Prove that ((sin 19θ)/(sin θ)) = cos(−18θ)+cos(−16θ)+... ...+ cos(−2θ)+1+cos(2θ)+....+cos(18θ).


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Question Number 54126 by rahul 19 last updated on 29/Jan/19

$P r o v e t h a t$ $\frac{\sin 19 θ}{\sin θ} = \cos (- 18 θ) + \cos (- 16 θ) + . . .$ $. . . + \cos (- 2 θ) + 1 + \cos (2 θ) + . . . . + \cos (18 θ) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19

	$S = [c o s (- 18 θ) + c o s (- 16 θ) + . . . + c o s (- 2 θ)] + 1 + [c o s 2 θ + c o s 4 θ + . . + c o s 18 θ]$ $S = S_{1} + 1 + S_{2} [S_{1} = s u m o f - v e s n g l e d t e r m s S_{2} f o r + v e a n g l e d t e r m s]$ $S_{1} = c o s (- 18 θ) + c o s (- 16 θ) + c o s (- 14 θ) + . . . + c o s (- 2 θ)$ $S_{1} = S_{2}$ $S = 2 S_{1} + 1$ $n o w$ $S_{1} = c o s 2 θ + c o s 4 θ + c o s 6 θ + . . . + c o s 18 θ$ $2 s i n θ c o s 2 θ = s i n 3 θ - s i n θ$ $2 s i n θ c o s 4 θ = s i n 5 θ - s i n 3 θ$ $2 s i n θ c o s 6 θ = s i n 7 θ - s i n 5 θ$ $. . .$ $. . . .$ $2 s i n θ c o s 18 θ = s i n 19 θ - s i n 17 θ$ $n o w a d d i t i o n g i v e \to L H S = 2 s i n θ \times S_{1}$ $d u r i n g a d d i t i o n t h e r e d m a r k e d t e r m s c a n c e l l e d$ $i n r i g h t h a n d s i d e . .$ $s o \to 2 s i n θ \times S_{1} = s i n 19 θ - s i n θ$ $S_{1} = \frac{s i n 19 θ}{2 s i n θ} - \frac{1}{2}$ $S = 2 S_{1} + 1$ $= 2 (\frac{s i n 19 θ}{2 s i n θ} - \frac{1}{2}) + 1$ $= \frac{s i n 19 θ}{s i n θ} - 1 + 1$ $= \frac{s i n 19 θ}{s i n θ} p r o v e d . .$
	Commented by Otchere Abdullai last updated on 29/Jan/19

	$w e l l d o n e s i r!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19

	$t h a n k y o u s i r . . .$
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