1.已知弧度数为2的圆心角多对的弦长也是2,则这个圆心角所对的弧长是___?2.设α角的终边上一点P的坐标是(cosπ/5,sinπ/5),则α等于___?3.f(x)=tanwx(w>0)的图象的相邻两支截直线y=π/8
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![1.已知弧度数为2的圆心角多对的弦长也是2,则这个圆心角所对的弧长是___?2.设α角的终边上一点P的坐标是(cosπ/5,sinπ/5),则α等于___?3.f(x)=tanwx(w>0)的图象的相邻两支截直线y=π/8](/uploads/image/z/2394701-53-1.jpg?t=1.%E5%B7%B2%E7%9F%A5%E5%BC%A7%E5%BA%A6%E6%95%B0%E4%B8%BA2%E7%9A%84%E5%9C%86%E5%BF%83%E8%A7%92%E5%A4%9A%E5%AF%B9%E7%9A%84%E5%BC%A6%E9%95%BF%E4%B9%9F%E6%98%AF2%2C%E5%88%99%E8%BF%99%E4%B8%AA%E5%9C%86%E5%BF%83%E8%A7%92%E6%89%80%E5%AF%B9%E7%9A%84%E5%BC%A7%E9%95%BF%E6%98%AF%EF%BC%BF%EF%BC%BF%EF%BC%BF%3F2.%E8%AE%BE%CE%B1%E8%A7%92%E7%9A%84%E7%BB%88%E8%BE%B9%E4%B8%8A%E4%B8%80%E7%82%B9P%E7%9A%84%E5%9D%90%E6%A0%87%E6%98%AF%EF%BC%88cos%CF%80%2F5%2Csin%CF%80%2F5%EF%BC%89%2C%E5%88%99%CE%B1%E7%AD%89%E4%BA%8E%EF%BC%BF%EF%BC%BF%EF%BC%BF%3F3.f%EF%BC%88x%EF%BC%89%3Dtanwx%EF%BC%88w%3E0%EF%BC%89%E7%9A%84%E5%9B%BE%E8%B1%A1%E7%9A%84%E7%9B%B8%E9%82%BB%E4%B8%A4%E6%94%AF%E6%88%AA%E7%9B%B4%E7%BA%BFy%3D%CF%80%2F8)
1.已知弧度数为2的圆心角多对的弦长也是2,则这个圆心角所对的弧长是___?2.设α角的终边上一点P的坐标是(cosπ/5,sinπ/5),则α等于___?3.f(x)=tanwx(w>0)的图象的相邻两支截直线y=π/8
1.已知弧度数为2的圆心角多对的弦长也是2,则这个圆心角所对的弧长是___?
2.设α角的终边上一点P的坐标是(cosπ/5,sinπ/5),则α等于___?
3.f(x)=tanwx(w>0)的图象的相邻两支截直线y=π/8所得线段长为π/8,则f(π/8)的值是___
1.已知弧度数为2的圆心角多对的弦长也是2,则这个圆心角所对的弧长是___?2.设α角的终边上一点P的坐标是(cosπ/5,sinπ/5),则α等于___?3.f(x)=tanwx(w>0)的图象的相邻两支截直线y=π/8
1 圆心角为 2/2π*360=360/π
半径R=1/SIN(180/π)
弧长L=R 360/π =360/π SIN(180/π)
2 R=COS^2π/5+SIN^2π/5
sinα=sinπ/5 / R=sinπ/5 α=π/5
3 π/8=tanwx
π/8=tanw(x+π/8)
tanwx=tanw(x+π/8)
w=8
f(π/8)=tanπ
ch
(4sin((π-2)/2)/sin2
π/5+2kπ(k∈Z)
0
1.已知偶函数y=f(x)在〖-1,0〗上为单调递减函数,又α。β为锐角三角形的两内角,则_________。
因为α+β>π/2
所以α>π/2-β
于是sinα>sin(π/2-β)即sinα>cosβ
又因为偶函数y=f(x)在〖-1,0〗上为单调递减函数
所以y=f(x)在〖0,1〗上为单调递增函数
所以f(sinα)...
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1.已知偶函数y=f(x)在〖-1,0〗上为单调递减函数,又α。β为锐角三角形的两内角,则_________。
因为α+β>π/2
所以α>π/2-β
于是sinα>sin(π/2-β)即sinα>cosβ
又因为偶函数y=f(x)在〖-1,0〗上为单调递减函数
所以y=f(x)在〖0,1〗上为单调递增函数
所以f(sinα)>f(cosβ)
2.已知cosθ=-3/5,且π<θ<3/2π,则tan(θ-π/4)=______________。
由cosθ=-3/5,且π<θ<3/2π,得sinθ=-4/5,tanθ=4/3
所以tan(θ-π/4)=tanθ-tanπ/4)/(1+tanθtanπ/4)=1/7
3.已知向量a=(2,-1)与向量b共线,且满足a*b=-10,则向量b=_______________.
设向量b=x(2,-1),由a*b=-10得x=-2,所以b=(-4,2),
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