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Question Number 136094 by mnjuly1970 last updated on 19/Mar/21

$$....nicecalculus....$$$$inA\stackrel{\mathrm{\Delta }}{BC}::cos\left(A\right)cos\left(B\right)cos\left(C\right)=\frac{1}{9}$$$$findthevalueof:$$$${cos}^2\left(A\right)+{cos}^2\left(B\right)+{cos}^2\left(C\right)=?$$

Commented by MJS_new last updated on 19/Mar/21

$$\cos \alpha \cos \beta \cos \gamma ⩽\frac{1}{8}\Rightarrow \mathrm{no}\mathrm{such}\mathrm{triangle}\mathrm{exists}$$

Commented by mr W last updated on 19/Mar/21

$$justbecause\cos A\cos B\cos C=\frac{1}{9}<\frac{1}{8},$$$$itshouldbepossible,ithink.$$$$$$$${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1,$$$$withA,B,C=({\cos }^{-1}\frac{1}{\sqrt{3}},{\cos }^{-1}\frac{1}{\sqrt{3}},{\cos }^{-1}\frac{1}{3})\equiv (54.735°,54.735°,70.529°)$$$$orA,B,C=({\cos }^{-1}\frac{1}{\sqrt{6}},{\cos }^{-1}\frac{1}{\sqrt{6}},{\cos }^{-1}\frac{2}{3})\equiv (65.905°,65.905°,48.190°)$$

Commented by MJS_new last updated on 19/Mar/21

$$\mathrm{the}\mathrm{constant}\mathrm{had}\mathrm{been}\frac{3}{8},\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{different}$$$$\mathrm{question}$$

Commented by mr W last updated on 19/Mar/21

$$ok.i{didn}^{\prime }tknowthat.$$

Commented by MJS_new last updated on 19/Mar/21

$$2{\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\frac{1}{3}\right)}^2=\frac{7}{9}\ne 1$$$$2{\left(\frac{1}{\sqrt{6}}\right)}^2+{\left(\frac{2}{3}\right)}^2=\frac{7}{9}\ne 1$$

Answered by MJS_new last updated on 19/Mar/21

$$\cos \alpha \cos \beta \cos (\pi -\alpha -\beta )=\frac{1}{9}$$$$-\cos \alpha \cos \beta \cos (\alpha +\beta )=\frac{1}{9}$$$$\sin \alpha \cos \alpha \sin \beta \cos \beta -{\cos }^2\alpha {\cos }^2\beta =\frac{1}{9}$$$$-\frac{1-\tan \alpha \tan \beta }{(1+{\tan }^2\alpha )(1+{\tan }^2\beta )}=\frac{1}{9}$$$$$$$${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2(\pi -\alpha -\beta )=$$$$={\cos }^2\alpha +{\cos }^2\beta +{\cos }^2(\alpha +\beta )=...$$$$...=1+\frac{2(1-\tan \alpha \tan \beta )}{(1+{\tan }^2\alpha )(1+{\tan }^2\beta )}=$$$$=1+2\times (-\frac{1}{9})=\frac{7}{9}$$

Commented by mr W last updated on 19/Mar/21

$$butigot\mathrm{\Sigma }{\cos }^{2}A=1withthegiven$$$$examplesabove.doesitmean$$$$that\mathrm{\Sigma }{\cos }^{2}A\ne constant?$$

Commented by MJS_new last updated on 19/Mar/21

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{constant}\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\frac{7}{9}\mathrm{instead}\mathrm{of}1$$

Commented by mr W last updated on 19/Mar/21

$$isee.imadeasimplemistake.$$$$thanksalot!$$

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