Question

已知sin(π+α)=-1/2,求cos(2π-α)，tan（α-7π）

Answer 1

sin(π+α)=-1/2
得出a是第一、二象限角
sina=1/2
cosa=±√3/2
tana=±√3/3

cos(2π-a)=cos(-a)=cosa=±√3/2
tan（α-7π）=tana=±√3/3

Answer 2

1.
tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)]
=tanα+[sin(α+π/2)/cos(-α+π/2)]
=tanα+[sin(α+π/2)/cos(-α+π/2)]
=tanα+[sin(π/2-α)/cos(π/2-α)]
=tanα+cosα/sinα
=(sinα/cosα)+(cosα/sinα)
=1/(sinαcosα)
cosα=±√(1-4/5)=√5/5,sinαcosα=±2/5
tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)]=±5/2
2.
(1)角α是第二象限角,cosα<0
tanα√[(1/sin^α)-1]
=tanα·√[(1-sin^α)/sin^α]
=tanα·√(cos^α/sin^a)
=tanα·(-cosα/sinα)
=-1
(2)
[√（1-2sin130°cos130°）]/[sin130°+√1-sin^2(130°)]
=[√（1+2sin50°cos50°）]/[sin50°+√cos^2(130°)]
=[√（sin50°+cos50°）^2]/(sin50°+cos50°)
=（sin50°+cos50°）/(sin50°+cos50°)
=1
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