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Question Number 227328 by cherokeesay last updated on 16/Jan/26
Answered by som(math1967) last updated on 17/Jan/26
sin^(10) x+cos^(10) x=((11)/(36)) let sin^2 x=a,cos^2 x=b a^5 +b^5 =((11)/(36)) (a+b)(a^4 −a^3 b+a^2 b^2 −ab^3 +b^4 )=((11)/(36)) 36(5a^2 b^2 −5ab+1)=11★ ⇒(180a^2 b^2 −180ab+25)=0 ⇒(6ab−1)(6ab−5)=0 ⇒ab=(1/6) or (5/6) (sin^2 x+cos^2 x)(sin^(10) x+cos^(10) x)=((11)/(36)) sin^(12) x+cos^(12) x+ sin^2 xcos^2 x(sin^8 x+cos^8 x)=((11)/(36))×1 sin^(12) x+cos^(12) x+(1/6){(1−2×(1/6))^2 −2×(1/(36))} =((11)/(36)) sin^(12) x+cos^(12) x+(1/6)×((4/9)−(1/(18)))=((11)/(36)) sin^(12) x+cos^(12) x=((11)/(36))−(7/(18×6)) sin^(12) x+cos^(12) x=((13)/(54)) m+n=67 when sin^2 xcos^2 x=(5/6) then .... ★ a+b=1 a^2 +b^2 =1−2ab a^4 +b^4 =(a^2 +b^2 )^2 −2a^2 b^2 =1−4ab+2a^2 b^2
$${sin}^{\mathrm{10}} {x}+{cos}^{\mathrm{10}} {x}=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$${let}\:{sin}^{\mathrm{2}} {x}={a},{cos}^{\mathrm{2}} {x}={b} \\ $$$$\:{a}^{\mathrm{5}} +{b}^{\mathrm{5}} =\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\left({a}+{b}\right)\left({a}^{\mathrm{4}} −{a}^{\mathrm{3}} {b}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{ab}^{\mathrm{3}} +{b}^{\mathrm{4}} \right)=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\:\mathrm{36}\left(\mathrm{5}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{5}{ab}+\mathrm{1}\right)=\mathrm{11}\bigstar \\ $$$$\Rightarrow\left(\mathrm{180}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{180}{ab}+\mathrm{25}\right)=\mathrm{0} \\ $$$$\:\Rightarrow\left(\mathrm{6}{ab}−\mathrm{1}\right)\left(\mathrm{6}{ab}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow{ab}=\frac{\mathrm{1}}{\mathrm{6}}\:\:{or}\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\:\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)\left({sin}^{\mathrm{10}} {x}+{cos}^{\mathrm{10}} {x}\right)=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$${sin}^{\mathrm{12}} {x}+{cos}^{\mathrm{12}} {x}+ \\ $$$${sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\left({sin}^{\mathrm{8}} {x}+{cos}^{\mathrm{8}} {x}\right)=\frac{\mathrm{11}}{\mathrm{36}}×\mathrm{1} \\ $$$${sin}^{\mathrm{12}} {x}+{cos}^{\mathrm{12}} {x}+\frac{\mathrm{1}}{\mathrm{6}}\left\{\left(\mathrm{1}−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} −\mathrm{2}×\frac{\mathrm{1}}{\mathrm{36}}\right\} \\ $$$$\:=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$${sin}^{\mathrm{12}} {x}+{cos}^{\mathrm{12}} {x}+\frac{\mathrm{1}}{\mathrm{6}}×\left(\frac{\mathrm{4}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{18}}\right)=\frac{\mathrm{11}}{\mathrm{36}}\: \\ $$$${sin}^{\mathrm{12}} {x}+{cos}^{\mathrm{12}} {x}=\frac{\mathrm{11}}{\mathrm{36}}−\frac{\mathrm{7}}{\mathrm{18}×\mathrm{6}} \\ $$$${sin}^{\mathrm{12}} {x}+{cos}^{\mathrm{12}} {x}=\frac{\mathrm{13}}{\mathrm{54}} \\ $$$${m}+{n}=\mathrm{67} \\ $$$${when}\:{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${then}\:…. \\ $$$$\bigstar\:{a}+{b}=\mathrm{1} \\ $$$$\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}−\mathrm{2}{ab} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\:\:\:=\mathrm{1}−\mathrm{4}{ab}+\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$
Commented by cherokeesay last updated on 17/Jan/26
so nice ! thank you sir.
$${so}\:{nice}\:!\:{thank}\:{you}\:{sir}. \\ $$
Answered by Ghisom_ last updated on 17/Jan/26
c^(10) +s^(10) =c^(10) +(1−c^2 )^5 =((11)/(36)) c^8 −2c^6 +2c^4 −c^2 +(5/(36))=0 (c^2 −(1/2))^4 +(1/2)(c^2 −(1/2))^2 −(7/(144))=0 c^2 −(1/2)=±((√3)/6) c^2 −(1/2)=±((√(21))/6)i c^2 =(1/2)±((√3)/6) c^2 =(1/2)±((√(21))/6)i c^(12) +s^(12) =c^(12) +(1−c^2 )^6 =(c^2 )^6 +(1−c^2 )^6 cos^(12) x +sin^(12) x = { ((13/54)),((59/54)) :} m+n=67∨m+n=113
$${c}^{\mathrm{10}} +{s}^{\mathrm{10}} ={c}^{\mathrm{10}} +\left(\mathrm{1}−{c}^{\mathrm{2}} \right)^{\mathrm{5}} =\frac{\mathrm{11}}{\mathrm{36}} \\ $$$${c}^{\mathrm{8}} −\mathrm{2}{c}^{\mathrm{6}} +\mathrm{2}{c}^{\mathrm{4}} −{c}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{36}}=\mathrm{0} \\ $$$$\left({c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{2}}\left({c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{144}}=\mathrm{0} \\ $$$${c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\:\:\:\:\:{c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}=\pm\frac{\sqrt{\mathrm{21}}}{\mathrm{6}}\mathrm{i} \\ $$$${c}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\:\:\:\:\:{c}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{21}}}{\mathrm{6}}\mathrm{i} \\ $$$$ \\ $$$${c}^{\mathrm{12}} +{s}^{\mathrm{12}} ={c}^{\mathrm{12}} +\left(\mathrm{1}−{c}^{\mathrm{2}} \right)^{\mathrm{6}} =\left({c}^{\mathrm{2}} \right)^{\mathrm{6}} +\left(\mathrm{1}−{c}^{\mathrm{2}} \right)^{\mathrm{6}} \\ $$$$\mathrm{cos}^{\mathrm{12}} \:{x}\:+\mathrm{sin}^{\mathrm{12}} \:{x}\:=\begin{cases}{\mathrm{13}/\mathrm{54}}\\{\mathrm{59}/\mathrm{54}}\end{cases} \\ $$$${m}+{n}=\mathrm{67}\vee{m}+{n}=\mathrm{113} \\ $$
Answered by Spillover last updated on 17/Jan/26
Answered by Spillover last updated on 17/Jan/26
Answered by Spillover last updated on 17/Jan/26

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