3-sin-cos-4-6-sin-cos-2-4-sin-6-cos-6-

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) = ?

$$\mathrm{3}\left(\mathrm{sin}\theta\:−\:\mathrm{cos}\theta\right)^{\mathrm{4}} \:+\:\mathrm{6}\left(\mathrm{sin}\theta\:+\:\mathrm{cos}\theta\right)^{\mathrm{2}} \\ $$$$+\:\mathrm{4}\left(\mathrm{sin}^{\mathrm{6}} \theta\:+\:\mathrm{cos}^{\mathrm{6}} \theta\right)\:=\:? \\ $$

Commented by A5T last updated on 17/May/24

$${Did}\:{you}\:{edit}\:{this}\:{question}\:{to}\:{change}\:{something}? \\ $$

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

$$\mathrm{3}\left[{sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta−\mathrm{2}{sin}\theta{cos}\theta\right]^{\mathrm{2}} +\mathrm{6}\left(\mathrm{1}+\mathrm{2}{sin}\theta{cos}\theta\right) \\ $$$$+\mathrm{4}\left[\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{3}} −\mathrm{3}{sin}^{\mathrm{2}} \theta{cos}^{\mathrm{2}} \theta\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right)\right] \\ $$$$=\mathrm{3}\left[\mathrm{1}−\mathrm{2}{sin}\theta{cos}\theta\right]^{\mathrm{2}} +\mathrm{6}+\mathrm{6}{sin}\mathrm{2}\theta+\mathrm{4}\left[\mathrm{1}−\mathrm{3}{sin}^{\mathrm{2}} \theta{cos}^{\mathrm{2}} \theta\right] \\ $$$$=\mathrm{3}\left[\mathrm{1}+\mathrm{4}{sin}^{\mathrm{2}} \theta{cos}^{\mathrm{2}} \theta−\mathrm{4}{sin}\theta{cos}\theta\right]+\mathrm{6}+\mathrm{4} \\ $$$$−\mathrm{12}{sin}^{\mathrm{2}} \theta{cos}^{\mathrm{2}} \theta+\mathrm{6}{sin}\mathrm{2}\theta \\ $$$$=\mathrm{13}−\mathrm{12}{sin}\theta{cos}\theta+\mathrm{6}{sin}\mathrm{2}\theta=\mathrm{13} \\ $$

Commented by Frix last updated on 17/May/24

$$\mathrm{6}\left(\mathrm{sin}\:\theta\:+\mathrm{cos}\:\theta\right)^{\mathrm{2}} \neq\mathrm{6}\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$

Commented by A5T last updated on 17/May/24

6(sin^2 θ+cos^2 θ)^2 was there initially, I suppose.

$$\mathrm{6}\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} \:{was}\:{there}\:{initially},\:{I}\:{suppose}. \\ $$

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...

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know},\:\mathrm{I}\:\mathrm{saw}\:\mathrm{it}\:\mathrm{like}\:\mathrm{it}\:\mathrm{is}. \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{believe}\:\mathrm{you},\:\mathrm{this}\:\mathrm{also}\:\mathrm{happened}\:\mathrm{to}\:\mathrm{me} \\ $$$$\mathrm{before}… \\ $$

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

$$\mathrm{Let}\:{t}=\mathrm{tan}\:\theta \\ $$$$…=\frac{\mathrm{3}\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{6}\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}\left({t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{13} \\ $$

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