Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

Commented by Frix last updated on 22/May/24

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

Answered by Berbere last updated on 22/May/24

$$cos\left(20\right)=a;cos(3.20)=\frac{1}{2}$$$$\Rightarrow 4a^3-3a=\frac{1}{2}\Rightarrow 8a^3-6a-1=0$$$$a^3=\frac{3a}{4}+\frac{1}{8}$$$$\iff 4{(2a^2-1)}^2-\frac{1}{a}=4(4a^4-4a^2+1)-\frac{1}{a}$$$$=a\left(16a^3\right)-16a^2+4-\frac{1}{a}=a(12a+2)-16a^2+4-\frac{1}{a}$$$$=-4a^2+2a+4-\frac{1}{a}=\frac{-4a^3+2a^2+4a-1}{a}$$$$=\frac{2a^2+a-\frac{3}{2}}{a}=\frac{4a^2-3+2a}{2a}t=\frac{4a^3-3a+2a^2}{2a^2}=1+\frac{1}{4a^2}$$$$cos\left(\frac{\pi }{9}\right)=\frac{1}{2}(e^{\frac{i\pi }{9}}+e^{-\frac{i\pi }{9}})=\frac{1}{2}(\sqrt[3]{(\frac{1}{2}+\frac{i\sqrt{3}}{2})}+\sqrt[3]{\frac{1}{2}-\frac{i\sqrt{3}}{2}})$$$$4{cos}^2\left(40\right)-\frac{1}{cos\left(20\right)}=1-\frac{1}{\sqrt[3]{4}.{(\sqrt[3]{1+i\sqrt{3}}+\sqrt[3]{1-i\sqrt{3}})}^2}$$$$$$$$$$

Commented by Frix last updated on 22/May/24

$$\mathrm{I}\mathrm{think}\mathrm{this}\mathrm{is}\mathrm{wrong}.\mathrm{Your}\mathrm{solution}\mathrm{gives}$$$$1.2660...\mathrm{but}{4\cos }^{2}40°-\frac{1}{\cos 20°}\approx 1.2831$$

Commented by Berbere last updated on 22/May/24

$$thereisasquareyesthankyou$$

Commented by Frix last updated on 22/May/24

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

Commented by Berbere last updated on 23/May/24

$$nicesolutionsir$$

Commented by hardmath last updated on 24/May/24

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com