求值:sin214π3+cos3π+tan5π4-cos2(-11π6)+sin(-7π6).
问题描述:
求值:sin2
+cos3π+tan14π 3
-cos2(-5π 4
)+sin(-11π 6
). 7π 6
答
原式=sin2(4π+
)+cos(2π+π)+tan(π+2π 3
)-cos2(π 4
-2π)-sin(π+π 6
)π 6
=sin2
+cosπ+tan2π 3
-cos2π 4
+sinπ 6
π 6
=
-1+1-3 4
+3 4
1 2
=
.1 2