given that sinA+cosA=3/4
find the value of sin3次方A+cos3次方A
additional mathematic
2007-12-22 9:13 pm
回答 (2)
2007-12-22 9:25 pm
✔ 最佳答案
(sinA + cosA)^2 = 9/16sin^2 A + 2 sinAcosA + cos^2 A = 9/16
2 sinAcosA + 1 = 9/16
2 sinAcosA = -7/16
sinAcosA = -7/32
sin^3 A + cos^3 A
= (sin A + cos A)(sin^2 A - sin A cos A+ cos^2 A)
= (sin A + cos A)(1 - sin A cos A)
= (3/4)(1 + 7/32)
= 117/128
2007-12-22 9:24 pm
sin(1)=0.017452
cos(1)=0.999848
0.017452A + 0.99848A=3/4
A=.737246
(0.017452)^3 x A +(0.0999848)^3 x A=???
=0.736913
Answer=0.736913
cos(1)=0.999848
0.017452A + 0.99848A=3/4
A=.737246
(0.017452)^3 x A +(0.0999848)^3 x A=???
=0.736913
Answer=0.736913
參考: myself
收錄日期: 2021-04-13 14:46:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071222000051KK01841