sin(kπ-α)•cos(kπ+α)∕sin[(k+1)π+α]•cos[(k+1)π-α]=1.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 00:19:27
![sin(kπ-α)•cos(kπ+α)∕sin[(k+1)π+α]•cos[(k+1)π-α]=1.](/uploads/image/z/13276482-42-2.jpg?t=sin%EF%BC%88k%CF%80%EF%BC%8D%CE%B1%EF%BC%89%26%238226%3Bcos%EF%BC%88k%CF%80%2B%CE%B1%EF%BC%89%E2%88%95sin%5B%EF%BC%88k%2B1%EF%BC%89%CF%80%2B%CE%B1%5D%26%238226%3Bcos%5B%EF%BC%88k%2B1%EF%BC%89%CF%80%EF%BC%8D%CE%B1%5D%3D1.)
sin(kπ-α)•cos(kπ+α)∕sin[(k+1)π+α]•cos[(k+1)π-α]=1.
sin(kπ-α)•cos(kπ+α)∕sin[(k+1)π+α]•cos[(k+1)π-α]=1.
sin(kπ-α)•cos(kπ+α)∕sin[(k+1)π+α]•cos[(k+1)π-α]=1.
证明:
左边=sina*(-cosa)/[-sina*(-cosa)]=-1=右边.
得证.
k为奇数
sin(kπ-α)•cos(kπ+α)∕sin[(k+1)π+α]•cos[(k+1)π-α]
=sina*(-cosa)/sina*cosa
=-1
k为偶数
sin(kπ-α)•cos(kπ+α)∕sin[(k+1)π+α]•cos[(k+1)π-α]
=-sina*cosa/(-sina)*(-cosa)
=-1