express (sinx)^6+(cosx)^6 in terms of A+Bcos4x
here is my steps, which one should i amend?
define sinx=s cosx=c
s^6+c^6=(s^4)-( sc )^2+c^4
=(1-2( sc )^2) - (sc)^2
=1-3(sc)^2
consider cos4x=cos( 2(2x) )=1-2( sin2x )^2=1-2( 2sc )^2=1-8(sc)^2
so s^6+c^6=1-3(sc)^2=cos4x+5(sc)^2= ???
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or show me your steps.
thz
(sinx)^6+(cosx)^6
2008-03-09 11:14 pm
回答 (2)
2008-03-13 12:52 pm
✔ 最佳答案
s^6+c^6=(s^4)-( sc )^2+c^4=(1-2( sc )^2) - (sc)^2
=1-3(sc)^2
consider cos4x=cos( 2(2x) )=1-2( sin2x )^2=1-2( 2sc )^2=1-8(sc)^2
8(sc)^2 = 1-cos4x
(sc)^2 =(1-cos4x)/8
Hence
s^6+c^6=1-3(sc)^2 =1-3(1-cos4x)/8 = (5+3cos4x)/8
2008-03-10 2:07 am
cos4x
=2cos^2(2x)-1
=2[[2cos^4x-1]^2]-1
=8cos^4x-8cos^2x+1
[(sin^2x)+(cos^2x)]^3
=sin^6x+3sin^4xcos^2x+3cos^4xsin^2x+cos^6x
So
(sinx)^6+(cosx)^6
=1-3sin^4xcos^2x+3cos^4xsin^2x
=1-3sin^2xcos^2x
=1-3(1-cos^2x)cos^2x
=1-3cos^2x+3cos^4x
=5/8+3/8(8cos^4x-8cos^2x+1)
=(5/8)+(3/8)cos4x
=2cos^2(2x)-1
=2[[2cos^4x-1]^2]-1
=8cos^4x-8cos^2x+1
[(sin^2x)+(cos^2x)]^3
=sin^6x+3sin^4xcos^2x+3cos^4xsin^2x+cos^6x
So
(sinx)^6+(cosx)^6
=1-3sin^4xcos^2x+3cos^4xsin^2x
=1-3sin^2xcos^2x
=1-3(1-cos^2x)cos^2x
=1-3cos^2x+3cos^4x
=5/8+3/8(8cos^4x-8cos^2x+1)
=(5/8)+(3/8)cos4x
收錄日期: 2021-04-26 13:04:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080309000051KK01772