1.已知α是第二象限角,f(α)=[sin(3π/2-α)·cos(2π-α)·tan(π-α)]/-cos(-π-α).(1)化简f(α)(2)若sin(π/2 -α)=-½,求f(α)的值.2.若π/2<α<π,化简√(1/2+1/2√1/2+1/2cos2α)3.已知0<α<(π/4),0<β<(π/4),
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![1.已知α是第二象限角,f(α)=[sin(3π/2-α)·cos(2π-α)·tan(π-α)]/-cos(-π-α).(1)化简f(α)(2)若sin(π/2 -α)=-½,求f(α)的值.2.若π/2<α<π,化简√(1/2+1/2√1/2+1/2cos2α)3.已知0<α<(π/4),0<β<(π/4),](/uploads/image/z/5189057-17-7.jpg?t=1.%E5%B7%B2%E7%9F%A5%CE%B1%E6%98%AF%E7%AC%AC%E4%BA%8C%E8%B1%A1%E9%99%90%E8%A7%92%2Cf%28%CE%B1%29%3D%5Bsin%283%CF%80%2F2-%CE%B1%29%C2%B7cos%282%CF%80-%CE%B1%29%C2%B7tan%28%CF%80-%CE%B1%29%5D%2F-cos%28-%CF%80-%CE%B1%29.%281%29%E5%8C%96%E7%AE%80f%28%CE%B1%29%282%29%E8%8B%A5sin%28%CF%80%2F2+-%CE%B1%29%3D-%26frac12%3B%2C%E6%B1%82f%28%CE%B1%29%E7%9A%84%E5%80%BC.2.%E8%8B%A5%CF%80%2F2%EF%BC%9C%CE%B1%EF%BC%9C%CF%80%2C%E5%8C%96%E7%AE%80%E2%88%9A%281%2F2%2B1%2F2%E2%88%9A1%2F2%2B1%2F2cos2%CE%B1%293.%E5%B7%B2%E7%9F%A50%EF%BC%9C%CE%B1%EF%BC%9C%EF%BC%88%CF%80%2F4%EF%BC%89%2C0%EF%BC%9C%CE%B2%EF%BC%9C%EF%BC%88%CF%80%2F4%EF%BC%89%2C)
1.已知α是第二象限角,f(α)=[sin(3π/2-α)·cos(2π-α)·tan(π-α)]/-cos(-π-α).(1)化简f(α)(2)若sin(π/2 -α)=-½,求f(α)的值.2.若π/2<α<π,化简√(1/2+1/2√1/2+1/2cos2α)3.已知0<α<(π/4),0<β<(π/4),
1.已知α是第二象限角,f(α)=[sin(3π/2-α)·cos(2π-α)·tan(π-α)]/-cos(-π-α).
(1)化简f(α)
(2)若sin(π/2 -α)=-½,求f(α)的值.
2.若π/2<α<π,化简√(1/2+1/2√1/2+1/2cos2α)
3.已知0<α<(π/4),0<β<(π/4),且3sinβ=sin(2α+β),4tan(α/2)=1-tan²(α/2),求α+β的值.
我数学很不好,遇见这种题没有思路,
1.已知α是第二象限角,f(α)=[sin(3π/2-α)·cos(2π-α)·tan(π-α)]/-cos(-π-α).(1)化简f(α)(2)若sin(π/2 -α)=-½,求f(α)的值.2.若π/2<α<π,化简√(1/2+1/2√1/2+1/2cos2α)3.已知0<α<(π/4),0<β<(π/4),
1.(1)f(α)=【-cosα·(-cosα)(-tanα)】/(-cosα)=sinα,看π/2的倍数,奇变偶不变,再判断符号
(2)cosα=-1/2,∴α=2π/3,f(α)=√3/2
2.sin(α/2)利用cos2x=1-2(sinx)^2=2(cosx)^2-1逐步化简就行了
3.tanα=1/2,∴sinα=√5/5,cosα=2√5/5,化简求解即得
这个化简应该很简单的,老师应该讲了口诀:奇变偶不变,符号看象限
所以
f(α)= -cosα *cosα * (-tanα)/cosα =sinα
(2)
sin(π/2 -α)=-?, 即 cosα=-1/2 α是第二象限角 α=2π/3
f(α)=sinα=sin2π/3=(根号3)/2
2.那些根号不明确不好做
3. ...
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这个化简应该很简单的,老师应该讲了口诀:奇变偶不变,符号看象限
所以
f(α)= -cosα *cosα * (-tanα)/cosα =sinα
(2)
sin(π/2 -α)=-?, 即 cosα=-1/2 α是第二象限角 α=2π/3
f(α)=sinα=sin2π/3=(根号3)/2
2.那些根号不明确不好做
3. 4tan(α/2)=1-tan2(α/2)
tanα=2tan(α/2)/[ 1-tan2(α/2) ] =1/2
求出sinα=(根号5)/5 cosα=2*(根号5)/5
3sinβ=sin(2α+β)
3sinβ=sin2α*cosβ+cos2α*sinβ
得tanβ=2/7
tan(α+β)=(tanα+tanβ)/(1-tanα*tanβ)=11/12
所以α+β=arctan11/12
收起
(sinα)方/cosα
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