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在△ABC中,A+B<π,则0cos(π-B).∴ cos A+cos B>0. (*)(*)式说明△ABC中任两角的余弦之和均大于零,利用三角形的这一性质解一类三角求值问题,既可以避免繁烦的角的范围讨论,又可以防止增解,达到迅速解题目的.下面举例说明.
In △ABC, A+B<π, then 0cos(π-B). ∴ cos A+cos B>0. (*)(*) Note that the sum of the cosines of any two angles in △ABC is greater than zero. Using this property of triangles to solve a type of trigonometric problem can avoid bothersomeness. The discussion of the range of angles can also prevent escalation and achieve rapid resolution of the problem. Here is an example.