(a) is (x+1)(4x^2 - 2x -1)
(b) 4cos^3 (theta) - 3cos(theta)
(c) if let x=cos(pi/5), then cos(3theta) = 4cos^3(pi/5) - 3cos(pi/5)=4x^3 - 3x
But then how to proceed? where does the factor x+1 come in?
更新1:
Hi, Gong, yes of course.
How to prove cos(3pi/5)=((1-sqrt5)/4)?
2021-01-13 7:01 pm
Already found
(a) is (x+1)(4x^2 - 2x -1)
(b) 4cos^3 (theta) - 3cos(theta)
(c) if let x=cos(pi/5), then cos(3theta) = 4cos^3(pi/5) - 3cos(pi/5)=4x^3 - 3x
But then how to proceed? where does the factor x+1 come in?
(a) is (x+1)(4x^2 - 2x -1)
(b) 4cos^3 (theta) - 3cos(theta)
(c) if let x=cos(pi/5), then cos(3theta) = 4cos^3(pi/5) - 3cos(pi/5)=4x^3 - 3x
But then how to proceed? where does the factor x+1 come in?
回答 (2)
2021-01-13 10:48 pm
✔ 最佳答案
昨晚的回答不知何故又被隱了,簡單回答吧.
令 x = cos(3π/5) 則
cos(9π/5) = 4x^3 - 3x
cos(6π/5) = 2x^2 - 1
cos(9π/5) + cos(6π/5) = 0
即
4x^3 + 2x^2 - 3x - 1
= (x+1)(4x^2 - 2x - 1 ) = 0.
因 π/2 < 3π/5 < π, 故
0 > cos(3π/5) > -1
故 cos(3π/5) = (1-√5)/4.
2021-01-14 9:37 am
用中文回答可以嗎?..........
收錄日期: 2021-04-24 08:26:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20210113110129AAwQg9g