已知sin x + cos x = 3/5, 且sin x 及 cos x 為2y^2 +py + q =0 的兩根
求p^2-8q之值
數學問題 (一題only)
2008-04-30 6:20 am
回答 (2)
2008-04-30 4:46 pm
✔ 最佳答案
Given sinx + cosx = 3/5(sinx+cosx)^2 = 9/25
sinx^2(x) + 2sinxcosx + cos^2(x) = 9/25
1+2sinxcosx = 9/25
sinxcosx = -8/25
Sum of root sinx + cos x = -p/2 = 3/5
Therefore, p = -6/5
Product of root sinxcosx = q/2 = -8/25
Therefore, q = -16/25
p^2-8q
= (-6/5)^2-8(-16/25)
=36/25 - 128/25
=-92/25
2008-04-30 6:25 am
用product root同 sum of root做...求出p and q 值...就能求出p^2-8q之值
2008-04-29 22:30:18 補充：
同咪a^2+b^2=(a+b)^2-2ab
sin^2x+cos^2x=1
2008-04-29 22:30:18 補充：
同咪a^2+b^2=(a+b)^2-2ab
sin^2x+cos^2x=1
參考: 自己諗
收錄日期: 2021-04-13 15:29:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080429000051KK02983