prove sin18×cos36=(1/4)


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Question Number 57474 by malwaan last updated on 05/Apr/19

$prove$ $s i n 18 \times c o s 36 = \frac{1}{4}$
Commented by Abdo msup. last updated on 05/Apr/19
$18→x 180→π ⇒180x =18π ⇒x =(π/(10)) ⇒36 =(π/5) for that let prove sin((π/(10)))cos((π/5))=(1/4) we have proved that cos((π/5))=((1+(√5))/4) sin((π/(10)))=(√((1−cos((π/5)))/2))=(√((1−((1+(√5))/4))/2)) =(√((3−(√5))/8))=((√(3−(√5)))/(2(√2))) we have { sin((π/(10)))cos((π/5))}^2 =((3−(√5))/8) ((6+2(√5))/(16)) =((3−(√5))/8) ((3+2(√5))/8) =((9−5)/(64)) =(4/(64)) =(1/(16)) ⇒ sin((π/(10))).cos((π/5)) =(1/4) .$
$18 \to x$ $180 \to π \Rightarrow 180 x = 18 π \Rightarrow x = \frac{π}{10} \Rightarrow 36 = \frac{π}{5}$ $f o r t h a t l e t p r o v e s i n (\frac{π}{10}) c o s (\frac{π}{5}) = \frac{1}{4}$ $w e h a v e p r o v e d t h a t c o s (\frac{π}{5}) = \frac{1 + \sqrt{5}}{4}$ $s i n (\frac{π}{10}) = \sqrt{\frac{1 - c o s (\frac{π}{5})}{2}} = \sqrt{\frac{1 - \frac{1 + \sqrt{5}}{4}}{2}}$ $= \sqrt{\frac{3 - \sqrt{5}}{8}} = \frac{\sqrt{3 - \sqrt{5}}}{2 \sqrt{2}}$ $w e h a v e {s i n (\frac{π}{10}) c o s (\frac{π}{5})}^{2}$ $= \frac{3 - \sqrt{5}}{8} \frac{6 + 2 \sqrt{5}}{16} = \frac{3 - \sqrt{5}}{8} \frac{3 + 2 \sqrt{5}}{8} = \frac{9 - 5}{64} = \frac{4}{64} = \frac{1}{16} \Rightarrow$ $s i n (\frac{π}{10}) . c o s (\frac{π}{5}) = \frac{1}{4} .$
Commented by malwaan last updated on 06/Apr/19

$t h a n k y o u s i r$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Apr/19

	$a = 18^{o}$ $5 a = 2 a + 3 a = 90^{o}$ $2 a = 90^{o} - 3 a$ $s i n 2 a = s i n (90^{o} - 3 a)$ $2 s i n a c o s a = c o s 3 a$ $2 s i n a c o s a = 4 {c o s}^{3} a - 3 c o s a$ $c o s a (2 s i n a - 4 {c o s}^{2} a + 3) = 0$ $c o s a \neq 0$ $2 s i n a - 4 (1 - {s i n}^{2} a) + 3 = 0$ $4 {s i n}^{2} a + 2 s i n a - 1 = 0$ $s i n 18^{o} > 0 [18^{o} l i e s i n f i r s t q u a d r a n t]$ $s i n a = \frac{- 2 + \sqrt{{(2)}^{2} - 4 \times 4 \times (- 1)}}{2 \times 4}$ $s i n 18^{o} = \frac{- 2 + 2 \sqrt{5}}{8}$ $s i n 18^{o} = \frac{\sqrt{5} - 1}{4}$ $c o s 36^{o} = 1 - 2 {s i n}^{2} 18^{o}$ $= 1 - 2 {(\frac{\sqrt{5} - 1}{4})}^{2}$ $= \frac{16 - 2 (5 - 2 \sqrt{5} + 1)}{16} = \frac{8 - (6 - 2 \sqrt{5})}{8}$ $= \frac{2 \sqrt{5} + 2}{8} \to \frac{\sqrt{5} + 1}{4}$ $s i n 18 \times c o s 36$ $= \frac{\sqrt{5} - 1}{4} \times \frac{\sqrt{5} + 1}{4}$ $= \frac{5 - 1}{16} = \frac{1}{4}$ $i h a v e d e r i v e d t h e v a l u e o f s i n 18^{o} a n d c o s 36^{o}$ $t h e n s o l v e d t h e p r o b l e m . .$ $f o r s h o r t c u t m e t h o d m e m o r i z e t h e v a l u e$ $o f s i n 18^{o} a n d c o s 36^{o} t h e n s o l v e i t . . .$
	Commented by malwaan last updated on 06/Apr/19

	$t h a n k y o u s o m u c h$
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