trigonometry Q4

1.(tan135 ﹣sin210) / (tan240 + sin270)
= [-tan45 ﹣(-sin30)] / [tan60 + (-1)]
= [-1 ﹣(-1/2)] / [√3 + (-1)]
= [-1 + (1/2)] / (√3 ﹣1)
= (-1/2) / (√3 ﹣1)
= (-1/2)[1 / (√3 ﹣1)]
= (-1/2)[1 / (√3 ﹣1) X (√3 + 1)/(√3 + 1)]
= (-1/2)[(√3 + 1) / 2]
= - (√3 + 1) / 4

2)[sinθ+cos(360﹣θ)]2 +[cosθ+sin(180+θ)]2
= (sinθ+ cosθ)2 + (cosθ﹣sinθ)2
= sin2θ+ 2sinθcosθ+ cos2θ + cos2θ﹣2sinθcosθ+ sin2θ
= sin2θ+ cos2θ + cos2θ+ sin2θ
= 1 + 1
= 2

3.[sin(360﹣θ) /tan(θ﹣180)] + 1/cos(360﹣θ)﹣cos(θ﹣360)

= (- sinθ / -tanθ) + (1/cosθ)﹣cosθ
= [- sinθ / -(sinθ/cosθ)] + (1/cosθ)﹣cosθ
= [sinθ(cosθ/ sinθ)] + (1/cosθ)﹣cosθ
= cosθ+ (1/cosθ)﹣cosθ
= 1/cosθ

4)sin(180﹣θ)/ [cos(180+θ)cos(180﹣θ)+sin(-θ)sin(180+θ)]˙1/tan(360﹣θ)
= sinθ/ (-cosθ-cosθ) + (-sinθ)(-sinθ)]˙(-1/tanθ)
= sinθ/ (cos2θ) + sin2θ˙(-1/tanθ)
= sinθ/ 1˙(-1/tanθ)
= sinθ˙[-1/(sinθ/ cosθ)
= sinθ˙(-cosθ/sinθ)
= -cosθ


2008-04-14 18:47:02 補充：
tan135° = -tan(180° ﹣135) = -tan45°
sin210° = -sin(180° + 30°) = -sin30°
tan240° = tan(180° + 60°) = tan60°
sin270° = -1