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Question-143635

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Question Number 143635 by SOMEDAVONG last updated on 16/Jun/21
Answered by TheHoneyCat last updated on 16/Jun/21
let x∈C _(or R if you want) let s=sin(x) and c=cos(x) s^4 +c^4 +s^2 c^2 =(s^2 )^2 +(c^2 )^2 +2(s^2 )(c^2 )−(sc)^2 =(s^2 +c^2 )−(sc)^2 =1−(sc)^2 also, c^8 +s^8 =c^2 c^6 +s^2 s^6 =(1−s^2 )c^6 +(1−c^2 )s^6 =c^6 +s^6 −s^2 c^6 −c^2 s^6 =(1−s^2 )c^4 +(1−c^2 )s^4 −c^2 s^2 (c^4 +s^4 ) =c^4 +s^4 −s^2 c^4 −c^2 s^4 −c^2 s^2 (c^4 +s^4 ) =(c^4 +s^4 )(1−(cs)^2 )−(cs)^2 (c^2 +s^2 ) =( (1−s^2 )c^2 + (1−c^2 )s^2 )(1−(cs)^2 ) − (cs)^2 =( c^2 +s^2 −2(cs)^2 )(1−(cs)^2 ) − (cs)^2 =(1−2(cs)^2 )(1−(cs)^2 )−(cs)^2 =1−4(cs)^2 +2(cs)^4 therefore: N=2−2(sc)^2 −1+4(sc)^2 −2(cs)^4 =1+2(sc)^2 −2(sc)^4 knowing that sin.cos(x)=(1/2)sin^2 (2x) N=1+(1/2)sin^2 (2x)−(1/8)sin^4 (2x)( ✠) =1+(1/2)sin^2 (2x)−(1/8)(1−cos^2 (2x))sin^2 (2x) =(1/8)(8+4sin^2 −sin^2 +(cos.sin)^2 )(2x) =(1/8)(8+3sin^2 (2x)+((1/2)sin(4x))^2 ) =(1/(64))(64+24sin^2 (2x)+2sin^2 (4x)) =(1/(64))(64+12+1 −12(1−sin^2 (2x)) −(1−2sin^2 (4x))) =(1/(64))(77−12cos(4x)−cos(8x)) Wich is the ′′simpliest′′ expression _((it is its Fourrier transformation)) thougth I do prefer (✠)
letx∈CorRifyouwantlets=sin(x)andc=cos(x)s4+c4+s2c2=(s2)2+(c2)2+2(s2)(c2)−(sc)2=(s2+c2)−(sc)2=1−(sc)2also,c8+s8=c2c6+s2s6=(1−s2)c6+(1−c2)s6=c6+s6−s2c6−c2s6=(1−s2)c4+(1−c2)s4−c2s2(c4+s4)=c4+s4−s2c4−c2s4−c2s2(c4+s4)=(c4+s4)(1−(cs)2)−(cs)2(c2+s2)=((1−s2)c2+(1−c2)s2)(1−(cs)2)−(cs)2=(c2+s2−2(cs)2)(1−(cs)2)−(cs)2=(1−2(cs)2)(1−(cs)2)−(cs)2=1−4(cs)2+2(cs)4therefore:N=2−2(sc)2−1+4(sc)2−2(cs)4=1+2(sc)2−2(sc)4knowingthatsin.cos(x)=12sin2(2x)N=1+12sin2(2x)−18sin4(2x)(✠)=1+12sin2(2x)−18(1−cos2(2x))sin2(2x)=18(8+4sin2−sin2+(cos.sin)2)(2x)=18(8+3sin2(2x)+(12sin(4x))2)=164(64+24sin2(2x)+2sin2(4x))=164(64+12+1−12(1−sin2(2x))−(1−2sin2(4x)))=164(77−12cos(4x)−cos(8x))Wichisthe″simpliest″expression(itisitsFourriertransformation)thougthIdoprefer(✠)

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