cos(π/5)=...? with solution pls


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

$\cos \frac{π}{5} = . . . ? with solution pls$
Commented by cortano last updated on 08/Oct/21

$\frac{π}{5} = 36 ° \Rightarrow \cos 72 ° = 2 \cos 36 ° - 1$ $\Rightarrow \cos 36 ° = \sqrt{\frac{1 + \cos 72 °}{2}} = \sqrt{\frac{1 + \sin 18 °}{2}}$
	Answered by puissant last updated on 08/Oct/21
	$a=cos(π/5) ; b=cos((2π)/5) and −a=cos((4π)/5) We have cos((2π)/5)= 2cos^2 (π/5) − 1 ⇒ { ((b=2a^2 −1)),((−a=2b^2 −1)) :} ⇒ a+b=2(a^2 −b^2 ) ⇒ a+b=2(a+b)(a−b) ⇒ 1=a−b ⇒ b=a−(1/2).. ⇒ a−(1/2) = 2a^2 −1 ⇒ 4a^2 −2a−1=0 Δ=4−4(−1)(4)=20 → (√Δ)=2(√5).. a_1 =((−2−2(√5))/8)=((−1−(√5))/4)<0 (impossible) a_2 =((2+2(√5))/8)=((1+(√5))/4)=cos(π/5) cos((2π)/5)=cos(π/5)−(1/2)=(((√5)−1)/4) ∴∵cos((π/5))=((1+(√5))/4) and cos(((2π)/5))=(((√5)−1)/4).. ..........Le puissant...........$
	$a = c o s \frac{π}{5}; b = c o s \frac{2 π}{5} a n d - a = c o s \frac{4 π}{5}$ $W e h a v e c o s \frac{2 π}{5} = 2 {c o s}^{2} \frac{π}{5} - 1$ $\Rightarrow {\begin{cases} b = 2 a^{2} - 1 \\ - a = 2 b^{2} - 1 \end{cases} \Rightarrow a + b = 2 (a^{2} - b^{2})$ $\Rightarrow a + b = 2 (a + b) (a - b)$ $\Rightarrow 1 = a - b \Rightarrow b = a - \frac{1}{2} . .$ $\Rightarrow a - \frac{1}{2} = 2 a^{2} - 1$ $\Rightarrow 4 a^{2} - 2 a - 1 = 0$ $Δ = 4 - 4 (- 1) (4) = 20 \to \sqrt{Δ} = 2 \sqrt{5} . .$ $a_{1} = \frac{- 2 - 2 \sqrt{5}}{8} = \frac{- 1 - \sqrt{5}}{4} < 0 (i m p o s s i b l e)$ $a_{2} = \frac{2 + 2 \sqrt{5}}{8} = \frac{1 + \sqrt{5}}{4} = c o s \frac{π}{5}$ $c o s \frac{2 π}{5} = c o s \frac{π}{5} - \frac{1}{2} = \frac{\sqrt{5} - 1}{4}$ $∴∵ c o s (\frac{π}{5}) = \frac{1 + \sqrt{5}}{4} a n d c o s (\frac{2 π}{5}) = \frac{\sqrt{5} - 1}{4} . .$ $. . . . . . . . . . L e p u i s s a n t . . . . . . . . . . .$
	Commented by VIDDD last updated on 09/Oct/21

	$t h a n k s$
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