Question

函数y=sin(x+π/3）sin(x+π/2)的最小正周期T=？

函数y=sin(x+π/3）sin(x+π/2)的最小正周期T=？
要过程

Answer 1

y=sin(x+π/3）sin(x+π/2)
=(sinx+√3cosx)cosx/2
=[sinxcosx+√3(cosx)^2]/2
=[sin2x/2+√3(cos2x+1)/2]/2
=(sin2x+√3cos2x+√3)/4
=2sin(2x+.....
划到这边可以了吧？？
周期是π

Answer 2

均是2pi.pi为圆周率

Answer 3

用积化和差：sin(x+π/3）sin(x+π/2)=-0.5{cos[(x+π/3）+(x+π/2)]-cos[(x+π/3）-(x+π/2)]}=-0.5[cos(2x+5/6π)+cos(-1/6π)],显然周期就是π了啊