3 |
1-cos2x |
2 |
| ||
2 |
1 |
2 |
π |
6 |
令2kπ+
π |
2 |
π |
6 |
3π |
2 |
解得:kπ+
π |
6 |
2π |
3 |
所以函数f(x)的单调递增区间为[kπ+
π |
6 |
2π |
3 |
(2)由f(A)+sin(2A-
π |
6 |
1 |
2 |
π |
6 |
π |
6 |
化简得:cos2A=-
1 |
2 |
又因为0<A<
π |
2 |
π |
3 |
由题意知:S△ABC=
1 |
2 |
3 |
又b+c=7,所以a2=b2+c2-2bccosA=(b+c)2-2bc(1+cosA)=49-2×8×(1+
1 |
2 |
∴a=5
m |
3 |
n |
m |
n |
π |
6 |
3 |
3 |
1-cos2x |
2 |
| ||
2 |
1 |
2 |
π |
6 |
π |
2 |
π |
6 |
3π |
2 |
π |
6 |
2π |
3 |
π |
6 |
2π |
3 |
π |
6 |
1 |
2 |
π |
6 |
π |
6 |
1 |
2 |
π |
2 |
π |
3 |
1 |
2 |
3 |
1 |
2 |