求值sin(-23/6π)+cos13/7πtan4π−cos13/3π=_.

问题描述:

求值sin(-

23
6
π)+cos
13
7
πtan4π−cos
13
3
π
=______.

原式=sin(-4π+

π
6
)+cos(2π-
π
7
)tan4π-cos(4π+
π
3
)=sin
π
6
+0-cos
π
3
=
1
2
+0-
1
2
=0.
故答案为:0