Math Question and Answers


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 104676 by qwertyu last updated on 23/Jul/20

Commented by Dwaipayan Shikari last updated on 23/Jul/20

$c o s 36 ° + c o s 108 ° = c o s 36 ° - c o s 72 °$ $= \frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4} = \frac{1}{2}$
Commented by OlafThorendsen last updated on 23/Jul/20

$36 ° and 72 ° are not$ $remarkable angles sir .$ $You should prove that$ $\cos 36 ° = \frac{\sqrt{5} + 1}{4} and \cos 72 ° = \frac{\sqrt{5} - 1}{4}$
Commented by Dwaipayan Shikari last updated on 23/Jul/20

$θ = 18 °$ $s i n 5 θ = 1$ $2 θ + 3 θ = \frac{π}{2}$ $s i n 2 θ = c o s 3 θ$ $2 s i n θ c o s θ = 4 {c o s}^{3} θ - 3 c o s θ$ $2 s i n θ = 4 {c o s}^{2} θ - 3$ $2 s i n θ = 1 - 4 {s i n}^{2} θ$ $4 {s i n}^{2} θ + 2 s i n θ - 1 = 0$ $s i n θ = \frac{- 2 + \sqrt{4 + 16}}{2.4} = \frac{\sqrt{5} - 1}{4}$ $s i n 18 ° = \frac{\sqrt{5} - 1}{4} = c o s 72 °$ $c o s 72 ° = {c o s}^{2} 36 ° - {s i n}^{2} 36 ° = 2 {c o s}^{2} 36 ° - 1$ $c o s 36 ° = \frac{\sqrt{5} + 1}{4}$
Commented by OlafThorendsen last updated on 23/Jul/20

$Waouh . Great method sir!$
	Answered by bemath last updated on 23/Jul/20

	$\cos a + \cos b = 2 \cos (\frac{a + b}{2}) \cos (\frac{a - b}{2})$ $\cos 36 ° + \cos 108 ° =$ $2 \cos 72 ° \cos 36 ° = 2 \sin 18 ° \cos 36 °$ $= 2 \sin 18 ° (1 - 2 \sin^{2} 18 °)$ $= 2 (\frac{\sqrt{5} - 1}{4}) (1 - 2 {(\frac{\sqrt{5} - 1}{4})}^{2})$ $= \frac{1}{2} (\sqrt{5} - 1) (1 - 2 (\frac{6 - 2 \sqrt{5}}{16}))$ $= \frac{1}{2} (\sqrt{5} - 1) (1 - (\frac{3 - \sqrt{5}}{4}))$ $= \frac{1}{8} (\sqrt{5} - 1) (\sqrt{5} + 1)$ $= \frac{1}{8} \times 4 = \frac{1}{2} ★$
	Commented by OlafThorendsen last updated on 23/Jul/20

	$you should explain why$ $\sin 18 ° is equal to \frac{\sqrt{5} - 1}{4} sir .$
	Commented by 6478 last updated on 23/Jul/20

	$\cos a + \cos b = 2 \cos (\frac{a + b}{2}) \cos (\frac{a - b}{2})$ $\cos 36 ° + \cos 108 ° =$ $2 \cos 72 ° \cos 36 ° = 2 \sin 18 ° \cos 36 °$ $= 2 \sin 18 ° (1 - 2 \sin^{2} 18 °)$ $= 2 (\frac{\sqrt{5} - 1}{4}) (1 - 2 {(\frac{\sqrt{5} - 1}{4})}^{2})$ $= \frac{1}{2} (\sqrt{5} - 1) (1 - 2 (\frac{6 - 2 \sqrt{5}}{16}))$ $= \frac{1}{2} (\sqrt{5} - 1) (1 - (\frac{3 - \sqrt{5}}{4}))$ $= \frac{1}{8} (\sqrt{5} - 1) (\sqrt{5} + 1)$ $= \frac{1}{8} \times 4 = \frac{1}{2} ★$
	Commented by bemath last updated on 23/Jul/20

	Answered by OlafThorendsen last updated on 23/Jul/20

	$a = \cos 36 °$ $b = \cos 108 °$ $a + b = \cos 36 ° + \cos 108 ° = \cos \frac{π}{5} + \cos \frac{3 π}{5}$ $a + b = \cos \frac{π}{5} - \cos \frac{2 π}{5}$ $a + b = \cos \frac{π}{5} - ({2 \cos}^{2} \frac{π}{5} - 1)$ $a + b = a - 2 a^{2} + 1 (1)$ $a + b = \cos \frac{π}{5} + \cos \frac{3 π}{5}$ $a + b = 2 \cos \frac{\frac{3 π}{5} + \frac{π}{5}}{2} \cos \frac{\frac{3 π}{5} - \frac{π}{5}}{2}$ $a + b = 2 \cos \frac{2 π}{5} \cos \frac{π}{5}$ $a + b = 2 (2 a^{2} - 1) a (2)$ $(1) and (2) :$ $2 (2 a^{2} - 1) a = a - 2 a^{2} + 1$ $4 a^{3} + 2 a^{2} - 3 a - 1 = 0$ $(a + 1) (4 a^{2} - 2 a - 1) = 0$ $a = \cos \frac{π}{5} = - 1 : impossible$ $4 a^{2} - 2 a - 1 = 0$ $a = \frac{1 \pm \sqrt{5}}{4}$ $a = \cos \frac{π}{5} = \frac{1 - \sqrt{5}}{4} < 0 : impossible$ $Finally a = \cos \frac{π}{5} = \frac{1 + \sqrt{5}}{4}$ $Then b = \cos \frac{3 π}{5} = - \cos \frac{2 π}{5} = - (2 a^{2} - 1)$ $b = 1 - 2 {(\frac{1 + \sqrt{5}}{4})}^{2} = \frac{1 - \sqrt{5}}{4}$ $a + b = \frac{1 + \sqrt{5}}{4} + \frac{1 - \sqrt{5}}{4} = \frac{1}{2}$
	Answered by 1549442205PVT last updated on 23/Jul/20

	$36 ° + 54 ° = 90 ° \Rightarrow \sin 36 ° = \cos 54 °$ $\Rightarrow 2 \sin 18 ° \cos 18 ° = \cos 3 . 18 ° = {4 \cos}^{3} 18 ° - 3 \cos 18 °$ $\Rightarrow 2 \sin 18 ° = {4 \cos}^{2} 18 ° - 3$ $\Rightarrow 2 \sin 18 ° = 4 (1 - \sin^{2} 18 °) - 3$ $\Leftrightarrow {4 \sin}^{2} 18 ° + 2 \sin 18 ° - 1 = 0$ $\Leftrightarrow {(2 \sin 18 ° + \frac{1}{2})}^{2} - \frac{5}{4} = 0 \Leftrightarrow 2 \sin 18 ° + \frac{1}{2} = \frac{\sqrt{5}}{2}$ $\Rightarrow \sin 18 ° = \frac{\sqrt{5} - 1}{4} \Rightarrow \cos 72 ° = \sin 18 ° = \frac{\sqrt{5} - 1}{4}$ $\cos 72 ° = {2 co}^{2} 36 ° - 1 = \frac{\sqrt{5} - 1}{4}$ $\Rightarrow \cos^{2} 36 ° = \frac{3 + \sqrt{5}}{8} = \frac{6 + 2 \sqrt{5}}{16} \Rightarrow \cos 36 ° = \frac{\sqrt{5} + 1}{4}$ $\cos 108 ° = \cos (90 ° + 18 °) = - \sin 18 ° = \frac{1 - \sqrt{5}}{4}$ $Therefore, \cos 36 ° + \cos 108 ° =$ $\frac{1 + \sqrt{5}}{4} + \frac{1 - \sqrt{5}}{4} = \frac{1}{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum