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Question Number 136094 by mnjuly1970 last updated on 19/Mar/21
.... nice calculus.... in AB^Δ C :: cos(A)cos(B)cos(C)=(1/9) find the value of: cos^2 (A)+cos^2 (B)+cos^2 (C)=?
$$\:\:\:\:\:\:\:\:\:\:\:….\:{nice}\:\:\:\:{calculus}…. \\ $$$$\:\:\:\:{in}\:{A}\overset{\Delta} {{B}C}\:::\:{cos}\left({A}\right){cos}\left({B}\right){cos}\left({C}\right)=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\:\:\:{find}\:\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}^{\mathrm{2}} \left({A}\right)+{cos}^{\mathrm{2}} \left({B}\right)+{cos}^{\mathrm{2}} \left({C}\right)=? \\ $$
Commented by MJS_new last updated on 19/Mar/21
cos α cos β cos γ ≤(1/8) ⇒ no such triangle exists
$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:\mathrm{cos}\:\gamma\:\leqslant\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow\:\mathrm{no}\:\mathrm{such}\:\mathrm{triangle}\:\mathrm{exists} \\ $$
Commented by mr W last updated on 19/Mar/21
just because cos A cos B cos C=(1/9)<(1/8), it should be possible, i think. cos^2 A+cos^2 B+cos^2 C=1, with A,B,C=(cos^(−1) (1/( (√3))), cos^(−1) (1/( (√3))), cos^(−1) (1/3))≡(54.735°,54.735°,70.529°) or A,B,C=(cos^(−1) (1/( (√6))), cos^(−1) (1/( (√6))), cos^(−1) (2/3))≡(65.905°,65.905°,48.190°)
$${just}\:{because}\:\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}=\frac{\mathrm{1}}{\mathrm{9}}<\frac{\mathrm{1}}{\mathrm{8}}, \\ $$$${it}\:{should}\:{be}\:{possible},\:{i}\:{think}. \\ $$$$ \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{A}+\mathrm{cos}^{\mathrm{2}} \:{B}+\mathrm{cos}^{\mathrm{2}} \:{C}=\mathrm{1}, \\ $$$${with}\:{A},{B},{C}=\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}},\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}},\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\right)\equiv\left(\mathrm{54}.\mathrm{735}°,\mathrm{54}.\mathrm{735}°,\mathrm{70}.\mathrm{529}°\right) \\ $$$${or}\:{A},{B},{C}=\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}},\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}},\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}\right)\equiv\left(\mathrm{65}.\mathrm{905}°,\mathrm{65}.\mathrm{905}°,\mathrm{48}.\mathrm{190}°\right) \\ $$
Commented by MJS_new last updated on 19/Mar/21
the constant had been (3/8), now it′s a different question
$$\mathrm{the}\:\mathrm{constant}\:\mathrm{had}\:\mathrm{been}\:\frac{\mathrm{3}}{\mathrm{8}},\:\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{different} \\ $$$$\mathrm{question} \\ $$
Commented by mr W last updated on 19/Mar/21
ok. i didn′t know that.
$${ok}.\:{i}\:{didn}'{t}\:{know}\:{that}. \\ $$
Commented by MJS_new last updated on 19/Mar/21
2((1/( (√3))))^2 +((1/3))^2 =(7/9)≠1 2((1/( (√6))))^2 +((2/3))^2 =(7/9)≠1
$$\mathrm{2}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{9}}\neq\mathrm{1} \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{9}}\neq\mathrm{1} \\ $$
Answered by MJS_new last updated on 19/Mar/21
cos α cos β cos (π−α−β) =(1/9) −cos α cos β cos (α+β) =(1/9) sin α cos α sin β cos β −cos^2 α cos^2 β =(1/9) −((1−tan α tan β)/((1+tan^2 α)(1+tan^2 β)))=(1/9) cos^2 α +cos^2 β +cos^2 (π−α−β) = =cos^2 α +cos^2 β +cos^2 (α+β) =... ...=1+((2(1−tan α tan β))/((1+tan^2 α)(1+tan^2 β)))= =1+2×(−(1/9))=(7/9)
$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:\mathrm{cos}\:\left(\pi−\alpha−\beta\right)\:=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$−\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:\mathrm{cos}\:\left(\alpha+\beta\right)\:=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\beta\:−\mathrm{cos}^{\mathrm{2}} \:\alpha\:\mathrm{cos}^{\mathrm{2}} \:\beta\:=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$−\frac{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta\right)}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$ \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\alpha\:+\mathrm{cos}^{\mathrm{2}} \:\beta\:+\mathrm{cos}^{\mathrm{2}} \:\left(\pi−\alpha−\beta\right)\:= \\ $$$$=\mathrm{cos}^{\mathrm{2}} \:\alpha\:+\mathrm{cos}^{\mathrm{2}} \:\beta\:+\mathrm{cos}^{\mathrm{2}} \:\left(\alpha+\beta\right)\:=… \\ $$$$…=\mathrm{1}+\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\right)}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\beta\right)}= \\ $$$$=\mathrm{1}+\mathrm{2}×\left(−\frac{\mathrm{1}}{\mathrm{9}}\right)=\frac{\mathrm{7}}{\mathrm{9}} \\ $$
Commented by mr W last updated on 19/Mar/21
but i got Σcos^2 A=1 with the given examples above. does it mean that Σcos^2 A≠constant?
$${but}\:{i}\:{got}\:\Sigma\mathrm{cos}^{\mathrm{2}} \:{A}=\mathrm{1}\:{with}\:{the}\:{given} \\ $$$${examples}\:{above}.\:{does}\:{it}\:{mean} \\ $$$${that}\:\Sigma\mathrm{cos}^{\mathrm{2}} \:{A}\neq{constant}? \\ $$
Commented by MJS_new last updated on 19/Mar/21
it′s constant but it′s (7/9) instead of 1
$$\mathrm{it}'\mathrm{s}\:\mathrm{constant}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\frac{\mathrm{7}}{\mathrm{9}}\:\mathrm{instead}\:\mathrm{of}\:\mathrm{1} \\ $$
Commented by mr W last updated on 19/Mar/21
i see. i made a simple mistake. thanks alot!
$${i}\:{see}.\:{i}\:{made}\:{a}\:{simple}\:{mistake}. \\ $$$${thanks}\:{alot}! \\ $$

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