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Question-62112

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Question Number 62112 by aliesam last updated on 15/Jun/19
Answered by MJS last updated on 15/Jun/19
sin^(10) x +cos^(10) x =((11)/(36)) (5/(128))cos 8x +((15)/(32))cos 4x +((63)/(128))=((11)/(36)) cos 8x +12cos 4x +((43)/9)=0 x=(1/2)arctan t −((56t^4 −32t^2 −160)/(9(t^2 +1)^2 ))=0 (((t^2 −2)(7t^2 +10))/((t^2 +1)^2 ))=0 ⇒ t=±(√2) sin^(12) x +cos^(12) x =(1/(1024))cos 12x +((33)/(512))cos 8x +((495)/(1024))cos 4x +((231)/(512))= =(((t^2 +2)(t^4 +16t^2 +16))/(32(t^2 +1)^3 ))=((13)/(54)) m+n=67
$$\mathrm{sin}^{\mathrm{10}} \:{x}\:+\mathrm{cos}^{\mathrm{10}} \:{x}\:=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\frac{\mathrm{5}}{\mathrm{128}}\mathrm{cos}\:\mathrm{8}{x}\:+\frac{\mathrm{15}}{\mathrm{32}}\mathrm{cos}\:\mathrm{4}{x}\:+\frac{\mathrm{63}}{\mathrm{128}}=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\mathrm{cos}\:\mathrm{8}{x}\:+\mathrm{12cos}\:\mathrm{4}{x}\:+\frac{\mathrm{43}}{\mathrm{9}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:{t} \\ $$$$−\frac{\mathrm{56}{t}^{\mathrm{4}} −\mathrm{32}{t}^{\mathrm{2}} −\mathrm{160}}{\mathrm{9}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\frac{\left({t}^{\mathrm{2}} −\mathrm{2}\right)\left(\mathrm{7}{t}^{\mathrm{2}} +\mathrm{10}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:{t}=\pm\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{sin}^{\mathrm{12}} \:{x}\:+\mathrm{cos}^{\mathrm{12}} \:{x}\:=\frac{\mathrm{1}}{\mathrm{1024}}\mathrm{cos}\:\mathrm{12}{x}\:+\frac{\mathrm{33}}{\mathrm{512}}\mathrm{cos}\:\mathrm{8}{x}\:+\frac{\mathrm{495}}{\mathrm{1024}}\mathrm{cos}\:\mathrm{4}{x}\:+\frac{\mathrm{231}}{\mathrm{512}}= \\ $$$$=\frac{\left({t}^{\mathrm{2}} +\mathrm{2}\right)\left({t}^{\mathrm{4}} +\mathrm{16}{t}^{\mathrm{2}} +\mathrm{16}\right)}{\mathrm{32}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{13}}{\mathrm{54}} \\ $$$${m}+{n}=\mathrm{67} \\ $$
Commented by MJS last updated on 15/Jun/19
please study the transformation formulas of the trigonometric functions. I′m too busy to type them all
$$\mathrm{please}\:\mathrm{study}\:\mathrm{the}\:\mathrm{transformation}\:\mathrm{formulas} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{functions}.\:\mathrm{I}'\mathrm{m}\:\mathrm{too}\:\mathrm{busy} \\ $$$$\mathrm{to}\:\mathrm{type}\:\mathrm{them}\:\mathrm{all} \\ $$
Commented by aliesam last updated on 15/Jun/19
ok i will and thank you sir
$${ok}\:{i}\:{will}\:{and}\:{thank}\:{you}\:{sir} \\ $$
Answered by mr W last updated on 15/Jun/19
sin^(10) x+cos^(10) x=((11)/(36)) (sin^2 x)^5 +(cos^2 x)^5 =((11)/(36)) (sin^2 x+cos^2 x)(sin^8 x−sin^6 x cos^2 x+sin^4 x cos^4 x−sin^2 x cos^6 x+cos^8 x)=((11)/(36)) (sin^4 x+cos^4 x)^2 −sin^6 x cos^2 x−sin^4 x cos^4 x−sin^2 x cos^6 x=((11)/(36)) (1−2 sin^2 x cos^2 x)^2 −sin^2 x cos^2 x(1−2 sin^2 x cos^2 x)−sin^4 x cos^4 x=((11)/(36)) let s=sin^2 x cos^2 x=((sin^2 2x)/4)≤(1/4) (1−2s)^2 −s(1−2s)−s^2 =((11)/(36)) ⇒s^2 −s+(5/(36))=0 ⇒s=(1/2)(1±(2/3))=(1/6), (5/6) ⇒s=(1/6)<(1/4) sin^(12) x+cos^(12) x =(sin^4 x)^3 +(cos^4 x)^3 =(sin^4 x+cos^4 x)(sin^8 x−sin^4 x cos^4 x+cos^8 x) =(1−2 sin^2 x cos^2 x)[(sin^4 x+cos^4 x)^2 −3 sin^4 x cos^4 x] =(1−2 sin^2 x cos^2 x)[(1−2 sin^2 x cos^2 x)^2 −3 sin^4 x cos^4 x] =(1−2s)[(1−2s)^2 −3s^2 ] =(1−(1/3))[(1−(1/3))^2 −3×(1/6^2 )] =(2/3)[(4/9)−(1/(12))] =((13)/(54))=(m/n) ⇒m+n=13+54=67 generally: if sin^(10) x+cos^(10) x=a ((1/(16))≤a≤1) then sin^(12) x+cos^(12) x=(1/2)(√((1+4a)/5))(((2a−7)/5)+3(√((1+4a)/5)))
$$\mathrm{sin}^{\mathrm{10}} \:{x}+\mathrm{cos}^{\mathrm{10}} \:{x}=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\left(\mathrm{sin}^{\mathrm{2}} \:{x}\right)^{\mathrm{5}} +\left(\mathrm{cos}^{\mathrm{2}} \:{x}\right)^{\mathrm{5}} =\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\left(\mathrm{sin}^{\mathrm{2}} \:{x}+\mathrm{cos}^{\mathrm{2}} \:{x}\right)\left(\mathrm{sin}^{\mathrm{8}} \:{x}−\mathrm{sin}^{\mathrm{6}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{sin}^{\mathrm{4}} \:{x}\:\mathrm{cos}^{\mathrm{4}} \:{x}−\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{6}} \:{x}+\mathrm{cos}^{\mathrm{8}} \:{x}\right)=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\left(\mathrm{sin}^{\mathrm{4}} \:{x}+\mathrm{cos}^{\mathrm{4}} \:{x}\right)^{\mathrm{2}} −\mathrm{sin}^{\mathrm{6}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{sin}^{\mathrm{4}} \:{x}\:\mathrm{cos}^{\mathrm{4}} \:{x}−\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{6}} \:{x}=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\right)−\mathrm{sin}^{\mathrm{4}} \:{x}\:\mathrm{cos}^{\mathrm{4}} \:{x}=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$${let}\:{s}=\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}=\frac{\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}}{\mathrm{4}}\leqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left(\mathrm{1}−\mathrm{2}{s}\right)^{\mathrm{2}} −{s}\left(\mathrm{1}−\mathrm{2}{s}\right)−{s}^{\mathrm{2}} =\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\Rightarrow{s}^{\mathrm{2}} −{s}+\frac{\mathrm{5}}{\mathrm{36}}=\mathrm{0} \\ $$$$\Rightarrow{s}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\pm\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{6}},\:\frac{\mathrm{5}}{\mathrm{6}}\:\Rightarrow{s}=\frac{\mathrm{1}}{\mathrm{6}}<\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{sin}^{\mathrm{12}} \:{x}+\mathrm{cos}^{\mathrm{12}} \:{x} \\ $$$$=\left(\mathrm{sin}^{\mathrm{4}} \:{x}\right)^{\mathrm{3}} +\left(\mathrm{cos}^{\mathrm{4}} \:{x}\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{sin}^{\mathrm{4}} \:{x}+\mathrm{cos}^{\mathrm{4}} \:{x}\right)\left(\mathrm{sin}^{\mathrm{8}} \:{x}−\mathrm{sin}^{\mathrm{4}} \:{x}\:\mathrm{cos}^{\mathrm{4}} \:{x}+\mathrm{cos}^{\mathrm{8}} \:{x}\right) \\ $$$$=\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\right)\left[\left(\mathrm{sin}^{\mathrm{4}} \:{x}+\mathrm{cos}^{\mathrm{4}} \:{x}\right)^{\mathrm{2}} −\mathrm{3}\:\mathrm{sin}^{\mathrm{4}} \:{x}\:\mathrm{cos}^{\mathrm{4}} \:{x}\right] \\ $$$$=\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\right)\left[\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} −\mathrm{3}\:\mathrm{sin}^{\mathrm{4}} \:{x}\:\mathrm{cos}^{\mathrm{4}} \:{x}\right] \\ $$$$=\left(\mathrm{1}−\mathrm{2}{s}\right)\left[\left(\mathrm{1}−\mathrm{2}{s}\right)^{\mathrm{2}} −\mathrm{3}{s}^{\mathrm{2}} \right] \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left[\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{3}×\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }\right] \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left[\frac{\mathrm{4}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{12}}\right] \\ $$$$=\frac{\mathrm{13}}{\mathrm{54}}=\frac{{m}}{{n}} \\ $$$$\Rightarrow{m}+{n}=\mathrm{13}+\mathrm{54}=\mathrm{67} \\ $$$$ \\ $$$${generally}: \\ $$$${if}\:\mathrm{sin}^{\mathrm{10}} \:{x}+\mathrm{cos}^{\mathrm{10}} \:{x}={a}\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{16}}\leqslant{a}\leqslant\mathrm{1}\right) \\ $$$${then} \\ $$$$\mathrm{sin}^{\mathrm{12}} \:{x}+\mathrm{cos}^{\mathrm{12}} \:{x}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}+\mathrm{4}{a}}{\mathrm{5}}}\left(\frac{\mathrm{2}{a}−\mathrm{7}}{\mathrm{5}}+\mathrm{3}\sqrt{\frac{\mathrm{1}+\mathrm{4}{a}}{\mathrm{5}}}\right) \\ $$

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