化简(高一)[sin^2(pi+a)cos(pi+a)]/[tan(pi+a)cos^3(-pi-a)tan(2pi+a
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化简(高一)
[sin^2(pi+a)cos(pi+a)]/[tan(pi+a)cos^3(-pi-a)tan(2pi+a)]
[sin^2(pi+a)cos(pi+a)]/[tan(pi+a)cos^3(-pi-a)tan(2pi+a)]
![化简(高一)[sin^2(pi+a)cos(pi+a)]/[tan(pi+a)cos^3(-pi-a)tan(2pi+a](/uploads/image/z/17158063-31-3.jpg?t=%E5%8C%96%E7%AE%80%28%E9%AB%98%E4%B8%80%29%5Bsin%5E2%28pi%2Ba%29cos%28pi%2Ba%29%5D%2F%5Btan%28pi%2Ba%29cos%5E3%28-pi-a%29tan%282pi%2Ba)
[sin^2(∏+a)cos(∏+a)]/[tan(∏+a)cos^3(-∏-a)tan(2∏+a)] =[sin^2a(-cosa)]/[tana(-cos^3)tana=-sin^2acosa/(-cos^3sin^2a/cos^2a=-sin^2acosa/=-sin^2acosa=1
做这种类型题在化间的时候要特别注意符号,可以结合象限来判断
做这种类型题在化间的时候要特别注意符号,可以结合象限来判断
化简间sin(a-2pi)cos(a+pi)tan(a-99pi)cos(pi-a)sin(3pi-a)sin(-a-p
化简:sin(2pi-a)sin(pi+a)cos(-pi-a)/sin(3pi-a)cos(pi-a)
f(a)=sin(pi-a)cos(2pi-a)tan(-a+3pi/2)/cos(-pi-a) 求 f(-31pi/3
tana=2求sin(pi-a)cos(2pi-a)sin(-a+3pi/2)/tan(-a-pi)sin(-pi-a)
sin^3(-a)cos(2pi+a)tan(-a-pi)化简
sin^2(-a)cos(2pi+a)tan(-a-pi)化简
cosa=2/3 a是第四象限角求sin(a-2Pi)+sin(-a-3Pi)cos(a-3Pi)/cos(Pi-a)-
cos(Pi+a)=-0.5,3/2Pi
是否存在a属于(-pi/2,pi/2),b属于(0,pi),使等式sin(3Pi-a)=根号2cos(pi/2)-b),
若cos(pi/6-a)=1/2,sin(a+pi/3)=?cos(2pi/3+2a)=?注:pi指圆周率
三角形ABC中sin(2Pi-A)=-根号2cos(3Pi/2+B)根号3cos(2Pi-A)=根号2sin(Pi/2+
化简三角函数[sin(α+pi)*cos(pi+α)*cos(α+2pi)]/[tan(pi+α)*cos^3(-α-p