1.已知sinα=(2√5)/5,求tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)] 2.(1)若角α是第二象限角,化简
1.已知sinα=(2√5)/5,求tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)]
2.(1)若角α是第二象限角,化简tan√[(1/sin^a)-1]
(2)化简:[√(1-2sin130°cos130°)]/[sin130°+√1-sin^2(130°)]
人气:457 ℃ 时间:2020-05-15 09:48:21
解答
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
推荐
- 已知α是第四象限角,且f(α)=sin(π-α)*cos(2π-α)*tan(-α+3π/2)/cos(-α-π),(1)化简f(a).
- 设θ为第二象限角,若tan(θ+π4)=12,则sinθ+cosθ=( ) A.105 B.−105 C.2105 D.−2105
- 已知a为第二象限角,f(a)=sin(a-π/2)cos(3π/2+a)tan(π-a)/tan(-a-π)sin(-a-π)(1)化简f(a)
- 已知sinα=4/5,且α是第二象限角,求cosα,tanα
- 已知α是第三象限角,f(α)=[sin(π-a)cos(2π-a)tan(1.5π-a)tan(-a-π)]/sin(-π-a),化简f(α)
- -What's the ___(关系) between you two? -Teacher and student.
- 当x=-1时,代数式5x+2的绝对值和代数式1-3x的值分别为m,n,则m,n之间的关系为什么?() A.M>N B.M=N
- 山顶洞人生活的集体是由( )结合起来的( )
猜你喜欢