求下列三角函数式的值:(1)sinπ4cos19π6tan21π4;(2)3sin(-1200°)tan19π6-cos585°tan(-37π

2024-12-25 02:46:52
推荐回答(1个)
回答1:

(1)sin

π
4
cos
19π
6
tan
21π
4
=
2
2
cos(π+
π
6
)tan
π
4
=
2
2
×(-
3
2
)×1=-
6
4

(2)
3
sin(-1200°)tan
19π
6
-cos585°tan(-
37π
4
).
=