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Question Number 207516 by MATHEMATICSAM last updated on 17/May/24
3(sinθ − cosθ)^4  + 6(sinθ + cosθ)^2   + 4(sin^6 θ + cos^6 θ) = ?
3(sinθcosθ)4+6(sinθ+cosθ)2+4(sin6θ+cos6θ)=?
Commented by A5T last updated on 17/May/24
Did you edit this question to change something?
Didyoueditthisquestiontochangesomething?
Answered by A5T last updated on 17/May/24
3[sin^2 θ+cos^2 θ−2sinθcosθ]^2 +6(1+2sinθcosθ)  +4[(sin^2 θ+cos^2 θ)^3 −3sin^2 θcos^2 θ(sin^2 θ+cos^2 θ)]  =3[1−2sinθcosθ]^2 +6+6sin2θ+4[1−3sin^2 θcos^2 θ]  =3[1+4sin^2 θcos^2 θ−4sinθcosθ]+6+4  −12sin^2 θcos^2 θ+6sin2θ  =13−12sinθcosθ+6sin2θ=13
3[sin2θ+cos2θ2sinθcosθ]2+6(1+2sinθcosθ)+4[(sin2θ+cos2θ)33sin2θcos2θ(sin2θ+cos2θ)]=3[12sinθcosθ]2+6+6sin2θ+4[13sin2θcos2θ]=3[1+4sin2θcos2θ4sinθcosθ]+6+412sin2θcos2θ+6sin2θ=1312sinθcosθ+6sin2θ=13
Commented by Frix last updated on 17/May/24
6(sin θ +cos θ)^2 ≠6(1)^2
6(sinθ+cosθ)26(1)2
Commented by A5T last updated on 17/May/24
6(sin^2 θ+cos^2 θ)^2  was there initially, I suppose.
6(sin2θ+cos2θ)2wasthereinitially,Isuppose.
Commented by Frix last updated on 17/May/24
I don′t know, I saw it like it is.  But I believe you, this also happened to me  before...
Idontknow,Isawitlikeitis.ButIbelieveyou,thisalsohappenedtomebefore
Answered by Frix last updated on 17/May/24
Let t=tan θ  ...=((3(t−1)^4 )/((t^2 +1)^2 ))+((6(t+1)^2 )/(t^2 +1))+((4(t^4 −t^2 +1))/((t^2 +1)^2 ))=13
Lett=tanθ=3(t1)4(t2+1)2+6(t+1)2t2+1+4(t4t2+1)(t2+1)2=13

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