Question

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

第二个是cos(a-3π)

Answer 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

Answer 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

Answer 3

＝tana*cosa*cosa/cosa*sina=1

Answer 4

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

Answer 5

-1