Forgot password?
 Register account
View 1454|Reply 1

[函数] 求证三角恒等式

[Copy link]

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

hbghlyj Posted 2020-1-20 18:45 |Read mode
Last edited by hbghlyj 2020-1-20 18:53${(\sin a + \sin b + \sin c + \sin d - \sin (a + b) - \sin (c + d))^2} + {( - 2 + \cos a + \cos b + \cos c + \cos d - \cos (a + b) - \cos (c + d))^2} = {(\sin a + \sin b - 2\sin (a + b) + \sin (a + b - c) + \sin (a + b - d) - \sin (a + b - c - d))^2} + {( - 1 + \cos a + \cos b - 2\cos (a + b) + \cos (a + b - c) + \cos (a + b - d) - \cos (a + b - c - d))^2}$

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2020-1-20 18:48
这是这道题的副产品。我本想通过模长均为1来导致矛盾,后来发现这是个恒等式

Mobile version|Discuz Math Forum

2025-5-31 11:07 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit