函数y=1/2 sin2x+sin^2x x属于R的值域是—
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解决时间 2021-02-27 14:48
- 提问者网友：星軌
- 2021-02-27 06:04
rt
最佳答案
- 五星知识达人网友：轻雾山林
- 2021-02-27 07:17
y=1/2sin2x+(1-cos2x)/2
=1/2(sin2x-cos2x)+1/2
=√2/2*(√2/2*sin2x+√2/2zos2x)+1/2
=√2/2*(sin2xzosπ/4+cos2xsinπ/4)+1/2
=√2/2*sin(2x+π/4)+1/2
-1<=sin(2x+π/4)<=1
所以值域[-√2/2+1/2,√2/2+1/2]
=1/2(sin2x-cos2x)+1/2
=√2/2*(√2/2*sin2x+√2/2zos2x)+1/2
=√2/2*(sin2xzosπ/4+cos2xsinπ/4)+1/2
=√2/2*sin(2x+π/4)+1/2
-1<=sin(2x+π/4)<=1
所以值域[-√2/2+1/2,√2/2+1/2]
全部回答
- 1楼网友：山君与见山
- 2021-02-27 07:32
y=1/2sin2x+1/2(1-cos2x)
=1/2(sin2x-cos2x+1)
=1/2[√2sin(2x-45º)+1]
值域[ 1/2(1-√2), 1/2(1+√2)]
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