Find-4-cos-2-40-1-cos-20-

Question Number 207641 by hardmath last updated on 21/May/24

$$\mathrm{Find}:\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}}\:\:=\:\:? \\ $$

Commented by Frix last updated on 22/May/24

4−2(√7)cos ((π+2sin^(−1) ((37(√7))/(98)))/6)

$$\mathrm{4}−\mathrm{2}\sqrt{\mathrm{7}}\mathrm{cos}\:\frac{\pi+\mathrm{2sin}^{−\mathrm{1}} \:\frac{\mathrm{37}\sqrt{\mathrm{7}}}{\mathrm{98}}}{\mathrm{6}} \\ $$

Answered by Berbere last updated on 22/May/24

cos(20)=a;cos(3.20)=(1/2) ⇒4a^3 −3a=(1/2)⇒8a^3 −6a−1=0 a^3 =((3a)/4)+(1/8) ⇔4(2a^2 −1)^2 −(1/a)=4(4a^4 −4a^2 +1)−(1/a) =a(16a^3 )−16a^2 +4−(1/a)=a(12a+2)−16a^2 +4−(1/a) =−4a^2 +2a+4−(1/a)=((−4a^3 +2a^2 +4a−1)/a) =((2a^2 +a−(3/2))/a)=((4a^2 −3+2a)/(2a))t=((4a^3 −3a+2a^2 )/(2a^2 ))=1+(1/(4a^2 )) cos((π/9))=(1/2)(e^((iπ)/9) +e^(−((iπ)/9)) )=(1/2)(((((1/2)+((i(√3))/2))))^(1/3) +(((1/2)−((i(√3))/2)))^(1/3) ) 4cos^2 (40)−(1/(cos(20)))=1−(1/( (4)^(1/3) .(((1+i(√3)))^(1/3) +((1−i(√3)))^(1/3) )^2 ))

$${cos}\left(\mathrm{20}\right)={a};{cos}\left(\mathrm{3}.\mathrm{20}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{3}} −\mathrm{3}{a}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{8}{a}^{\mathrm{3}} −\mathrm{6}{a}−\mathrm{1}=\mathrm{0} \\ $$$${a}^{\mathrm{3}} =\frac{\mathrm{3}{a}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Leftrightarrow\mathrm{4}\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{{a}}=\mathrm{4}\left(\mathrm{4}{a}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)−\frac{\mathrm{1}}{{a}} \\ $$$$={a}\left(\mathrm{16}{a}^{\mathrm{3}} \right)−\mathrm{16}{a}^{\mathrm{2}} +\mathrm{4}−\frac{\mathrm{1}}{{a}}={a}\left(\mathrm{12}{a}+\mathrm{2}\right)−\mathrm{16}{a}^{\mathrm{2}} +\mathrm{4}−\frac{\mathrm{1}}{{a}} \\ $$$$=−\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{4}−\frac{\mathrm{1}}{{a}}=\frac{−\mathrm{4}{a}^{\mathrm{3}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{4}{a}−\mathrm{1}}{{a}} \\ $$$$=\frac{\mathrm{2}{a}^{\mathrm{2}} +{a}−\frac{\mathrm{3}}{\mathrm{2}}}{{a}}=\frac{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{3}+\mathrm{2}{a}}{\mathrm{2}{a}}{t}=\frac{\mathrm{4}{a}^{\mathrm{3}} −\mathrm{3}{a}+\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${cos}\left(\frac{\pi}{\mathrm{9}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\frac{{i}\pi}{\mathrm{9}}} +{e}^{−\frac{{i}\pi}{\mathrm{9}}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}−\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}}\right) \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} \left(\mathrm{40}\right)−\frac{\mathrm{1}}{{cos}\left(\mathrm{20}\right)}=\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}.\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+{i}\sqrt{\mathrm{3}}}+\sqrt[{\mathrm{3}}]{\mathrm{1}−{i}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$

Commented by Frix last updated on 22/May/24

I think this is wrong. Your solution gives 1.2660... but 4cos^2 40° −(1/(cos 20°))≈1.2831

$$\mathrm{I}\:\mathrm{think}\:\mathrm{this}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{Your}\:\mathrm{solution}\:\mathrm{gives} \\ $$$$\mathrm{1}.\mathrm{2660}…\:\mathrm{but}\:\mathrm{4cos}^{\mathrm{2}} \:\mathrm{40}°\:−\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\approx\mathrm{1}.\mathrm{2831} \\ $$

Commented by Berbere last updated on 22/May/24

$${there}\:{is}\:{a}\:{square}\:{yes}\:{thank}\:{you} \\ $$

Commented by Frix last updated on 22/May/24

Look at my question 207434, I found an interesting equation...

$$\mathrm{Look}\:\mathrm{at}\:\mathrm{my}\:\mathrm{question}\:\mathrm{207434},\:\mathrm{I}\:\mathrm{found}\:\mathrm{an} \\ $$$$\mathrm{interesting}\:\mathrm{equation}… \\ $$

Commented by Berbere last updated on 23/May/24

$${nice}\:{solution}\:{sir} \\ $$

Commented by hardmath last updated on 24/May/24

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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