3/(sin20^2)-1/(cos20^2)+64sin20^2的值是?cos20cos40cos80的值是?已知a,b均为锐角,且满足3sina^2+2sinb^2=1,3sin2a-2sin2b=0,求证a+2b=90度
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![3/(sin20^2)-1/(cos20^2)+64sin20^2的值是?cos20cos40cos80的值是?已知a,b均为锐角,且满足3sina^2+2sinb^2=1,3sin2a-2sin2b=0,求证a+2b=90度](/uploads/image/z/4343799-39-9.jpg?t=3%2F%28sin20%5E2%29-1%2F%28cos20%5E2%29%2B64sin20%5E2%E7%9A%84%E5%80%BC%E6%98%AF%3Fcos20cos40cos80%E7%9A%84%E5%80%BC%E6%98%AF%3F%E5%B7%B2%E7%9F%A5a%2Cb%E5%9D%87%E4%B8%BA%E9%94%90%E8%A7%92%2C%E4%B8%94%E6%BB%A1%E8%B6%B33sina%5E2%2B2sinb%5E2%3D1%2C3sin2a-2sin2b%3D0%2C%E6%B1%82%E8%AF%81a%2B2b%3D90%E5%BA%A6)
3/(sin20^2)-1/(cos20^2)+64sin20^2的值是?cos20cos40cos80的值是?已知a,b均为锐角,且满足3sina^2+2sinb^2=1,3sin2a-2sin2b=0,求证a+2b=90度
3/(sin20^2)-1/(cos20^2)+64sin20^2的值是?
cos20cos40cos80的值是?
已知a,b均为锐角,且满足3sina^2+2sinb^2=1,3sin2a-2sin2b=0,求证a+2b=90度
3/(sin20^2)-1/(cos20^2)+64sin20^2的值是?cos20cos40cos80的值是?已知a,b均为锐角,且满足3sina^2+2sinb^2=1,3sin2a-2sin2b=0,求证a+2b=90度
1.3/sin^2 20-1/cos^2 20+64sin20
=[3(cos20)^2-(sin20)^2]/(sin20)^2(cos20)^2+64(sin20)^2
=[(√3cos20+sin20)(√3cos20-sin20)]/(sin20cos20)^2+64(sin20)^2
=[2sin(60+20)*2sin(60-20)]/[(sin40)/2]^2+64(sin20)^2
=16sin80sin40/sin40*sin40+64(sin20)^2
=16sin80/sin40+64(sin20)^2
=32sin40cos40/sin40+64(sin20)^2
=32cos40+64(sin20)^2
=32[1-2(sin20)^2]+64(sin20)^2
=32
2.=sin20cos20cos40cos80/sin20=0.5sin40cos40cos80/sin20=0.25sin80cos80/sin20=0.125sin160/sin20=0.125
3..因为3(sinA)^2=1-2(sinB)^2=cos2B
3sin2A/2=sin2B
(cos2B)^2+(sin2B)^2=1
所以[3(sinA)^2]^2+(3sin2A/2)^2=1
-->9(sinA)^4+9(sinAcosA)^2=1
-->9(sinA)^4+9(sinA)^2[1-(sinA)^2]=1
设(sinA)^2=t,则9t^2+9t(1-t)=1,
解得t=1/9,从而sinA=1/3,cosA=2*根号2/3.
因为2B∈(0,2π)且cos2B=3(sinA)^2=1/3>0,所以2B∈(0,π).
因为A∈(0,π/2),所以(π/2-A)∈(0,π/2).
由cos2B=sinA=cos(π/2-A)且余弦函数在(0,π/2)有单调性,
可得2B=π/2-A
即A+2B=π/2.证毕.