y=sin(π/3-2a)+cos(π/3+2a)的最大值和周期
问题描述：
y=sin(π/3-2a)+cos(π/3+2a)的最大值和周期
答
y=sin(π/3-2a)+cos(π/3+2a)
=sin(π/3-2a)+cos(π/2-π/6+2a)
=sin(π/3-2a)+cos(π/2-(π/6-2a))
=sin(π/3-2a)+sin(π/6-2a)
=2sin(π/3-2a+π/6-2a)/2cos(π/3-2a-(π/6-2a))/2
=2sin(π/2-4a)/2cos(π/3-2a-π/6+2a)/2
=2sin(π/4-2a)cos(π/6)/2
=2cos(π/12)sin(π/4-2a)
所以最大值是：2cos(π/12),周期是: 2兀/2=兀
答
最大值gen2
答
用和差化积公式.
y=sin(π/3-2a)+cos(π/3+2a)
=sin(π/3-2a)+sin(π/3+2a+π/2)
=2sin(π/3+π/4)cos(2a-π/4)
所以最大值是2sin(7π/12)=(√6-√2)/2,周期是π