求数学题第三大题
1个回答
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3
1)=2sin(90+30)tan(360-60)
=-2cos30tan60
=-2*(√3)/2*(√3)
=-3
2)=3cos(180+60)-2tan((180+60))
=-3cos(60)-2tan(60)
=-3*1/2-2*(√3)
=-3/2-2*(√3)
3)=2sin(2pi+pi/4)-cos(-4pi+pi/6-tan(pi+pi/3)
=2sin(pi/4)-cos(pi/6)-tan(pi/3)
=2*(√2)/2-(√3)/2-(√3)
=(√2)-3(√3)/2
4)=8*1/2*1/2*tan(pi+pi/3)cot(pi+pi/6)
=2*(√3)*(√3)
=6
5)=cot(2pi-pi/3)tan(-4pi+pi/3)-2cos(-4pi-pi/4)sin(-4pi+pi+pi/4)
=-cot(pi/3)tan(pi/3)-2cos(-pi/4)sin(pi+pi/4)
=-1-2*(√2)/2*[-(√2)/2]
=-1+1=0
1)=2sin(90+30)tan(360-60)
=-2cos30tan60
=-2*(√3)/2*(√3)
=-3
2)=3cos(180+60)-2tan((180+60))
=-3cos(60)-2tan(60)
=-3*1/2-2*(√3)
=-3/2-2*(√3)
3)=2sin(2pi+pi/4)-cos(-4pi+pi/6-tan(pi+pi/3)
=2sin(pi/4)-cos(pi/6)-tan(pi/3)
=2*(√2)/2-(√3)/2-(√3)
=(√2)-3(√3)/2
4)=8*1/2*1/2*tan(pi+pi/3)cot(pi+pi/6)
=2*(√3)*(√3)
=6
5)=cot(2pi-pi/3)tan(-4pi+pi/3)-2cos(-4pi-pi/4)sin(-4pi+pi+pi/4)
=-cot(pi/3)tan(pi/3)-2cos(-pi/4)sin(pi+pi/4)
=-1-2*(√2)/2*[-(√2)/2]
=-1+1=0
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