已知函数f(x)=a(2cos^2x/2+sinx)+b 当a=1时,求函数f(x)的单调递增区间;

f(x)=a(2cos²x/2+sinx)+b=a(1+cosx+sinx)+b=a(cosx+sinx)+a+b=√2a(sinπ/4cosx+cosπ/4sinx)+a+b=√2asin(π/4+x)+a+b∴单调递增区间为[2kπ+3π/2-π/4,(2k+1)π+π/2-π/4]即：[2kπ+5π/4,(2k+1)π+π/4]...a(1+cosx+sinx)+b这一步怎么来cosx=2cos²x/2-1
2cos²x/2=cosx+1