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Question Number 101107 by PengagumRahasiamu last updated on 30/Jun/20

Commented by Dwaipayan Shikari last updated on 30/Jun/20

$s i n 6 ° - s i n 66 ° + s i n 78 ° - s i n 42 °$ $= - 2 c o s 36 ° s i n 30 ° + 2 c o s 60 ° s i n 18 °$ $= s i n 18 ° - c o s 36 ° = \frac{\sqrt{5} - 1}{4} - \frac{\sqrt{5} + 1}{4} = - \frac{1}{2}$ $\frac{}{}$
Commented by Dwaipayan Shikari last updated on 01/Jul/20

$\frac{s i n 66 ° - s i n 6 °}{s i n 6 ° s i n 66 °} + \frac{1}{s i n 78 °} - \frac{1}{s i n 42 °}$ $= \frac{2 c o s 36 ° s i n 30 °}{s i n 6 ° s i n 66 °} + \frac{s i n 42 ° - s i n 78 °}{s i n 78 ° s i n 42 °}$ $= \frac{c o s 36 °}{s i n 6 ° s i n 66 °} - \frac{s i n 18 °}{s i n 78 s i n 42}$ $= \frac{2 c o s 36 °}{c o s 60 ° - c o s 72 °} - \frac{2 s i n 18 °}{c o s 36 - c o s 120 °}$ $= \frac{\frac{\sqrt{5} + 1}{2}}{\frac{1}{2} - \frac{\sqrt{5} - 1}{4}} + \frac{\frac{\sqrt{5} - 1}{2}}{\frac{1}{2} - \frac{\sqrt{5} + 1}{4}} = 8 (y o u w i l l g e t y o u r a n s w e r a f t e r s i m p l i f y i n g)$
	Answered by Ar Brandon last updated on 30/Jun/20
	$a\S=sin6°−sin42°−sin66°+sin78° =(sin6°−sin66°)+(sin78°−sin42°) =−2cos36°sin30°+2cos60°sin18° =sin18°−cos36°=(((√5)−1)/4)−(((√5)+1)/4)=−(1/2)$
	$a ∖ S = \sin 6 ° - \sin 42 ° - \sin 66 ° + \sin 78 °$ $= (\sin 6 ° - \sin 66 °) + (\sin 78 ° - \sin 42 °)$ $= - 2 \cos 36 ° \sin 30 ° + 2 \cos 60 ° \sin 18 °$ $= \sin 18 ° - \cos 36 ° = \frac{\sqrt{5} - 1}{4} - \frac{\sqrt{5} + 1}{4} = - \frac{1}{2}$
	Answered by Ar Brandon last updated on 30/Jun/20
	$b\T=(1/(sin6°))−(1/(sin66°))+(1/(sin78°))−(1/(sin42°))=((sin66°−sin6°)/(sin66°sin6°))+((sin42°−sin78°)/(sin42°sin78°)) =((2cos36°sin30°)/(((−1)/2)(cos72°−cos60°)))+((−2cos60°sin18°)/(((−1)/2)(cos120°−cos36°))) =((−2cos36°)/((sin18°−(1/2))))+((2sin18°)/((−(1/2)−cos36°)))=((−2∙((((√5)+1))/4))/(((((√5)−1)/4)−(1/2))))+((2∙(((√5)−1)/4))/((((−1)/2)−(((√5)+1)/4)))) =((−2((√5)+1))/((√5)−1−2))+((2((√5)−1))/((−2−(√5)−1)))=((2((√5)+1))/(3−(√5)))+((2(1−(√5)))/(3+(√5))) =((2[(1+(√5))(3+(√5))+(1−(√5))(3−(√5))])/((3−(√5))(3+(√5))))=((2(8+4(√5)+8−4(√5)))/4) (1/(sin6°))−(1/(sin66°))+(1/(sin78°))−(1/(sin42°))=8$
	$b ∖ T = \frac{1}{\sin 6 °} - \frac{1}{\sin 66 °} + \frac{1}{\sin 78 °} - \frac{1}{\sin 42 °} = \frac{\sin 66 ° - \sin 6 °}{\sin 66 ° \sin 6 °} + \frac{\sin 42 ° - \sin 78 °}{\sin 42 ° \sin 78 °}$ $= \frac{2 \cos 36 ° \sin 30 °}{\frac{- 1}{2} (\cos 72 ° - \cos 60 °)} + \frac{- 2 \cos 60 ° \sin 18 °}{\frac{- 1}{2} (\cos 120 ° - \cos 36 °)}$ $= \frac{- 2 \cos 36 °}{(\sin 18 ° - \frac{1}{2})} + \frac{2 \sin 18 °}{(- \frac{1}{2} - \cos 36 °)} = \frac{- 2 \cdot \frac{(\sqrt{5} + 1)}{4}}{(\frac{\sqrt{5} - 1}{4} - \frac{1}{2})} + \frac{2 \cdot \frac{\sqrt{5} - 1}{4}}{(\frac{- 1}{2} - \frac{\sqrt{5} + 1}{4})}$ $= \frac{- 2 (\sqrt{5} + 1)}{\sqrt{5} - 1 - 2} + \frac{2 (\sqrt{5} - 1)}{(- 2 - \sqrt{5} - 1)} = \frac{2 (\sqrt{5} + 1)}{3 - \sqrt{5}} + \frac{2 (1 - \sqrt{5})}{3 + \sqrt{5}}$ $= \frac{2 [(1 + \sqrt{5}) (3 + \sqrt{5}) + (1 - \sqrt{5}) (3 - \sqrt{5})]}{(3 - \sqrt{5}) (3 + \sqrt{5})} = \frac{2 (8 + 4 \sqrt{5} + 8 - 4 \sqrt{5})}{4}$ $\frac{1}{\sin 6 °} - \frac{1}{\sin 66 °} + \frac{1}{\sin 78 °} - \frac{1}{\sin 42 °} = 8$
	Commented by PengagumRahasiamu last updated on 30/Jun/20
	Thank you, Sir ��
	Commented by Ar Brandon last updated on 30/Jun/20
	You're welcome mate ! ��
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