熱心網友
|x|=-π/4=-√2/2= |x|≤π/4, == -π/4≤x≤π/4, == -√2/2≤sinx≤√2/2f(x)=(cosx)^2+sinx=1-(sinx)^2+sinx=(5/4)-[sinx-(1/2)]^2所以：fmin=(5/4)-[-√2/2-(1/2)]^2,既fmin=(2-√2)/4熱心網友
|x|=-π/4=-√2/2= |x|≤π/4, == -π/4≤x≤π/4, == -√2/2≤sinx≤√2/2f(x)=(cosx)^2+sinx=1-(sinx)^2+sinx=(5/4)-[sinx-(1/2)]^2所以：fmin=(5/4)-[-√2/2-(1/2)]^2,既fmin=(2-√2)/4熱心網友