f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(2x+π/4)=√2sin(2x+π/2)=√2cos2xf(x)=√2cos2x的最小值即cos2x=-1时f(x) 的最小值为 -√2递增区间为[-π/2+kπ,kπ] (k∈Z),不是tan4/π=b/a=1吗,为啥是√2sin(2x+2/
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 02:44:36
![f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(2x+π/4)=√2sin(2x+π/2)=√2cos2xf(x)=√2cos2x的最小值即cos2x=-1时f(x) 的最小值为 -√2递增区间为[-π/2+kπ,kπ] (k∈Z),不是tan4/π=b/a=1吗,为啥是√2sin(2x+2/](/uploads/image/z/7156491-51-1.jpg?t=f%28x%29%3D1-2sin%5E2%28x%2B%CF%80%2F8%29%2B2sin%28x%2B%CF%80%2F8%29%2Acos%28x%2B%CF%80%2F8%29%3Dcos%282x%2B%CF%80%2F4%29%2Bsin%282x%2B%CF%80%2F4%29%3D%E2%88%9A2sin%282x%2B%CF%80%2F2%29%3D%E2%88%9A2cos2xf%28x%29%3D%E2%88%9A2cos2x%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E5%8D%B3cos2x%3D-1%E6%97%B6f%28x%29+%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA+-%E2%88%9A2%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%E4%B8%BA%5B-%CF%80%2F2%2Bk%CF%80%2Ck%CF%80%5D+%28k%E2%88%88Z%29%2C%E4%B8%8D%E6%98%AFtan4%2F%CF%80%3Db%2Fa%3D1%E5%90%97%2C%E4%B8%BA%E5%95%A5%E6%98%AF%E2%88%9A2sin%282x%2B2%2F)
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(2x+π/4)=√2sin(2x+π/2)=√2cos2xf(x)=√2cos2x的最小值即cos2x=-1时f(x) 的最小值为 -√2递增区间为[-π/2+kπ,kπ] (k∈Z),不是tan4/π=b/a=1吗,为啥是√2sin(2x+2/
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(2x+π/4)=√2sin(2x+π/2)=√2cos2xf(x)=√2cos2x的最小值即cos2x=-1时f(x) 的最小值为 -√2递增区间为[-π/2+kπ,kπ] (k∈Z),不是tan4/π=b/a=1吗,为啥是√2sin(2x+2/π)呢?
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(2x+π/4)=√2sin(2x+π/2)=√2cos2xf(x)=√2cos2x的最小值即cos2x=-1时f(x) 的最小值为 -√2递增区间为[-π/2+kπ,kπ] (k∈Z),不是tan4/π=b/a=1吗,为啥是√2sin(2x+2/
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)cos(x+π/8)
=cos(2x+π/4)+sin(2x+π/4)
=√2*[√2/2*cos(2x+π/4)+√2/2*sin(2x+π/4)]
=√2*[sinπ/4*cos(2x+π/4)+cosπ/4*sin(2x+π/4)]
=√2*sin(π/4+2x+π/4)
=√2*sin(π/2+2x)
=√2*cos2x
所以f(x)最大值为:√2,
单调增区间:2kπ-π