作业帮[cos(a+π)sin²（2+3π）]/[tan(a+π)cos³（-a-π）]打错了一个字，应该是：[cos(a+π)sin²（a+3π）]/[tan(a+π)cos³（-a-π）]
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/28 22:27:45
![[cos(a+π)sin²（2+3π）]/[tan(a+π)cos³（-a-π）]打错了一个字，应该是：[cos(a+π)sin²（a+3π）]/[tan(a+π)cos³（-a-π）]](/uploads/image/z/1105025-41-5.jpg?t=%5Bcos%28a%2B%CF%80%29sin%26%23178%3B%EF%BC%882%2B3%CF%80%EF%BC%89%5D%2F%5Btan%28a%2B%CF%80%29cos%26%23179%3B%EF%BC%88-a-%CF%80%EF%BC%89%5D%E6%89%93%E9%94%99%E4%BA%86%E4%B8%80%E4%B8%AA%E5%AD%97%EF%BC%8C%E5%BA%94%E8%AF%A5%E6%98%AF%EF%BC%9A%5Bcos%28a%2B%CF%80%29sin%26%23178%3B%EF%BC%88a%2B3%CF%80%EF%BC%89%5D%2F%5Btan%28a%2B%CF%80%29cos%26%23179%3B%EF%BC%88-a-%CF%80%EF%BC%89%5D)
[cos(a+π)sin²（2+3π）]/[tan(a+π)cos³（-a-π）]打错了一个字，应该是：[cos(a+π)sin²（a+3π）]/[tan(a+π)cos³（-a-π）]
[cos(a+π)sin²（2+3π）]/[tan(a+π)cos³（-a-π）]
打错了一个字，应该是：
[cos(a+π)sin²（a+3π）]/[tan(a+π)cos³（-a-π）]
[cos(a+π)sin²（2+3π）]/[tan(a+π)cos³（-a-π）]打错了一个字，应该是：[cos(a+π)sin²（a+3π）]/[tan(a+π)cos³（-a-π）]
设a+π=x,则原式可化为[cosxsin²x]/[（sinx/cosx）cos³（-x）]
因为cos(-x)=cosx,则原式=[cosxsin²x]/[sinxcos²x]=sinx/cosx=tanx,
即原式=tan(a+π)