f(a)=[sin(a-π/2)cos(3π/2+a)tan(π-a)]除以tan(-a-π)sin(-a-π)已知a为第二象限角,1、化简fa2、若cos(a-3π/2)=1/5,求fa的值?
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![f(a)=[sin(a-π/2)cos(3π/2+a)tan(π-a)]除以tan(-a-π)sin(-a-π)已知a为第二象限角,1、化简fa2、若cos(a-3π/2)=1/5,求fa的值?](/uploads/image/z/3709799-71-9.jpg?t=f%EF%BC%88a%EF%BC%89%3D%5Bsin%28a-%CF%80%2F2%29cos%EF%BC%883%CF%80%2F2%2Ba%EF%BC%89tan%EF%BC%88%CF%80-a%EF%BC%89%5D%E9%99%A4%E4%BB%A5tan%EF%BC%88-a-%CF%80%EF%BC%89sin%EF%BC%88-a-%CF%80%EF%BC%89%E5%B7%B2%E7%9F%A5a%E4%B8%BA%E7%AC%AC%E4%BA%8C%E8%B1%A1%E9%99%90%E8%A7%92%2C1%E3%80%81%E5%8C%96%E7%AE%80fa2%E3%80%81%E8%8B%A5cos%EF%BC%88a-3%CF%80%2F2%EF%BC%89%3D1%2F5%2C%E6%B1%82fa%E7%9A%84%E5%80%BC%3F)
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f(a)=[sin(a-π/2)cos(3π/2+a)tan(π-a)]除以tan(-a-π)sin(-a-π)已知a为第二象限角,1、化简fa2、若cos(a-3π/2)=1/5,求fa的值?
f(a)=[sin(a-π/2)cos(3π/2+a)tan(π-a)]除以tan(-a-π)sin(-a-π)
已知a为第二象限角,
1、化简fa
2、若cos(a-3π/2)=1/5,求fa的值?
f(a)=[sin(a-π/2)cos(3π/2+a)tan(π-a)]除以tan(-a-π)sin(-a-π)已知a为第二象限角,1、化简fa2、若cos(a-3π/2)=1/5,求fa的值?
1、f(a)=[sin(a-π/2)cos(3π/2+a)tan(π-a)]/tan(-a-π)sin(-a-π)
=[-cosa*sina*(-tana)]/-tana*sina=-cosa
2、因为cos(a-3π/2)=-sina=1/5,所以sina=-1/5
又已知a为第二象限角,所以cosa=-2√6/5.
故f(a)=-cosa=2√6/5