已知tan (π+ α)=-2/1,求下列各式的值(1)[2cos(π-α)-3sin(π+α)]/[4cos(α-2π)+sin(4π-α)](2)sin(α-2π)xcos(α+5π)

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/04 19:04:06
已知tan (π+ α)=-2/1,求下列各式的值(1)[2cos(π-α)-3sin(π+α)]/[4cos(α-2π)+sin(4π-α)](2)sin(α-2π)xcos(α+5π)
x){}KK47h+ۨikolcӓO;?tOY-OhjF%Ug恴ٱ& su7hjdL b54A\TD)mTO:tD/n ZQS`kC Oxf͓]=/6.|9sX74n]jglΧ0}ST@"0(6$ف 

已知tan (π+ α)=-2/1,求下列各式的值(1)[2cos(π-α)-3sin(π+α)]/[4cos(α-2π)+sin(4π-α)](2)sin(α-2π)xcos(α+5π)
已知tan (π+ α)=-2/1,求下列各式的值
(1)[2cos(π-α)-3sin(π+α)]/[4cos(α-2π)+sin(4π-α)]
(2)sin(α-2π)xcos(α+5π)

已知tan (π+ α)=-2/1,求下列各式的值(1)[2cos(π-α)-3sin(π+α)]/[4cos(α-2π)+sin(4π-α)](2)sin(α-2π)xcos(α+5π)
tan(pai+alpha) = -2,tan(alpha) = -2,alpha 在第二象限
cos(pai-alpha) = -cos(alpha) = 1/根号(5),sin(pai+alpha) = -sin(alpha) =-2/根号(5)
...