已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z) 求f(π/2010)+f(502π/1005)已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z)求f(π/2010)+f(502π/1005)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 18:11:29
![已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z) 求f(π/2010)+f(502π/1005)已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z)求f(π/2010)+f(502π/1005)](/uploads/image/z/3757023-63-3.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3Dcos%26%23178%3B%28n%CF%80%2Bx%29sin%26%23178%3B%28n%CF%80-x%29%2Fcos%26%23178%3B%5B%282n%2B1%29%CF%80-x%5D%28n%E2%88%88Z%29+%E6%B1%82f%28%CF%80%2F2010%29%2Bf%28502%CF%80%2F1005%29%E5%B7%B2%E7%9F%A5f%28x%29%3Dcos%26%23178%3B%28n%CF%80%2Bx%29sin%26%23178%3B%28n%CF%80-x%29%2Fcos%26%23178%3B%5B%282n%2B1%29%CF%80-x%5D%28n%E2%88%88Z%29%E6%B1%82f%28%CF%80%2F2010%29%2Bf%28502%CF%80%2F1005%29)
已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z) 求f(π/2010)+f(502π/1005)已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z)求f(π/2010)+f(502π/1005)
已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z) 求f(π/2010)+f(502π/1005)
已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z)
求f(π/2010)+f(502π/1005)
已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z) 求f(π/2010)+f(502π/1005)已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z)求f(π/2010)+f(502π/1005)
f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x]=cos²(x)sin²(x)/cos²(x)=sin²(x)
∴f(π/2010)+f(502π/1005)=sin²(π/2010)+sin²(502π/1005)
=[1-cos(π/2010)]/2+[1-cos(1004π/1005)]/2
=1-[cos(π/2010)+cos(1004π/1005]/2
=1