前苏联有一道三角竞赛题: 设锐角α、β满足sin2α+sin2β=sin(α+β). 求证:α+β=π/2. 一般学生容易从求sin(α+β)=1去入手,但经过一番苦思冥想之后,由条件无法求解出sin(α+β)=1这个结果.遇到这种情况,不妨改变一下解题方向,观察条件等式sin2α+sin2β=sin(α+β),即可从sin(α+β)≤1入手,可顺利地证明此题. 证明由sin(α+β)≤1得 sin2α+sin2β≤sin2α+cos2α① sin2α+sin2β≤cos2β+sin2β② In the former Soviet Union there was a trigonometric contest: Let the acute angles α and β satisfy sin2α+sin2β=sin(α+β). Proof: α+β=π/2. It is easy for average students to start with sin(α+β)=1. However, after some meditation, the condition cannot solve the result of sin(α+β)=1. In this case, you may wish to change the direction of the problem and observe the condition equation sin2α+sin2β=sin(α+β ), you can start from sin (α + β) ≤ 1 to successfully prove this problem. Prove by sin (α + β) ≤ 1 sin2α + sin2β ≤ sin2α + cos2α1 sin2α + sin2β ≤ cos2β + sin2β2