已知sin(a-兀/6)=4/5,0≤a≤2兀/3,求sin[(5兀/6)-a]+cos[(5兀/6)-a]1/已知sin(a-π/6)=4/5,0≤a≤2π/3,求sin[(5π/6)-a]+cos[(5π/6)-a]2/已知sina、cosa是关于x的二次方程4x^2-4mx+2m-1=0的两个根,3π/2<a<2π,求m,a的值3/
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![已知sin(a-兀/6)=4/5,0≤a≤2兀/3,求sin[(5兀/6)-a]+cos[(5兀/6)-a]1/已知sin(a-π/6)=4/5,0≤a≤2π/3,求sin[(5π/6)-a]+cos[(5π/6)-a]2/已知sina、cosa是关于x的二次方程4x^2-4mx+2m-1=0的两个根,3π/2<a<2π,求m,a的值3/](/uploads/image/z/1063658-2-8.jpg?t=%E5%B7%B2%E7%9F%A5sin%28a-%E5%85%80%2F6%29%3D4%2F5%2C0%E2%89%A4a%E2%89%A42%E5%85%80%2F3%2C%E6%B1%82sin%5B%285%E5%85%80%2F6%29-a%5D%2Bcos%5B%285%E5%85%80%2F6%29-a%5D1%2F%E5%B7%B2%E7%9F%A5sin%28a-%CF%80%2F6%29%3D4%2F5%2C0%E2%89%A4a%E2%89%A42%CF%80%2F3%2C%E6%B1%82sin%5B%285%CF%80%2F6%29-a%5D%2Bcos%5B%285%CF%80%2F6%29-a%5D2%2F%E5%B7%B2%E7%9F%A5sina%E3%80%81cosa%E6%98%AF%E5%85%B3%E4%BA%8Ex%E7%9A%84%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8B4x%5E2-4mx%2B2m-1%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%EF%BC%8C3%CF%80%2F2%EF%BC%9Ca%EF%BC%9C2%CF%80%EF%BC%8C%E6%B1%82m%2Ca%E7%9A%84%E5%80%BC3%2F)
已知sin(a-兀/6)=4/5,0≤a≤2兀/3,求sin[(5兀/6)-a]+cos[(5兀/6)-a]1/已知sin(a-π/6)=4/5,0≤a≤2π/3,求sin[(5π/6)-a]+cos[(5π/6)-a]2/已知sina、cosa是关于x的二次方程4x^2-4mx+2m-1=0的两个根,3π/2<a<2π,求m,a的值3/
已知sin(a-兀/6)=4/5,0≤a≤2兀/3,求sin[(5兀/6)-a]+cos[(5兀/6)-a]
1/已知sin(a-π/6)=4/5,0≤a≤2π/3,求sin[(5π/6)-a]+cos[(5π/6)-a]
2/已知sina、cosa是关于x的二次方程4x^2-4mx+2m-1=0的两个根,3π/2<a<2π,求m,a的值
3/设函数f(x)=[1/(2^x)+1]+m 是奇函数,a=f(log2^5)
求/⑴m
/⑵a
/⑶tana=a,求1+cos[(π/2)+a]sin[(3π/2)-a]2cos^2(π+a)的值
已知sin(a-兀/6)=4/5,0≤a≤2兀/3,求sin[(5兀/6)-a]+cos[(5兀/6)-a]1/已知sin(a-π/6)=4/5,0≤a≤2π/3,求sin[(5π/6)-a]+cos[(5π/6)-a]2/已知sina、cosa是关于x的二次方程4x^2-4mx+2m-1=0的两个根,3π/2<a<2π,求m,a的值3/
1、已知sin(a-π/6)=4/5,0≤a≤2π/3,求sin[(5π/6)-a]+cos[(5π/6)-a]
∵0≤a≤2π/3,sin(a-π/6)=4/5,
∴0<a-π/6≤π/2,cos(a-π/6)=√[1-(4/5)²]=3/5
sin(π/6-a)=-sin(a-π/6)=-4/5,cos(π/6-a)=cos(a-π/6)=3/5
∴sin[(5π/6)-a]+cos[(5π/6)-a]=sin[(π/6-a)+2π/3]+cos[(π/6-a)+2π/3]
=sin(π/6-a)cos(2π/3)+cos(π/6-a)sin(2π/3)+cos(π/6-a)cos(2π/3)-sin(π/6-a)sin(2π/3)
=cos(2π/3)[sin(π/6-a)+cos(π/6-a)]+sin(2π/3)[cos(π/6-a)-sin(π/6-a)]
=(-1/2)(-4/5+3/5)+[(√3)/2][3/5-(-4/5)]
=(1+7√3)/10
2、已知sina、cosa是关于x的二次方程4x²-4mx+2m-1=0的两个根,3π/2<a<2π,求m,a的值
方程4x²-4mx+2m-1=0可变为(2x-1)(2x-2m+1)=0
于是二次方程的两个根为:x1=1/2,x2=m-1/2
由sina、cosa是这两个根,且3π/2<a<2π知:sina<0,cosa>0
∴cosa=x1=1/2,sina=x2=-(√3)/2,m=x2+1/2=1/2-(√3)/2
3、设函数f(x)=1/[(2^x)+1]+m 是奇函数,a=f(log2^5)
求:⑴m ⑵a ⑶tanα=a,求1+cos[(π/2)+α]sin[(3π/2)-α]2cos²(π+α)的值
⑴∵f(x)是奇函数
∴f(-x)=-f(x),即1/[(2^-x)+1]+m=-1/[(2^x)+1]-m
整理,得:2m(2^-x+2^x+2)+2^-x+2^x+2=0
∴m=-1/2
⑵代m=-1/2可得f(x)=1/[(2^x)+1]-1/2
∴a=f(㏒2^5)=1/(5+1)-1/2=-1/3
⑶这问应该是“1+cos[(π/2)+α]sin[(3π/2)-α]/cos²(π+α)”吧,不然没法算啊
=1+[-sinα][-cosα]/cos²α
=1+tanα
=2/3
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