosθ/sinθ) ‧ cosθ/-sinθ = -sinθ
8)LHS = sin^2(θ - 270°) + cos^2(90° + θ ) + tan^2 (θ - 360°)
=sin^2(180°+90°-θ) + (-sinθ)^2 + tan^2 (360°-θ)
=(cosθ)^2 + (sinθ)^2 + (tanθ)^2
=1 + (tanθ)^2 = (secθ)^2 = RHS
9)LHS = (sec(-θ) + sin (-θ - 90°) ) / ((csc(540° - θ ) - cos(270° + θ)
=(secθ - sin (θ + 90°))/((csc(360°+180° - θ ) - cos(180°+90° + θ))
=(secθ - cosθ)/(cscθ + sinθ)
=(1-(cosθ)^2)sinθ/(1-(sinθ)^2)cosθ
=(sinθ)^2 sinθ / (cosθ)^2 cosθ = (tan θ)^3