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Question Number 132006 by bramlexs22 last updated on 10/Feb/21

$$\mathrm{Given}\{\begin{array}{l}\sqrt{2}\cos \mathrm{A}=\cos \mathrm{B}+\cos {}^3\mathrm{B}\\ \sqrt{2}\sin \mathrm{A}=\sin \mathrm{B}-\sin {}^3\mathrm{B}\end{array}$$$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}\sin (\mathrm{A}-\mathrm{B}).$$

Answered by EDWIN88 last updated on 10/Feb/21

$$\{\begin{array}{l}2\cos {}^2\mathrm{A}=\cos {}^2\mathrm{B}+2\cos {}^4\mathrm{B}+\cos {}^6\mathrm{B}\\ 2\sin {}^2\mathrm{A}=\sin {}^2\mathrm{B}-2\sin {}^4\mathrm{B}+\sin {}^6\mathrm{B}\end{array}$$$$\left(1\right)+\left(2\right)\Rightarrow 2=1+2\cos 2\mathrm{B}+\cos {}^6\mathrm{B}+\sin {}^6\mathrm{B}$$$$\iff 1=2\cos 2\mathrm{B}+{\left(\cos {}^2\mathrm{B}\right)}^3+{\left(\sin {}^2\mathrm{B}\right)}^3$$$$\iff 1=2\cos 2\mathrm{B}+1-3\sin {}^2\mathrm{Bcos}{}^2\mathrm{B}$$$$\iff 0=2\cos 2\mathrm{B}-\frac{3}{4}{\left(2\sin \mathrm{Bcos}\mathrm{B}\right)}^2$$$$\iff 0=8\cos 2\mathrm{B}-3\sin {}^{2}2\mathrm{B}$$$$\iff 0=8\cos 2\mathrm{B}-3(1-\cos {}^{2}2\mathrm{B})$$$$\iff 3\cos {}^{2}2\mathrm{B}+8\cos 2\mathrm{B}-3=0$$$$(3\cos 2\mathrm{B}-1)(\cos 2\mathrm{B}+3)=0$$$$\cos 2\mathrm{B}=\frac{1}{3}\Rightarrow 2\cos {}^2\mathrm{B}=\frac{4}{3};\cos {}^2\mathrm{B}=\frac{2}{3}$$$$\mathrm{then}\sqrt{2}\cos \mathrm{A}=\cos \mathrm{B}(1+\cos {}^2\mathrm{B})$$$$\sqrt{2}\cos \mathrm{A}=\frac{\sqrt{2}}{\sqrt{3}}\left(\frac{5}{3}\right)\Rightarrow \cos \mathrm{A}=\frac{5}{3\sqrt{3}}\wedge \sin \mathrm{A}=\frac{\sqrt{2}}{3\sqrt{3}}$$$$\sin (\mathrm{A}-\mathrm{B})=\sin \mathrm{Acos}\mathrm{B}-\cos \mathrm{Asin}\mathrm{B}$$$$=\frac{\sqrt{2}}{3\sqrt{3}}\frac{\sqrt{2}}{\sqrt{3}}-\frac{5}{3\sqrt{3}}\frac{1}{\sqrt{3}}=-\frac{3}{3\times 3}=-\frac{1}{3}$$

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