find sin3x/sinx in terms of k
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amaths問題(20分)
回答 (2)
2008-04-18 9:26 pm
2008-04-16 10:44 pm
tan3x=(tan2x+tanx)/(1-tan2xtanx)
={[2tanx/(1-tan^2x)]+tanx}/{1-tanx[2tanx/(1-tan^2x)]}
=(2tanx+tanx-tan^3x)/(1-tan^2x-2tan^2x)
=(3tanx-tan^3x)/(1-3tan^2x)
tan3x/tanx=(3-tan^3x)(1-3tan^2x)
(1-3tan^2x)k=3-tan^3x
(1-3k)tan^2x=3-k
tan^2x=(3-k)/(1-3k)
1/(tan^2x+1)=cos^2x
=1-sin^2x
sin^2x=1-1/(tan^2x+1)
=1-(1-3k)/(3-k+1-3k)
=(4-4k-1+3k)/(4-4k)
=(3-k)/(4-4k)
sin3x/sinx=(3sinx-4sin^3x)/sinx
=-4sin^2x-3
=-4(3-k)(4-4k)-3
=-(3-k)/(1-k)-3
=(3-k)/(k-1)-3
={[2tanx/(1-tan^2x)]+tanx}/{1-tanx[2tanx/(1-tan^2x)]}
=(2tanx+tanx-tan^3x)/(1-tan^2x-2tan^2x)
=(3tanx-tan^3x)/(1-3tan^2x)
tan3x/tanx=(3-tan^3x)(1-3tan^2x)
(1-3tan^2x)k=3-tan^3x
(1-3k)tan^2x=3-k
tan^2x=(3-k)/(1-3k)
1/(tan^2x+1)=cos^2x
=1-sin^2x
sin^2x=1-1/(tan^2x+1)
=1-(1-3k)/(3-k+1-3k)
=(4-4k-1+3k)/(4-4k)
=(3-k)/(4-4k)
sin3x/sinx=(3sinx-4sin^3x)/sinx
=-4sin^2x-3
=-4(3-k)(4-4k)-3
=-(3-k)/(1-k)-3
=(3-k)/(k-1)-3
收錄日期: 2021-04-26 11:54:26
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https://hk.answers.yahoo.com/question/index?qid=20080416000051KK00703