f(x)=sin(x+2π-π/4)+cos(x-3π/4)
=sin(x-π/4)+cos(x-3π/4)
=sin(x-π/4)+cos(x-π/4-π/2) \公式 cos(x-π/2)=sinx\
=sin(x-π/4)+sin(x-π/4)
=2sin(β-π/4)
所以最小正周期为2π,函数最小值为 -2
(2)f(β)=2sin(β-π/4)
∴【f(β)】方=4sin?(β-π/4) \二倍角公式\
=2cos(2β-π/2) +2
=2 sin(2β)+2 \公式 cos(x-π/2)=sinx\
下面求sin(2β):
根据条件:cos(β-α)=4/5,cos(β+α)= - 4/5
sin(2β)=sin [(β-α)+(β+α) ]
=sin(β-α)cos(β+α) + cos(β-α)sin(β+α)
=3/5×4/5 -4/5×3/5=0 \这里这种方法很常见,也是很喜欢考的技巧之一,就是已知
(β-α)和(β+α)的正弦或者余弦,求2β或者2α的
正弦或者余弦\
【其中,∵0<α<β≤2分之π,∴0<β-α<π/2 ∴sin(β-α)>0 , ∴ sin(β-α)=3/5
∵ cos(β+α)=-5分之4