已知f(1+sinx)=2+sinx+cos2x,求f(x)令u=1+sinx,则sinx=u-1 (0≤u≤2),则f(u)=-u2+3u+1 (0≤u≤2)故f(x)=-x2+3x+1 (0≤u≤2)【f(u)=-u2+3u+1】是怎么得来的?求教!
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 19:39:07
![已知f(1+sinx)=2+sinx+cos2x,求f(x)令u=1+sinx,则sinx=u-1 (0≤u≤2),则f(u)=-u2+3u+1 (0≤u≤2)故f(x)=-x2+3x+1 (0≤u≤2)【f(u)=-u2+3u+1】是怎么得来的?求教!](/uploads/image/z/10395002-2-2.jpg?t=%E5%B7%B2%E7%9F%A5f%281%2Bsinx%29%3D2%2Bsinx%2Bcos2x%2C%E6%B1%82f%28x%29%E4%BB%A4u%3D1%2Bsinx%2C%E5%88%99sinx%3Du-1+%280%E2%89%A4u%E2%89%A42%29%2C%E5%88%99f%28u%29%3D-u2%2B3u%2B1+%280%E2%89%A4u%E2%89%A42%29%E6%95%85f%28x%29%3D-x2%2B3x%2B1+%280%E2%89%A4u%E2%89%A42%29%E3%80%90f%28u%29%3D-u2%2B3u%2B1%E3%80%91%E6%98%AF%E6%80%8E%E4%B9%88%E5%BE%97%E6%9D%A5%E7%9A%84%3F%E6%B1%82%E6%95%99%21)
已知f(1+sinx)=2+sinx+cos2x,求f(x)令u=1+sinx,则sinx=u-1 (0≤u≤2),则f(u)=-u2+3u+1 (0≤u≤2)故f(x)=-x2+3x+1 (0≤u≤2)【f(u)=-u2+3u+1】是怎么得来的?求教!
已知f(1+sinx)=2+sinx+cos2x,求f(x)
令u=1+sinx,则sinx=u-1 (0≤u≤2),则f(u)=-u2+3u+1 (0≤u≤2)
故f(x)=-x2+3x+1 (0≤u≤2)
【f(u)=-u2+3u+1】是怎么得来的?
求教!
已知f(1+sinx)=2+sinx+cos2x,求f(x)令u=1+sinx,则sinx=u-1 (0≤u≤2),则f(u)=-u2+3u+1 (0≤u≤2)故f(x)=-x2+3x+1 (0≤u≤2)【f(u)=-u2+3u+1】是怎么得来的?求教!
你好
解:
令u=1+sinx,则
sinx=u-1 (0≤u≤2),
cos2x=1-2sin²x=1-2(u-1)²=1-2(u²-2u+1)=1-2u²+4u-2=-2u²+4u-1
则f(u)=2+(u-1 )+(-2u²+4u-1)=-2u²+5u
不对呀
f(1+sinx)=2+sinx+cos²x
cos²x=1-sin²x=1-(u-1)²=1-u²+2u-1=-u²+2u
f(u)=2+(u-1 )+(-u2+2u) (0≤u≤2)
=2+u-1--u2+2u
=-u²+3u+1
故f(x)=-x²+3x+1 (0≤x≤2)
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