1 |
2 |
1−cos2x |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
| ||
2 |
π |
4 |
1 |
2 |
∴最小正周期T=
2π |
2 |
令2kπ−
π |
2 |
π |
4 |
π |
2 |
解得kπ−
π |
8 |
3π |
8 |
∴f(x)的单调递增区间为[kπ−
π |
8 |
3π |
8 |
(2)∵A是锐角三角形的一个内角,
∴0<A<
π |
2 |
π |
4 |
π |
4 |
3 |
4 |
∴f(A)=
| ||
2 |
π |
4 |
1 |
2 |
| ||
2 |
1 |
2 |
1−cos2x |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
| ||
2 |
π |
4 |
1 |
2 |
2π |
2 |
π |
2 |
π |
4 |
π |
2 |
π |
8 |
3π |
8 |
π |
8 |
3π |
8 |
π |
2 |
π |
4 |
π |
4 |
3 |
4 |
| ||
2 |
π |
4 |
1 |
2 |
| ||
2 |