f(x)=sin^2(x-π /4-acos(x-3 π/4)
来源:百度知道 编辑:UC知道 时间:2024/09/20 18:47:17
f(x)=sin^2(x-π /4)-acos(x-3 π/4),x取值【π/2,π】。①求f(x)的最小值g(a);②若a=sinb,b取值【0,2π),且有g(sinb)=1/2-根号下6/4,求∠b
①
f(x)=sin^2(x-π /4)-acos(x-3 π/4)
=sin^2(x-π /4)-asin(x-π /4);
=[sin(x-π /4) -a/2]^2 -a^2/4
则可知:
当-1<a/2<1时,f(x)≥0-a^2/4=-a^2/4;
当a/2≤-1时,取sin(x-π /4)=-1则f(x)最小;此时f(x)≥(-1)^2-a·(-1)=1+a;
当a/2≥1时,取sin(x-π /4)=1则f(x)最小;此时f(x)≥1-a;
综上所述,
f(x)的最小值g(a)={
1+a, a≤-2;
-a^2/4, -2<a<2;
1-a, a≥2.
②
若a=sinb,则-2<a<2;则g(a)=-a^2/4;
即g(sinb)= -sin^2 b /4;
即 -sin^2 b /4 = 1/2-√6/4;
则 sin^2 b = √6 -2;
1-2sin^2 b = 5-2√6;
cos2b= 5-2√6;
则 2b= arccos(5-2√6);
∠b=(1/2)·arccos(5-2√6)
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