化简:tan(5π-a)﹡cos(a-3)π*sin(-a-3π⼀2)⼀cos(-a-π)*sin(a-π)

第二个是cos(a-3π)
2025-03-02 08:15:43
推荐回答(5个)
回答1:

首先。。tan的周期是π,sin和cos周期都是2π,
sin(π-α)= sinα
cos(π-α)= -cosα
tan(π-α)= -tanα
sin(π/2+α)= cosα
cos(π/2+α)= -sinα

化简:原式=tan(π-a)*cos(a+π)*sin(-a+0.5π)/cos(a+π)*sin(a-π)
=-tan(a)*cos(a)/-sin(a)
=-tan(a)/-tan(a)=1

回答2:

解:
tan(5π-a)﹡cos(a-3π)*sin(-a-3π/2)/cos(-a-π)*sin(a-π)
=tan(4π+(π-a))﹡cos(a-2π-π)*(-1)sin(a+3π/2)/cos(a+π)*(-1)sin(π-a)
=tan(π-a)﹡cos(2π+π-a)*(-1)sin(a+3π/2)/cos(a+π)*(-1)sin(π-a)
=tan(π-a)﹡cos(π-a)*cos(a)/-sin(a)*(-1)cos(a)
=tana*cosa*cosa/sina*cosa
=tana*cosa/sina
=tana*ctana
=1

回答3:

=tana*cosa*cosa/cosa*sina=1

回答4:

=tan(-a)*(-cosa)*cos(-a)/-cos(-a)*-sina=-tana*cos^2a/cosa*sina=-1

回答5:

-1