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Question Number 156126 by VIDDD last updated on 08/Oct/21

cos(π/5)=...? with solution pls

$$\:\:\:\mathrm{cos}\frac{\pi}{\mathrm{5}}=...?\:\:\mathrm{with}\:\mathrm{solution}\:\mathrm{pls} \\ $$

Commented by cortano last updated on 08/Oct/21

(π/5)=36° ⇒cos 72°=2cos 36°−1 ⇒cos 36°=(√((1+cos 72°)/2))=(√((1+sin 18°)/2))

$$\frac{\pi}{\mathrm{5}}=\mathrm{36}°\:\Rightarrow\mathrm{cos}\:\mathrm{72}°=\mathrm{2cos}\:\mathrm{36}°−\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{36}°=\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{72}°}{\mathrm{2}}}=\sqrt{\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{18}°}{\mathrm{2}}} \\ $$

Answered by puissant last updated on 08/Oct/21

a=cos(π/5) ; b=cos((2π)/5) and −a=cos((4π)/5) We have cos((2π)/5)= 2cos^2 (π/5) − 1 ⇒ { ((b=2a^2 −1)),((−a=2b^2 −1)) :} ⇒ a+b=2(a^2 −b^2 ) ⇒ a+b=2(a+b)(a−b) ⇒ 1=a−b ⇒ b=a−(1/2).. ⇒ a−(1/2) = 2a^2 −1 ⇒ 4a^2 −2a−1=0 Δ=4−4(−1)(4)=20 → (√Δ)=2(√5).. a_1 =((−2−2(√5))/8)=((−1−(√5))/4)<0 (impossible) a_2 =((2+2(√5))/8)=((1+(√5))/4)=cos(π/5) cos((2π)/5)=cos(π/5)−(1/2)=(((√5)−1)/4) ∴∵cos((π/5))=((1+(√5))/4) and cos(((2π)/5))=(((√5)−1)/4).. ..........Le puissant...........

$${a}={cos}\frac{\pi}{\mathrm{5}}\:\:;\:\:{b}={cos}\frac{\mathrm{2}\pi}{\mathrm{5}}\:\:{and}\:\:−{a}={cos}\frac{\mathrm{4}\pi}{\mathrm{5}} \\ $$$${We}\:{have}\:{cos}\frac{\mathrm{2}\pi}{\mathrm{5}}=\:\mathrm{2}{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{5}}\:−\:\mathrm{1} \\ $$$$\Rightarrow\begin{cases}{{b}=\mathrm{2}{a}^{\mathrm{2}} −\mathrm{1}}\\{−{a}=\mathrm{2}{b}^{\mathrm{2}} −\mathrm{1}}\end{cases}\:\:\Rightarrow\:\:{a}+{b}=\mathrm{2}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{a}+{b}=\mathrm{2}\left({a}+{b}\right)\left({a}−{b}\right)\: \\ $$$$\Rightarrow\:\mathrm{1}={a}−{b}\:\Rightarrow\:{b}={a}−\frac{\mathrm{1}}{\mathrm{2}}.. \\ $$$$\Rightarrow\:{a}−\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{2}{a}^{\mathrm{2}} −\mathrm{1}\: \\ $$$$\Rightarrow\:\mathrm{4}{a}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\mathrm{4}−\mathrm{4}\left(−\mathrm{1}\right)\left(\mathrm{4}\right)=\mathrm{20}\:\rightarrow\:\sqrt{\Delta}=\mathrm{2}\sqrt{\mathrm{5}}.. \\ $$$${a}_{\mathrm{1}} =\frac{−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}<\mathrm{0}\:\left({impossible}\right) \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}={cos}\frac{\pi}{\mathrm{5}} \\ $$$${cos}\frac{\mathrm{2}\pi}{\mathrm{5}}={cos}\frac{\pi}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$\therefore\because{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:\:{and}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:..........\mathscr{L}{e}\:{puissant}........... \\ $$

Commented by VIDDD last updated on 09/Oct/21

thanks

$$\:\:{thanks}\: \\ $$

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