tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3t+a)cos^2(2/5π+2)=-2/1求cosa*|tana|
问题描述:
tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3t+a)cos^2(2/5π+2)=-2/1
求cosa*|tana|
答
tan(a-4π)=-tanacos(π+a)=-cosasin^2(a+3π)=sin^2atan(3π+a)=-tanacos^2(5π/2+a)=sin^2a方程左边=-cosa∴cosa=1/2|tana|=根号3所以答案=二分之根号三这位朋友打字错误太多.5/2写成2/5,1/2写成2/1.π写成t.最后...