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Question Number 62112 by aliesam last updated on 15/Jun/19

Answered by MJS last updated on 15/Jun/19

$${\sin }^{10}x+{\cos }^{10}x=\frac{11}{36}$$$$\frac{5}{128}\cos 8x+\frac{15}{32}\cos 4x+\frac{63}{128}=\frac{11}{36}$$$$\cos 8x+12\cos 4x+\frac{43}{9}=0$$$$x=\frac{1}{2}\arctan t$$$$-\frac{56t^4-32t^2-160}{9{(t^2+1)}^2}=0$$$$\frac{(t^2-2)(7t^2+10)}{{(t^2+1)}^2}=0$$$$\Rightarrow t=\pm \sqrt{2}$$$$$$$${\sin }^{12}x+{\cos }^{12}x=\frac{1}{1024}\cos 12x+\frac{33}{512}\cos 8x+\frac{495}{1024}\cos 4x+\frac{231}{512}=$$$$=\frac{(t^2+2)(t^4+16t^2+16)}{32{(t^2+1)}^3}=\frac{13}{54}$$$$m+n=67$$

Commented by MJS last updated on 15/Jun/19

$$\mathrm{please}\mathrm{study}\mathrm{the}\mathrm{transformation}\mathrm{formulas}$$$$\mathrm{of}\mathrm{the}\mathrm{trigonometric}\mathrm{functions}.{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{too}\mathrm{busy}$$$$\mathrm{to}\mathrm{type}\mathrm{them}\mathrm{all}$$

Commented by aliesam last updated on 15/Jun/19

$$okiwillandthankyousir$$

Answered by mr W last updated on 15/Jun/19

$${\sin }^{10}x+{\cos }^{10}x=\frac{11}{36}$$$${\left({\sin }^{2}x\right)}^5+{\left({\cos }^{2}x\right)}^5=\frac{11}{36}$$$$({\sin }^{2}x+{\cos }^{2}x)({\sin }^{8}x-{\sin }^{6}x{\cos }^{2}x+{\sin }^{4}x{\cos }^{4}x-{\sin }^{2}x{\cos }^{6}x+{\cos }^{8}x)=\frac{11}{36}$$$${({\sin }^{4}x+{\cos }^{4}x)}^2-{\sin }^{6}x{\cos }^{2}x-{\sin }^{4}x{\cos }^{4}x-{\sin }^{2}x{\cos }^{6}x=\frac{11}{36}$$$${(1-2{\sin }^{2}x{\cos }^{2}x)}^2-{\sin }^{2}x{\cos }^{2}x(1-2{\sin }^{2}x{\cos }^{2}x)-{\sin }^{4}x{\cos }^{4}x=\frac{11}{36}$$$$lets={\sin }^{2}x{\cos }^{2}x=\frac{{\sin }^{2}2x}{4}⩽\frac{1}{4}$$$${(1-2s)}^2-s(1-2s)-s^2=\frac{11}{36}$$$$\Rightarrow s^2-s+\frac{5}{36}=0$$$$\Rightarrow s=\frac{1}{2}(1\pm \frac{2}{3})=\frac{1}{6},\frac{5}{6}\Rightarrow s=\frac{1}{6}<\frac{1}{4}$$$$$$$${\sin }^{12}x+{\cos }^{12}x$$$$={\left({\sin }^{4}x\right)}^3+{\left({\cos }^{4}x\right)}^3$$$$=({\sin }^{4}x+{\cos }^{4}x)({\sin }^{8}x-{\sin }^{4}x{\cos }^{4}x+{\cos }^{8}x)$$$$=(1-2{\sin }^{2}x{\cos }^{2}x)[{({\sin }^{4}x+{\cos }^{4}x)}^2-3{\sin }^{4}x{\cos }^{4}x]$$$$=(1-2{\sin }^{2}x{\cos }^{2}x)[{(1-2{\sin }^{2}x{\cos }^{2}x)}^2-3{\sin }^{4}x{\cos }^{4}x]$$$$=(1-2s)[{(1-2s)}^2-3s^2]$$$$=(1-\frac{1}{3})[{(1-\frac{1}{3})}^2-3\times \frac{1}{6^2}]$$$$=\frac{2}{3}[\frac{4}{9}-\frac{1}{12}]$$$$=\frac{13}{54}=\frac{m}{n}$$$$\Rightarrow m+n=13+54=67$$$$$$$$generally:$$$$if{\sin }^{10}x+{\cos }^{10}x=a(\frac{1}{16}⩽a⩽1)$$$$then$$$${\sin }^{12}x+{\cos }^{12}x=\frac{1}{2}\sqrt{\frac{1+4a}{5}}(\frac{2a-7}{5}+3\sqrt{\frac{1+4a}{5}})$$

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