已知点A(2,0),B(0,2)C(cosα,sinα)(其中0<α<π),O为坐标原点
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解:
函数f(x)=向量om*向量oc
=(cosα,cosα)·(cosα,sinα)
=cos²α+sinαcosα
=1/2(1+cos2α)+1/2sin2α
=√2/2(√2/2*sin2a+√2/2*cos2a)+1/2
=√2/2sin(2a+π/4)+1/2,
因为-1≤sin(2a+π/4)≤1,
所以-√2/2+1/2≤√2/2sin(2a+π/4)+1/2≤√2/2+1/2,
即函数f(x)的值域为[-√2/2+1/2,2/2+1/2]
O(∩_∩)O~
函数f(x)=向量om*向量oc
=(cosα,cosα)·(cosα,sinα)
=cos²α+sinαcosα
=1/2(1+cos2α)+1/2sin2α
=√2/2(√2/2*sin2a+√2/2*cos2a)+1/2
=√2/2sin(2a+π/4)+1/2,
因为-1≤sin(2a+π/4)≤1,
所以-√2/2+1/2≤√2/2sin(2a+π/4)+1/2≤√2/2+1/2,
即函数f(x)的值域为[-√2/2+1/2,2/2+1/2]
O(∩_∩)O~
追问
[-√2/2+1/2,2/2+1/2]这边写错了哦
应该是[-√2/2+1/2,√2/2+1/2]
谢谢~~
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