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Question Number 127944 by liberty last updated on 03/Jan/21

$$\mathrm{If}\sin \mathrm{x}+\cos \mathrm{x}=\frac{1}{3}\mathrm{then}\sin {}^3\mathrm{x}+\cos {}^3\mathrm{x}=?$$

Answered by mr W last updated on 03/Jan/21

$$p_n={\sin }^{n}x+{\cos }^{n}x$$$$p_1=\frac{1}{3}=e_1$$$$p_2=e_1^2-2e_2=\frac{1}{9}-2e_2=1\Rightarrow e_2=-\frac{4}{9}$$$$p_3=e_1^3-3e_1{e}_2=\frac{1}{27}+3\times \frac{1}{3}\times \frac{4}{9}=\frac{13}{27}$$$$i.e.\sin {}^3\mathrm{x}+\cos {}^3\mathrm{x}=\frac{13}{27}$$

Commented by bramlexs22 last updated on 04/Jan/21

$$\mathrm{if}{\mathrm{p}}_{\mathrm{n}}=\sin {}^{\mathrm{n}}\left(\mathrm{x}\right)-\cos {}^{\mathrm{n}}\left(\mathrm{x}\right)$$$$\mathrm{how}\mathrm{to}\mathrm{get}\sin {}^3\left(\mathrm{x}\right)+\cos {}^3\left(\mathrm{x}\right)\mathrm{sir}?$$

Commented by mr W last updated on 04/Jan/21

$$with{\mathrm{p}}_{\mathrm{n}}=\sin {}^{\mathrm{n}}\left(\mathrm{x}\right)-\cos {}^{\mathrm{n}}\left(\mathrm{x}\right)youget$$$$usingthismethodcorespondingly$$$$\sin {}^3\left(\mathrm{x}\right)-\cos {}^3\left(\mathrm{x}\right).$$

Answered by bramlexs22 last updated on 03/Jan/21

Commented by mr W last updated on 03/Jan/21

$$inquestion:\sin x+\cos x=\frac{1}{3},not$$$$\sin x-\cos x=\frac{1}{3}.$$

Commented by bramlexs22 last updated on 03/Jan/21

������ misread...

Commented by liberty last updated on 03/Jan/21

$$\mathrm{hahahaha}$$

Answered by mathmax by abdo last updated on 03/Jan/21

$${\sin }^3\mathrm{x}+{\cos }^3\mathrm{x}=(\mathrm{sinx}+\mathrm{cosx})({\sin }^2\mathrm{x}-\mathrm{sinx}\mathrm{cosx}+{\cos }^2\mathrm{x})$$$$=\frac{1}{3}(1-\mathrm{sinx}\mathrm{cosx})\mathrm{we}\mathrm{have}{(\mathrm{cosx}+\mathrm{sinx})}^2=\frac{1}{9}\Rightarrow$$$$1+2\mathrm{cosx}\mathrm{sinx}=\frac{1}{9}\Rightarrow 2\mathrm{cosx}\mathrm{sinx}=\frac{1}{9}-1=-\frac{8}{9}\Rightarrow \mathrm{cosx}\mathrm{sinx}=-\frac{4}{9}$$$$\Rightarrow {\cos }^3\mathrm{x}+{\sin }^3\mathrm{x}=\frac{1}{3}(1+\frac{4}{9})=\frac{1}{3}\times \frac{13}{9}=\frac{13}{27}$$

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