1.在△abc中,已知sinA=1/2,求角a,已知cosB=-根号3/2,求角B
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1.在△abc中,已知sinA=1/2,求角a,已知cosB=-根号3/2,求角B
2.已知f(α)=[sin(π-α)cos(2π-α)tan(-α+π)]/[-tan(-α-π)sin(-π-α)]
(1)化简(2)若α是第三象限角,且cos(α-3π/2)=1/5,求cosα以及f(α)的值
3.已知sin(π-α)-cos(π+α)=根号2/3(π/2<α<π),求:
(1)sinα-cosα(2)sin三次方(2π-α)+cos三次方(2π-α)的值
2.已知f(α)=[sin(π-α)cos(2π-α)tan(-α+π)]/[-tan(-α-π)sin(-π-α)]
(1)化简(2)若α是第三象限角,且cos(α-3π/2)=1/5,求cosα以及f(α)的值
3.已知sin(π-α)-cos(π+α)=根号2/3(π/2<α<π),求:
(1)sinα-cosα(2)sin三次方(2π-α)+cos三次方(2π-α)的值
1、由于∠A,∠B为三角形的内角,
所以∠A=30°;∠B=120°
2、(1)
sin(π-α)=sin(α)
cos(2π-α)=cos(α)
tan(π-α)=-tan(α)
-tan(-α-π)=tan(α)
sin(-π-α)=sin(α)
f(α)=[-sin(α)cos(α)tan(α)]/[tan(α)sin(α)]
=-cos(α)
(2)
由于 cos(-α)=cos(α);及 cos(3π/2-α)=-sin(α)
所以 cos(α-3π/2)=cos(3π/2-α)=-sin(α)=1/5
即sin(α)=-1/5
α是第三象限角
所以cos(α)=-√(1-1/25)=-(2√6)/5
f(α)=-cos(α)=(2√6)/5
3、(1)
由于 sin(π-α)=sin(α)
cos(π+α)=-cos(α)
所以 sin(π-α)-cos(π+α)
=sin(α)+cos(α)=√2/3
且 sin(α)^2+cos(α)^2=1
令sin(α)-cos(α)=k
那么 sin(α)^2-cos(α)^2=(k√2)/3
sin(α)^2=[1+(k√2)/3]/2
cos(α)^2=[1-(k√2)/3]/2
因为 π/2<α<π
sin(α)=√{[1+(k√2)/3]/2}
cos(α)=-√{[1-(k√2)/3]/2}
因而 sin(α)-cos(α)=√{[1+(k√2)/3]/2}+√{[1-(k√2)/3]/2}=k
[1+(k√2)/3]/2+[1-(k√2)/3]/2+2×√{[1+(k√2)/3]/2}×√{[1-(k√2)/3]/2}=k^2
1+√{1-(2k^2)/9}=k^2
1-(2k^2)/9=k^4-2k^2+1
k^2=16/9
k=4/3
所以 sin(α)-cos(α)=4/3
(2)
sin(2π-α)=-sin(α)
cos(2π-α)=cos(α)
sin(2π-α)^3+cos(2π-α)^3=cos(α)^3-sin(α)^3
=[cos(α)-sin(α)][cos(α)^2-cos(α)sin(α)+sin(α)^2]
=[cos(α)-sin(α)][1-cos(α)sin(α)]
由 sin(α)-cos(α)=4/3
得 cos(α)-sin(α)=-4/3
cos(α)^2-2cos(α)sin(α)+sin(α)^2=16/9
2cos(α)sin(α)=1-16/9=-7/9
cos(α)sin(α)=-7/18
所以 sin(2π-α)^3+cos(2π-α)^3=[-4/3]×[1+7/18]
=-4/3×25/18
=-50/27
所以∠A=30°;∠B=120°
2、(1)
sin(π-α)=sin(α)
cos(2π-α)=cos(α)
tan(π-α)=-tan(α)
-tan(-α-π)=tan(α)
sin(-π-α)=sin(α)
f(α)=[-sin(α)cos(α)tan(α)]/[tan(α)sin(α)]
=-cos(α)
(2)
由于 cos(-α)=cos(α);及 cos(3π/2-α)=-sin(α)
所以 cos(α-3π/2)=cos(3π/2-α)=-sin(α)=1/5
即sin(α)=-1/5
α是第三象限角
所以cos(α)=-√(1-1/25)=-(2√6)/5
f(α)=-cos(α)=(2√6)/5
3、(1)
由于 sin(π-α)=sin(α)
cos(π+α)=-cos(α)
所以 sin(π-α)-cos(π+α)
=sin(α)+cos(α)=√2/3
且 sin(α)^2+cos(α)^2=1
令sin(α)-cos(α)=k
那么 sin(α)^2-cos(α)^2=(k√2)/3
sin(α)^2=[1+(k√2)/3]/2
cos(α)^2=[1-(k√2)/3]/2
因为 π/2<α<π
sin(α)=√{[1+(k√2)/3]/2}
cos(α)=-√{[1-(k√2)/3]/2}
因而 sin(α)-cos(α)=√{[1+(k√2)/3]/2}+√{[1-(k√2)/3]/2}=k
[1+(k√2)/3]/2+[1-(k√2)/3]/2+2×√{[1+(k√2)/3]/2}×√{[1-(k√2)/3]/2}=k^2
1+√{1-(2k^2)/9}=k^2
1-(2k^2)/9=k^4-2k^2+1
k^2=16/9
k=4/3
所以 sin(α)-cos(α)=4/3
(2)
sin(2π-α)=-sin(α)
cos(2π-α)=cos(α)
sin(2π-α)^3+cos(2π-α)^3=cos(α)^3-sin(α)^3
=[cos(α)-sin(α)][cos(α)^2-cos(α)sin(α)+sin(α)^2]
=[cos(α)-sin(α)][1-cos(α)sin(α)]
由 sin(α)-cos(α)=4/3
得 cos(α)-sin(α)=-4/3
cos(α)^2-2cos(α)sin(α)+sin(α)^2=16/9
2cos(α)sin(α)=1-16/9=-7/9
cos(α)sin(α)=-7/18
所以 sin(2π-α)^3+cos(2π-α)^3=[-4/3]×[1+7/18]
=-4/3×25/18
=-50/27
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