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Question Number 162649 by tounghoungko last updated on 31/Dec/21

$$\sin {}^{10}\left(x\right)+\cos {}^{10}\left(x\right)=\frac{11}{36}$$$$\sin {}^{12}\left(x\right)+\cos {}^{12}\left(x\right)=?$$

Answered by Ar Brandon last updated on 31/Dec/21

$${\mathrm{S}}^{12}+{\mathrm{C}}^{12}={\mathrm{S}}^{10}(1-{\mathrm{C}}^2)+{\mathrm{C}}^{10}(1-{\mathrm{S}}^2)$$$$={\mathrm{S}}^{10}+{\mathrm{C}}^{10}-{\mathrm{S}}^{10}{\mathrm{C}}^2-{\mathrm{C}}^{10}{\mathrm{S}}^2$$$$={\mathrm{S}}^{10}+{\mathrm{C}}^{10}-{\mathrm{S}}^2{\mathrm{C}}^2({\mathrm{S}}^8+{\mathrm{C}}^8)$$$$={\mathrm{S}}^{10}+{\mathrm{C}}^{10}-{\mathrm{S}}^2{\mathrm{C}}^2[{({\mathrm{S}}^4+{\mathrm{C}}^4)}^2-{2\mathrm{S}}^4{\mathrm{C}}^4]$$$$={\mathrm{S}}^{10}+{\mathrm{C}}^{10}-{\mathrm{S}}^2{\mathrm{C}}^2[{({({\mathrm{S}}^2+{\mathrm{C}}^2)}^2-{2\mathrm{S}}^2{\mathrm{C}}^2)}^2-{2\mathrm{S}}^4{\mathrm{C}}^4]$$$$={\mathrm{S}}^{10}+{\mathrm{C}}^{10}-{\mathrm{S}}^2{\mathrm{C}}^2[1-{4\mathrm{S}}^2{\mathrm{C}}^2+{2\mathrm{S}}^4{\mathrm{C}}^4]$$$$=...$$

Answered by mindispower last updated on 31/Dec/21

$$c^2+s^2=1$$$$c^{10}+s^{10}=\frac{11}{36}$$$${(c^2+s^2)}^5=c^{10}+s^{10}+5c^8{s}^2+5c^2{s}^8+10c^4{s}^6+10s^4{c}^6$$$$1=\frac{11}{36}+5c^2{s}^2(c^6+s^6)+10c^4{s}^4....\left(E\right)$$$$c^6+s^6=1-3s^2{c}^2,X={\left(sc\right)}^2..\left(E\right)\iff$$$$\frac{25}{36}=5X(1-3X)+10X^2$$$$5X^2-5X+\frac{25}{36}=0$$$$X=25-\frac{125}{9}=\frac{100}{9}$$$$\frac{5+\frac{10}{3}}{10}2p=\frac{25}{30}>\frac{1}{4},\frac{5-\frac{10}{3}}{10}=\frac{5}{30}=\frac{1}{6}<\frac{1}{4}$$$${\left(sc\right)}^2=\frac{{sin}^2\left(2x\right)}{4}⩽\frac{1}{4}$$$${\left(sc\right)}^2=\frac{1}{6}$$$${(s^2+c^2)}^6=S^{12}+C^{12}+6S^2{C}^2(S^8+C^8)+15S^4{C}^4(S^4+C^4)$$$$+20C^6{S}^6$$$$S^8+C^8={(s^2+c^2)}^4-4s^2{c}^2(s^4+c^4)-6s^4{s}^4$$$$=1-\frac{2}{3}.\frac{2}{3}-\frac{1}{6}=\frac{7}{18}$$$$(S^4+C^4)={(s^2+c^2)}^2-2c^2{s}^2=1-\frac{1}{3}=\frac{2}{3}$$$$c^{12}+s^{12}=1-\frac{7}{18}-\frac{15}{36}.\frac{2}{3}-20.\frac{1}{216}$$$$tchekcalculusihaveproblemeswithemy$$$$phonehorribletousetactil$$$$$$$$$$$$$$

Answered by MJS_new last updated on 31/Dec/21

$$s^{10}+c^{10}-\frac{11}{36}=0$$$$[c=\sqrt{1-s^2}]$$$$5s^8-10s^6+10s^4-5s^2+\frac{25}{36}=0$$$${\left(s^2\right)}^4-2{\left(s^2\right)}^3+2{\left(s^2\right)}^2-\left(s^2\right)+\frac{5}{36}=0$$$$({\left(s^2\right)}^2-\left(s^2\right)+\frac{1}{6})({\left(s^2\right)}^2-\left(s^2\right)+\frac{5}{6})=0$$$$\left(s^2\right)=\frac{1}{2}\pm \frac{\sqrt{3}}{6}$$$${\left(s^2\right)}^6+{\left(\sqrt{1-\left(s^2\right)}\right)}^{12}=\frac{13}{54}$$

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