A) By using the sums and products formulae and cos36cos72 =1/4,
Find the value of cos36 + cos 72 in sure form.
B) By using formula (a+b)^2 =(a-b)^2 + 4ab and the above results,
find the value of cos36 + cos 72 in sure form.
C) hence, find the value of cos36 and cos72 in sure form.
三角學問題6/04/2013
2013-04-07 4:58 am
回答 (2)
2013-04-07 9:20 am
✔ 最佳答案
我睇你條問題應該a part 係搵 cos 36 - cos 72 吧 不然a part 就撞 b part 了cos 36 cos 72 = 1/4
[ cos (72-36) + cos (72+36) ] / 2 = 1/4
cos 36 + cos 108 = 1/2
cos 36 + cos (90+18) = 1/2
cos 36 -cos (90-18) = 1/2
cos 36 - cos 72 =1/2
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(cos 36 + cos 72)^2 = (cos 36 - cos 72)^2 + 4 cos 36 cos 72
(cos 36 + cos 72)^2 = (1/2)^2 + 4(1/4)
(cos 36 + cos 72)^2 = 1/4 + 1
(cos 36 + cos 72)^2 = 5/4
cos 36 + cos 72 = √5 / 2 // or - √5 / 2 (rej. ∵ cos 36 > cos 72 > 0)
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cos 36 - cos 72 =1/2 ---------------(1)
cos 36 + cos 72 = √5 / 2 ---------(2)
(1)+(2) : 2cos 36 =√5 + 1 / 2
cos 36 = √5 + 1 / 4 //
(2)-(1) : 2cos 72 = √5 - 1 / 2
cos 72 = √5 - 1 / 4 //
2013-04-08 4:03 am
a)
cos36cos72=1/4
2cos36cos72=1/2
(cos36)^2+2cos36cos72+(cos72)^2-(cos36)^2-(cos72)^2=1/2
(cos36+cos72)^2-[(cos36)^2+(cos72)^2]=1/2
(cos36+cos72)^2=5/4
cos36+cos72=(√5)/2 i
b)
(cos36+cos72)^2=5/4
(cos36-cos72)^2-4cos36cos72=5/4
(cos36-cos72)^2-1=5/4
cos36-cos72=1/2 ii
c)
i+ii, 2cos36=(√5)/2+1/2
cos36=[(√5)+1]/4
i-ii, 2cos72=(√5)/2-1/2
cos72=[(√5)-1]/4
cos36cos72=1/4
2cos36cos72=1/2
(cos36)^2+2cos36cos72+(cos72)^2-(cos36)^2-(cos72)^2=1/2
(cos36+cos72)^2-[(cos36)^2+(cos72)^2]=1/2
(cos36+cos72)^2=5/4
cos36+cos72=(√5)/2 i
b)
(cos36+cos72)^2=5/4
(cos36-cos72)^2-4cos36cos72=5/4
(cos36-cos72)^2-1=5/4
cos36-cos72=1/2 ii
c)
i+ii, 2cos36=(√5)/2+1/2
cos36=[(√5)+1]/4
i-ii, 2cos72=(√5)/2-1/2
cos72=[(√5)-1]/4
收錄日期: 2021-04-13 19:25:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130406000051KK00329