求y=sin^2(x/2)+4sinx的值域

题目

求y=sin^2(x/2)+4sinx的值域

答案

y=(1-cosx)/2+4sinx
=4sinx-cosx/2+1/2;
=√(16+1/4)(sinx×4/√(16+1/4)-cosx×(1/2)/√(16+1/4))+1/2;
=√65/2sin(x-φ)+1/2;(sinφ=1/2/(√65/2)=√65/65;)
∴-1≤sin(x-φ)≤1
∴-√65/2≤√65/2sin(x-φ)≤√65/2;
值域为[（1-√65）/2,(1+√65)/2]
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