1.求值sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π2.若sinθ=2cosθ,求sincosθ.3.若sin(-4π+a)=1/2,求sin(2π-a)/cos(π-a)
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![1.求值sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π2.若sinθ=2cosθ,求sincosθ.3.若sin(-4π+a)=1/2,求sin(2π-a)/cos(π-a)](/uploads/image/z/6866001-9-1.jpg?t=1.%E6%B1%82%E5%80%BCsin%EF%BC%88-4%2F5%CF%80%EF%BC%89cos%EF%BC%88-5%2F3%CF%80%EF%BC%89-tan%EF%BC%88-11%2F6%CF%80%EF%BC%89-%EF%BC%88cos7%2F6+%CF%80%C3%973tan4%2F3+%CF%80%EF%BC%89%2Fsin2%2F3+%CF%802.%E8%8B%A5sin%CE%B8%3D2cos%CE%B8%2C%E6%B1%82sincos%CE%B8.3.%E8%8B%A5sin%EF%BC%88-4%CF%80%2Ba%EF%BC%89%3D1%2F2%2C%E6%B1%82sin%EF%BC%882%CF%80-a%EF%BC%89%2Fcos%EF%BC%88%CF%80-a%EF%BC%89)
1.求值sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π2.若sinθ=2cosθ,求sincosθ.3.若sin(-4π+a)=1/2,求sin(2π-a)/cos(π-a)
1.求值sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π
2.若sinθ=2cosθ,求sincosθ.
3.若sin(-4π+a)=1/2,求sin(2π-a)/cos(π-a)
1.求值sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π2.若sinθ=2cosθ,求sincosθ.3.若sin(-4π+a)=1/2,求sin(2π-a)/cos(π-a)
1、sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π =-sin(4/5π)cos(5/3π)+tan(11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π=-sin(1/5π)cos(1/3π)-tan(1/6π)+(cos1/6 π×3tan1/3 π)/sin1/3 π=-(1/2)sin(1/5π)-3/√3+(√3/2X3X√3)/(√3/2)=-(1/2)sin(1/5π)
2、sinacosa=2/5
3、sin(2π-a)/cos(π-a)=-sina/-cosa=tana sin(-4π+a)=1/2 sina=1/2 tana=√3/3
1、sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π =-sin(4/5π)cos(5/3π)+tan(11/6π)+(cos7/6 π×3tan4/3 π)/sin2/3 π=sin(1/5π)cos(1/3π)+tan(1/6π)-(cos1/6 π×3tan1/3 π)/sin1/3 π=(1/2)sin(1/5π)...
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1、sin(-4/5π)cos(-5/3π)-tan(-11/6π)-(cos7/6 π×3tan4/3 π)/sin2/3 π =-sin(4/5π)cos(5/3π)+tan(11/6π)+(cos7/6 π×3tan4/3 π)/sin2/3 π=sin(1/5π)cos(1/3π)+tan(1/6π)-(cos1/6 π×3tan1/3 π)/sin1/3 π=(1/2)sin(1/5π)+3/√3-(√3/2X3X√3)/(√3/2)=(1/2)sin(1/5π)
2、
3、sin(2π-a)/cos(π-a)=-sina/-cosa=tana sin(-4π+a)=1/2 sina=1/2 tan=√3/3
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