cos sin tan

2007-01-29 5:06 am
3.(secθ - cos θ)(1+cotθ + tanθ)=(sec^2 θ)/(cosec θ) + (sec θ)/(cosec^2 θ)
6.sin( (7π/2)/(2) + θ)
http://hk.knowledge.yahoo.com/question/?qid=7007012804145<<一樣

回答 (1)

2007-01-29 5:27 am
✔ 最佳答案
3.(secθ - cos θ)(1+cotθ + tanθ)=(sec^2 θ)/(cosec θ) + (sec θ)/(cosec^2 θ)
LHS
=(secθ - cos θ)(1+cotθ + tanθ)
=secθ+secθcotθ+secθtanθ-cosθ-cosθcotθ-cosθtanθ
=secθ+cscθ+sinθ/cos^2θ-cosθ-cos^2θ/sinθ-sinθ
=(1-cos^2θ)/sinθ+(cosθ+sinθ)/cos^2θ-(cosθ+sinθ)
=(cosθ+sinθ)/cos^2θ-cosθ
=(cosθ+sinθ-cos^3θ)/cos^2θ
=[cosθ(1-cos^2θ)+sinθ]/cos^2θ
=(sin θ)(1+cos θsin θ)/(cos^2 θ)
RHS
=(sec^2 θ)/(cosec θ) + (sec θ)/(cosec^2 θ)
=sin θ/cos^2 θ+sin^2 θ/cos θ
=(sin θ)(1+cos θsin θ)/(cos^2 θ)
LHS=RHS
6
sin( (7π/2)/(2) + θ)
=sin(7π/4+θ)
=sin(2π-(π/4-θ))
=-sin(π/4-θ)
=-[sin(π/4)cosθ-sinθcosπ/4]
=-(√2/2cosθ-√2/2sinθ)
=√2/2(sinθ-cosθ)


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