cos2x=1-2sin^2x
所以
f(x)=1-2sin^2x+asinx
令sinx=t
t的取值范圍是[-1,1]
f(t)=-2t²+at+1
當a=2時
當t=-b/2a=1/2時取最大值 f(t)=3/2
當t=-1時取最大值 f(t)=-3
所以值域是[-3,3/2]
函數(shù)最小值與a的取值有關(guān)（也就是與對稱軸位置有關(guān)）
當 -b/2a=a/4=0 即a=0時,最小值在t=1或-1處取得
f(1)=a-1=-6
a=-5這與a=0矛盾.
當-b/2a=a/4>0 即a>0時,最小值在t=-1處取得
f(-1)=-1-a=-6
a=5
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