设0小于等于a小于等于π,P=sin2a+sina-cosa1.若t=sina-cisa,用t的式子表示P2.确定t的取值范围,并求出t的最大值

Question

设0小于等于a小于等于π,P=sin2a+sina-cosa
1.若t=sina-cisa,用t的式子表示P
2.确定t的取值范围,并求出t的最大值

Answer 1

1、P=sin2a+sina-cosa =2sinacosa+sina-cosa
=-1+2sinacosa+sina-cosa +1
=-(sina-cosa )^2+(sina-cosa)+1
=-t^2+t+1
2、P=-t^2+t+1=-（t-1/2)^2+5/4
t=sina-cosa=√2*sin(a-π/4)
a属于[0,π],a-π/4 属于[-π/4,3π/4],
sin(a-π/4)属于[-√2/2,1]
t属于[-1,√2]
当t=1/2时p有最大值5/4