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Question Number 86878 by jagoll last updated on 01/Apr/20
If sin^(10) (x) + cos^(10) (x) = ((11)/(36)) find sin^(12) (x) + cos ^(12) (x) = ?
$$\mathrm{If}\:\mathrm{sin}\:^{\mathrm{10}} \:\left(\mathrm{x}\right)\:+\:\mathrm{cos}\:^{\mathrm{10}} \:\left(\mathrm{x}\right)\:=\:\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\mathrm{find}\:\mathrm{sin}\:^{\mathrm{12}} \:\left(\mathrm{x}\right)\:+\:\mathrm{cos}\:\:^{\mathrm{12}} \left(\mathrm{x}\right)\:=\:? \\ $$
Commented by Prithwish Sen 1 last updated on 01/Apr/20
I got sin^2 xcos^2 x=(5/6) or(1/6) now sin^(12) x+cos^(12) x = 1−6sin^2 xcos^2 x+9sin^4 xcos^4 x −2sin^6 xcos^6 x
$$\mathrm{I}\:\mathrm{got} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{xcos}^{\mathrm{2}} \mathrm{x}=\frac{\mathrm{5}}{\mathrm{6}}\:\mathrm{or}\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{now} \\ $$$$\mathrm{sin}^{\mathrm{12}} \mathrm{x}+\mathrm{cos}^{\mathrm{12}} \mathrm{x}\:=\:\mathrm{1}−\mathrm{6sin}^{\mathrm{2}} \mathrm{xcos}^{\mathrm{2}} \mathrm{x}+\mathrm{9sin}^{\mathrm{4}} \mathrm{xcos}^{\mathrm{4}} \mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2sin}^{\mathrm{6}} \mathrm{xcos}^{\mathrm{6}} \mathrm{x} \\ $$
Answered by mr W last updated on 01/Apr/20
let s=sin^2 x, c=cos^2 x sin^(10) x+cos^(10) x=((11)/(36)) ⇒s^5 +c^5 =((11)/(36)) s+c=1 (s+c)^5 =1 s^5 +5s^4 c+10s^3 c^2 +10s^2 c^3 +5sc^4 +c^5 =1 s^5 +c^5 +5sc(s^3 +c^3 )+10s^2 c^2 (s+c)=1 s^5 +c^5 +5sc(s+c)[(s+c)^2 −3sc]+10(sc)^2 (s+c)=1 let λ=sc ((11)/(36))+5λ[1−3λ]+10λ^2 =1 λ^2 −λ+(5/(36))=0 (λ−(1/6))(λ−(5/6))=0 ⇒λ=(1/6), λ=(5/6)>(1/4) ⇒not suitable λ=(sin x cos x)^2 =(((sin 2x)/2))^2 ≤(1/4) ⇒λ=(1/6) (s+c)^6 =1 s^6 +6s^5 c+15s^4 c^2 +20s^3 c^3 +15s^2 c^4 +6sc^5 +c^6 =1 s^6 +c^6 +6sc(s^4 +c^4 )+15(sc)^2 (s^2 +c^2 )+20(sc)^3 =1 s^6 +c^6 +6sc[((s+c)^2 −2sc)^2 −2s^2 c^2 ]+15(sc)^2 [(s+c)^2 −2sc]+20(sc)^3 =1 s^6 +c^6 +6λ[(1−2λ)^2 −2λ^2 ]+15λ^2 [1−2λ]+20λ^3 =1 s^6 +c^6 +6λ−9λ^2 +2λ^3 =1 s^6 +c^6 =1−λ(2λ^2 −9λ+6) s^6 +c^6 =1−(1/6)((1/(18))−(3/2)+6)=((13)/(54)) ⇒sin^(12) x+cos^(12) x=((13)/(54))
$${let}\:{s}=\mathrm{sin}^{\mathrm{2}} \:{x},\:{c}=\mathrm{cos}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{sin}^{\mathrm{10}} \:{x}+\mathrm{cos}^{\mathrm{10}} \:{x}=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\Rightarrow{s}^{\mathrm{5}} +{c}^{\mathrm{5}} =\frac{\mathrm{11}}{\mathrm{36}} \\ $$$${s}+{c}=\mathrm{1} \\ $$$$\left({s}+{c}\right)^{\mathrm{5}} =\mathrm{1} \\ $$$${s}^{\mathrm{5}} +\mathrm{5}{s}^{\mathrm{4}} {c}+\mathrm{10}{s}^{\mathrm{3}} {c}^{\mathrm{2}} +\mathrm{10}{s}^{\mathrm{2}} {c}^{\mathrm{3}} +\mathrm{5}{sc}^{\mathrm{4}} +{c}^{\mathrm{5}} =\mathrm{1} \\ $$$${s}^{\mathrm{5}} +{c}^{\mathrm{5}} +\mathrm{5}{sc}\left({s}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)+\mathrm{10}{s}^{\mathrm{2}} {c}^{\mathrm{2}} \left({s}+{c}\right)=\mathrm{1} \\ $$$${s}^{\mathrm{5}} +{c}^{\mathrm{5}} +\mathrm{5}{sc}\left({s}+{c}\right)\left[\left({s}+{c}\right)^{\mathrm{2}} −\mathrm{3}{sc}\right]+\mathrm{10}\left({sc}\right)^{\mathrm{2}} \left({s}+{c}\right)=\mathrm{1} \\ $$$${let}\:\lambda={sc} \\ $$$$\frac{\mathrm{11}}{\mathrm{36}}+\mathrm{5}\lambda\left[\mathrm{1}−\mathrm{3}\lambda\right]+\mathrm{10}\lambda^{\mathrm{2}} =\mathrm{1} \\ $$$$\lambda^{\mathrm{2}} −\lambda+\frac{\mathrm{5}}{\mathrm{36}}=\mathrm{0} \\ $$$$\left(\lambda−\frac{\mathrm{1}}{\mathrm{6}}\right)\left(\lambda−\frac{\mathrm{5}}{\mathrm{6}}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{6}},\:\lambda=\frac{\mathrm{5}}{\mathrm{6}}>\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow{not}\:{suitable} \\ $$$$\lambda=\left(\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\right)^{\mathrm{2}} =\left(\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$$$\left({s}+{c}\right)^{\mathrm{6}} =\mathrm{1} \\ $$$${s}^{\mathrm{6}} +\mathrm{6}{s}^{\mathrm{5}} {c}+\mathrm{15}{s}^{\mathrm{4}} {c}^{\mathrm{2}} +\mathrm{20}{s}^{\mathrm{3}} {c}^{\mathrm{3}} +\mathrm{15}{s}^{\mathrm{2}} {c}^{\mathrm{4}} +\mathrm{6}{sc}^{\mathrm{5}} +{c}^{\mathrm{6}} =\mathrm{1} \\ $$$${s}^{\mathrm{6}} +{c}^{\mathrm{6}} +\mathrm{6}{sc}\left({s}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)+\mathrm{15}\left({sc}\right)^{\mathrm{2}} \left({s}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\mathrm{20}\left({sc}\right)^{\mathrm{3}} =\mathrm{1} \\ $$$${s}^{\mathrm{6}} +{c}^{\mathrm{6}} +\mathrm{6}{sc}\left[\left(\left({s}+{c}\right)^{\mathrm{2}} −\mathrm{2}{sc}\right)^{\mathrm{2}} −\mathrm{2}{s}^{\mathrm{2}} {c}^{\mathrm{2}} \right]+\mathrm{15}\left({sc}\right)^{\mathrm{2}} \left[\left({s}+{c}\right)^{\mathrm{2}} −\mathrm{2}{sc}\right]+\mathrm{20}\left({sc}\right)^{\mathrm{3}} =\mathrm{1} \\ $$$${s}^{\mathrm{6}} +{c}^{\mathrm{6}} +\mathrm{6}\lambda\left[\left(\mathrm{1}−\mathrm{2}\lambda\right)^{\mathrm{2}} −\mathrm{2}\lambda^{\mathrm{2}} \right]+\mathrm{15}\lambda^{\mathrm{2}} \left[\mathrm{1}−\mathrm{2}\lambda\right]+\mathrm{20}\lambda^{\mathrm{3}} =\mathrm{1} \\ $$$${s}^{\mathrm{6}} +{c}^{\mathrm{6}} +\mathrm{6}\lambda−\mathrm{9}\lambda^{\mathrm{2}} +\mathrm{2}\lambda^{\mathrm{3}} =\mathrm{1} \\ $$$${s}^{\mathrm{6}} +{c}^{\mathrm{6}} =\mathrm{1}−\lambda\left(\mathrm{2}\lambda^{\mathrm{2}} −\mathrm{9}\lambda+\mathrm{6}\right) \\ $$$${s}^{\mathrm{6}} +{c}^{\mathrm{6}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{1}}{\mathrm{18}}−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{6}\right)=\frac{\mathrm{13}}{\mathrm{54}} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{12}} \:{x}+\mathrm{cos}^{\mathrm{12}} \:{x}=\frac{\mathrm{13}}{\mathrm{54}} \\ $$
Answered by MJS last updated on 01/Apr/20
sin x =±((tan x)/( (√(1+tan^2 x)))) cos x =±(1/( (√(1+tan^2 x)))) let tan x =t sin^(10) x +cos^(10) x =((t^8 −t^6 +t^4 −t^2 +1)/((t^2 +1)^4 )) sin^(12) x +cos^(12) x =((t^(12) +1)/((t^2 +1)^6 )) ((t^8 −t^6 +t^4 −t^2 +1)/((t^2 +1)^4 ))=((11)/(36)) t^8 −((16)/5)t^6 −(6/5)t^4 −((16)/5)t^2 +1=0 (t^4 −4t^2 +1)(t^4 +(4/5)t^2 +1)=0 ⇒ t^2 =2±(√3) sin^(12) x +cos^(12) x =((t^(12) +1)/((t^2 +1)^6 ))=((13)/(54))
$$\mathrm{sin}\:{x}\:=\pm\frac{\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}} \\ $$$$\mathrm{cos}\:{x}\:=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}} \\ $$$$\mathrm{let}\:\mathrm{tan}\:{x}\:={t} \\ $$$$\mathrm{sin}^{\mathrm{10}} \:{x}\:+\mathrm{cos}^{\mathrm{10}} \:{x}\:=\frac{{t}^{\mathrm{8}} −{t}^{\mathrm{6}} +{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\mathrm{sin}^{\mathrm{12}} \:{x}\:+\mathrm{cos}^{\mathrm{12}} \:{x}\:=\frac{{t}^{\mathrm{12}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{6}} } \\ $$$$\frac{{t}^{\mathrm{8}} −{t}^{\mathrm{6}} +{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$${t}^{\mathrm{8}} −\frac{\mathrm{16}}{\mathrm{5}}{t}^{\mathrm{6}} −\frac{\mathrm{6}}{\mathrm{5}}{t}^{\mathrm{4}} −\frac{\mathrm{16}}{\mathrm{5}}{t}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} +\frac{\mathrm{4}}{\mathrm{5}}{t}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} =\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\mathrm{sin}^{\mathrm{12}} \:{x}\:+\mathrm{cos}^{\mathrm{12}} \:{x}\:=\frac{{t}^{\mathrm{12}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{6}} }=\frac{\mathrm{13}}{\mathrm{54}} \\ $$
Commented by Prithwish Sen 1 last updated on 01/Apr/20
Excellent. Thank you sir.
$$\mathrm{Excellent}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by jagoll last updated on 01/Apr/20
waww....super easy
$$\mathrm{waww}….\mathrm{super}\:\mathrm{easy}\: \\ $$

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