cos^-1(sin (16pi/5) find the exact value. how to do this? my answer is 6pi/10 but it is wrong...?
回答 (2)
2015-04-19 2:06 pm
16*pi/5 has the same terminal side as 6*pi/5, a 3rd quadrant angle. Its sine is the negative of sin(pi/5). And sin(pi/5) = cos(pi/2 - pi/5) = cos(3*pi/10). However, that's in the 1st quadrant. Angles whose cosine are the negative of cos(3*pi/10) will be 7*pi/10 and 13*pi/10. If this is something you have to enter in a machine, I'd go with 7*pi/10.
2015-04-20 12:34 am
sin(x) = cos(pi/2-x)
sin(16pi/5) = cos(pi/2 - 16pi/5) = cos(pi(1/2-16/5)) = cos(-27pi/10)= cos(2pi-27pi/10)=cos(-7pi/10)=cos(7pi/10)
cos^-1(sin(16pi/5)) = cos^-1(cos(7pi/10)) = 7pi/10
sin(16pi/5) = cos(pi/2 - 16pi/5) = cos(pi(1/2-16/5)) = cos(-27pi/10)= cos(2pi-27pi/10)=cos(-7pi/10)=cos(7pi/10)
cos^-1(sin(16pi/5)) = cos^-1(cos(7pi/10)) = 7pi/10
收錄日期: 2021-04-21 01:39:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150419055631AAIJgZz