If cos a=1/k and 0<a<90, then tan(a-270)=?
回答 (1)
2016-11-12 4:11 am
cos a=1/k
(sin a)^2 +(cos a)^2=1
sin a=[1-(1/k)^0.5]^0.5
tan(a-270)
=sin(a-270)/cos(a-270)
=[(sin a)(cos 270)-(cos a)(sin 270)]/[(cos a)(cos 270)+(sin a)(sin 270)]
=(cos a)/(-sin a)
=-(1/k)/[1-(1/k)^2]^0.5
=-(1/k)/[(k^2 -1)/k^2]^0.5
=-1/(k^2 -1)^0.5
(sin a)^2 +(cos a)^2=1
sin a=[1-(1/k)^0.5]^0.5
tan(a-270)
=sin(a-270)/cos(a-270)
=[(sin a)(cos 270)-(cos a)(sin 270)]/[(cos a)(cos 270)+(sin a)(sin 270)]
=(cos a)/(-sin a)
=-(1/k)/[1-(1/k)^2]^0.5
=-(1/k)/[(k^2 -1)/k^2]^0.5
=-1/(k^2 -1)^0.5
收錄日期: 2021-04-18 15:32:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160902231706AAl4bN6