(sinx)^6+(cosx)^6

2008-03-09 11:14 pm
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

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


收錄日期: 2021-04-26 13:04:18
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