If p = cos θ + i sinθ and q = cos φ + i sin φ Show that: (i) ((p − q)/(p + q)) = i tan (((θ − φ)/2)) (ii) (((p + q)(pq − 1))/((p − q)(pq + 1))) = ((sin θ + sin φ)/(sin θ − sin φ))


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Question Number 51905 by Tawa1 last updated on 01/Jan/19

$If p = \cos θ + i \sin θ and q = \cos ϕ + i \sin ϕ$ $Show that :$ $(i) \frac{p - q}{p + q} = i \tan (\frac{θ - ϕ}{2})$ $(ii) \frac{(p + q) (pq - 1)}{(p - q) (pq + 1)} = \frac{\sin θ + \sin ϕ}{\sin θ - \sin ϕ}$
	Answered by Kunal12588 last updated on 01/Jan/19

	$\frac{p - q}{p + q} = \frac{c o s θ + i s i n θ - c o s ϕ - i s i n ϕ}{c o s θ + i s i n θ + c o s ϕ + i s i n ϕ}$ $= \frac{(c o s θ - c o s ϕ) + (s i n θ - s i n ϕ) i}{(c o s θ + c o s ϕ) + (s i n θ + s i n ϕ) i} = \frac{- s i n \frac{θ + ϕ}{2} s i n \frac{θ - ϕ}{2} + c o s \frac{θ + ϕ}{2} s i n \frac{θ - ϕ}{2} i}{c o s \frac{θ + ϕ}{2} c o s \frac{θ - ϕ}{2} + s i n \frac{θ + ϕ}{2} c o s \frac{θ - ϕ}{2} i}$ $= \frac{s i n \frac{θ - ϕ}{2} (c o s \frac{θ + ϕ}{2} i - s i n \frac{θ + ϕ}{2}) (s i n \frac{θ + ϕ}{2} i - c o s \frac{θ + ϕ}{2})}{c o s \frac{θ - ϕ}{2} (s i n \frac{θ + ϕ}{2} i + c o s \frac{θ + ϕ}{2}) (s i n \frac{θ + ϕ}{2} i - c o s \frac{θ + ϕ}{2})}$ $= - t a n \frac{θ - ϕ}{2} (- {c o s}^{2} \frac{θ + ϕ}{2} - {s i n}^{2} \frac{θ + ϕ}{2}) i$ $= i t a n \frac{θ - ϕ}{2}$
	Commented by malwaan last updated on 01/Jan/19

	$no (ii)$
	Commented by Tawa1 last updated on 01/Jan/19

	$God bless you sir$
	Answered by $@ty@m last updated on 01/Jan/19
	$(p/q)=((cos θ+isin θ)/(cos ∅+isin φ)) =(cos θ+isin θ)(cos ∅−isin φ) =cos (θ−∅)+isin (θ−∅) (i) ((p − q)/(p + q)) = (((p/q)−1)/((p/q)+1)) =((cos (θ−∅)+isin (θ−∅)−1)/(cos (θ−∅)+isin (θ−∅)+1)) =((isin (θ−∅)−{1−cos (θ−∅)})/(isin (θ−∅)+{1+cos (θ−∅)})) =((i.2sin (((θ−∅))/2)cos (((θ−∅))/2)−2sin^2 (((θ−∅) )/2))/( i.2sin (((θ−∅))/2)cos (((θ−∅))/2)+2cos ^2 (((θ−∅) )/2))) =((2sin (((θ−∅))/2){icos (((θ−∅))/2)−sin (((θ−∅) )/2)})/(2cos (((θ−∅))/2){i.sin (((θ−∅))/2)+cos (((θ−∅))/2)})) =((i.sin (((θ−∅))/2){cos (((θ−∅))/2)+isin (((θ−∅) )/2)})/(cos (((θ−∅))/2){i.sin (((θ−∅))/2)+cos (((θ−∅))/2)})) =itan (((θ−∅))/2)$
	$\frac{p}{q} = \frac{\cos θ + i \sin θ}{\cos \emptyset + i \sin ϕ}$ $= (\cos θ + i \sin θ) (\cos \emptyset - i \sin ϕ)$ $= \cos (θ - \emptyset) + i \sin (θ - \emptyset)$ $(i) \frac{p - q}{p + q} = \frac{\frac{p}{q} - 1}{\frac{p}{q} + 1}$ $= \frac{\cos (θ - \emptyset) + i \sin (θ - \emptyset) - 1}{\cos (θ - \emptyset) + i \sin (θ - \emptyset) + 1}$ $= \frac{i \sin (θ - \emptyset) - {1 - \cos (θ - \emptyset)}}{i \sin (θ - \emptyset) + {1 + \cos (θ - \emptyset)}}$ $= \frac{i . 2 \sin \frac{(θ - \emptyset)}{2} \cos \frac{(θ - \emptyset)}{2} - 2 \sin^{2} \frac{(θ - \emptyset)}{2}}{i . 2 \sin \frac{(θ - \emptyset)}{2} \cos \frac{(θ - \emptyset)}{2} + 2 \cos^{2} \frac{(θ - \emptyset)}{2}}$ $= \frac{2 \sin \frac{(θ - \emptyset)}{2} {i \cos \frac{(θ - \emptyset)}{2} - \sin \frac{(θ - \emptyset)}{2}}}{2 \cos \frac{(θ - \emptyset)}{2} {i . \sin \frac{(θ - \emptyset)}{2} + \cos \frac{(θ - \emptyset)}{2}}}$ $= \frac{i . \sin \frac{(θ - \emptyset)}{2} {\cos \frac{(θ - \emptyset)}{2} + i \sin \frac{(θ - \emptyset)}{2}}}{\cos \frac{(θ - \emptyset)}{2} {i . \sin \frac{(θ - \emptyset)}{2} + \cos \frac{(θ - \emptyset)}{2}}}$ $= i \tan \frac{(θ - \emptyset)}{2}$
	Commented by Tawa1 last updated on 01/Jan/19

	$God bless you sir . Seen$
	Commented by $@ty@m last updated on 01/Jan/19
	$(ii) (((p+q)(pq−1))/((p−q)(pq+1))) We have, ((p−q)/(p+q))=−icot((θ−∅)/2) ....(1) already proved in (i). Now, ((pq−1)/(pq+1))=((cos (θ+φ)+isin (θ+∅)−1)/(cos (θ+∅)+isin (θ+∅)+1)) =((isin (θ+∅)−{1−cos (θ+φ)})/(isin (θ+∅)+{1+cos (θ+φ)})) =((i.2sin((θ+∅)/2)cos((θ+∅)/2) −2sin^2 ((θ+∅)/2))/(i.2sin ((θ+∅)/2)cos ((θ+∅)/2)+2cos^2 ((θ+∅)/2))) =((2i.sin((θ+∅)/2){cos((θ+∅)/2) +isin((θ+∅)/2)})/(2cos ((θ+∅)/2){cos((θ+∅)/2) +isin((θ+∅)/2)})) =i.tan ((θ+∅)/2) ...(2) by (1)×(2), we get (((p+q)(pq−1))/((p−q)(pq+1)))=−icot((θ−∅)/2)×itan ((θ+∅)/2) =((sin ((θ+∅)/2).cos ((θ−∅)/2))/(cos ((θ+∅)/2).sin ((θ−∅)/2))) =((2sin ((θ+∅)/2).cos ((θ−∅)/2))/(2cos ((θ+∅)/2).sin ((θ−∅)/2))) =((sin θ+sin ∅)/(sin θ−sin ∅)) Hence the result.$
	$(i i) \frac{(p + q) (p q - 1)}{(p - q) (p q + 1)}$ $W e h a v e,$ $\frac{p - q}{p + q} = - i c o t \frac{θ - \emptyset}{2} . . . . (1) a l r e a d y p r o v e d i n (i) .$ $N o w,$ $\frac{p q - 1}{p q + 1} = \frac{\cos (θ + ϕ) + i \sin (θ + \emptyset) - 1}{\cos (θ + \emptyset) + i \sin (θ + \emptyset) + 1}$ $= \frac{i \sin (θ + \emptyset) - {1 - \cos (θ + ϕ)}}{i \sin (θ + \emptyset) + {1 + \cos (θ + ϕ)}}$ $= \frac{i . 2 \sin \frac{θ + \emptyset}{2} \cos \frac{θ + \emptyset}{2} - 2 \sin^{2} \frac{θ + \emptyset}{2}}{i . 2 \sin \frac{θ + \emptyset}{2} \cos \frac{θ + \emptyset}{2} + 2 \cos^{2} \frac{θ + \emptyset}{2}}$ $= \frac{2 i . \sin \frac{θ + \emptyset}{2} {\cos \frac{θ + \emptyset}{2} + i \sin \frac{θ + \emptyset}{2}}}{2 \cos \frac{θ + \emptyset}{2} {\cos \frac{θ + \emptyset}{2} + i \sin \frac{θ + \emptyset}{2}}}$ $= i . \tan \frac{θ + \emptyset}{2} . . . (2)$ $b y (1) \times (2), w e g e t$ $\frac{(p + q) (p q - 1)}{(p - q) (p q + 1)} = - i c o t \frac{θ - \emptyset}{2} \times i \tan \frac{θ + \emptyset}{2}$ $= \frac{\sin \frac{θ + \emptyset}{2} . \cos \frac{θ - \emptyset}{2}}{\cos \frac{θ + \emptyset}{2} . \sin \frac{θ - \emptyset}{2}}$ $= \frac{2 \sin \frac{θ + \emptyset}{2} . \cos \frac{θ - \emptyset}{2}}{2 \cos \frac{θ + \emptyset}{2} . \sin \frac{θ - \emptyset}{2}}$ $= \frac{\sin θ + \sin \emptyset}{\sin θ - \sin \emptyset}$ $H e n c e t h e r e s u l t .$
	Commented by $@ty@m last updated on 01/Jan/19

	$I w a s i n h u r r y i n t h e m o r n i n g . S o,$ $t h e r e w e r e f e w t y p i n g m i s t a k e c r e p t$ $i n s o l u t i o n o f (i), w h i c h I h a v e$ $c o r r e c t e d .$
	Commented by peter frank last updated on 01/Jan/19

	$n i c e m e t h o d s i r$
	Commented by Tawa1 last updated on 01/Jan/19

	$God bless you sir$
	Commented by Tawa1 last updated on 01/Jan/19

	$Have tried but cannot simplify, please help if you have time sir$
	Commented by $@ty@m last updated on 01/Jan/19

	$d o n e a b o v e . . .$
	Commented by malwaan last updated on 05/Jan/19

	$thank you sir$
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