1.已知函数y=1+3sin(2x+派/4)
(1)求函数的周期?写出函数的单调递增区间?
(2) 求函数最大值所对应的x的集合?
(1)解析：∵f(x)=1+3sin(2x+π/4),∴T=2π/2=π
∵正弦函数的单调增区间为[2kπ-π/2,2kπ+π/2]
∴2kπ-π/2≤2x+π/4≤2kπ+π/2
==>2kπ-3π/4≤2x≤2kπ+π/4
==>kπ-3π/8≤x≤kπ+π/8
∴函数f(x)的单调递增区间kπ-3π/8≤x≤kπ+π/8
(2)由（1）可知,函数最大值所对应的x的集合为{x|x=kπ+π/8,k∈Z}
2.已知α属于（0,派/2）β属于（派/2,派）且sin（α+β）=33/65,cosβ=5/13,求sin=?
解析：∵α属于（0,派/2）β属于（派/2,派）且sin（α+β）=33/65,cosβ=5/13
sinβ=12/13
Sin(α+β)=sinαcosβ+cosαsinβ=33/65
sinα*5/13+cosα*12/13=33/65?sinα*5+cosα*12=33/5
又(sinα)^2+(cosα)^2=1
cosα=√(1-(sinα)^2)
√(1-(sinα)^2)=(33-25sinα)/60
60^2-60^2(sinα)^2=33^2+25^2(sinα)^2-2*33*25 sinα
(60^2+25^2) (sinα)^2-2*33*25 sinα+33^2-60^2=0
4225 (sinα)^2-1650 sinα-2511=0
sinα≈0.99053
3.已知函数f(x)=sin(派-x)cosx.求f(x)的最小正周期?求f(x)在区间〔-派/6,派/2〕上的最大值和最小值?
解析：∵函数f(x)=sin(派-x)cosx=-(cosx)^2=-1/2cos2x-1/2
∴f(x)的最小正周期为π
∵π/2-(-π/6)=2π/3
