prove that 1/cos2x+cosx+1=((sin((5x)/2))/(2sin(x/2)))+(1/2) 2/((cos(x)+isin(x)−1)/(cos(x)+isin(x)+1))=−i tan(x) 3/((cos(5x)+isin(5x)+1)/(cos(5x)−isin(x)+1))=cos(5x)+isin(5x)


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 86737 by M±th+et£s last updated on 30/Mar/20

$p r o v e t h a t$ $1 / c o s 2 x + c o s x + 1 = \frac{s i n \frac{5 x}{2}}{2 s i n \frac{x}{2}} + \frac{1}{2}$ $2 / \frac{c o s (x) + i s i n (x) - 1}{c o s (x) + i s i n (x) + 1} = - i t a n (x)$ $3 / \frac{c o s (5 x) + i s i n (5 x) + 1}{c o s (5 x) - i s i n (x) + 1} = c o s (5 x) + i s i n (5 x)$
Commented by som(math1967) last updated on 31/Mar/20

$2 / \frac{(c o s x + i s i n x - 1) (c o s x - i s i n x + 1)}{(c o s x + i s i n x + 1) (c o s x - i s i n x + 1)}$ $= \frac{{c o s}^{2} x - {(1 - i s i n x)}^{2}}{{(c o s x + 1)}^{2} - {(i s i n x)}^{2}}$ $= \frac{{c o s}^{2} x + {s i n}^{2} x - 1 + 2 i s i n x}{{c o s}^{2} x + {s i n}^{2} x + 1 + 2 c o s x}$ $= \frac{2 i s i n x}{2 (1 + c o s x)} = \frac{2 i s i n \frac{x}{2} c o s \frac{x}{2}}{2 {c o s}^{2} \frac{x}{2}}$ $= i t a n \frac{x}{2} ? ? ? ? ?$
Commented by Prithwish Sen 1 last updated on 31/Mar/20

$2 . \frac{e^{ix} - 1}{e^{ix} + 1} = \frac{\frac{e^{i \frac{x}{2}} - e^{- i \frac{x}{2}}}{2}}{i (\frac{e^{i \frac{x}{2}} + e^{- i \frac{x}{2}}}{2 i})} = \frac{\sin \frac{x}{2}}{icos \frac{x}{2}} = - itan \frac{x}{2}$ $please check .$
Commented by M±th+et£s last updated on 31/Mar/20

$y e s s i r i t s \frac{x}{2}$
Commented by M±th+et£s last updated on 31/Mar/20

$y o u a r e r i g h t$
	Answered by jagoll last updated on 31/Mar/20

	Commented by M±th+et£s last updated on 31/Mar/20

	$t h a n k y o u s i r$
	Answered by som(math1967) last updated on 31/Mar/20
	$3/(({cos(5x)+isin(5x)+1}^2 )/({cos(5x)+1}^2 −{isin(5x)}^2 )) =((cos^2 (5x)−sin^2 (5x)+1+2cos(5x)+2isin(5x)+2isin(5x)cos(5x))/(cos^2 (5x)+sin^2 (5x)+1+2cos(5x))) [∵ i^2 =−1] =((2cos^2 (5x)+2cos(5x)+2isin(5x){1+cos(5x)})/(2{1+cos(5x)})) =((2cos(5x){1+cos(5x)}+2isin(5x){1+cos(5x)})/(2{1+cos(5x)})) =((2{1+cos(5x)}{cos(5x)+isin(5x)})/(2{1+cos(5x)})) =cos(5x)+isin(5x)$
	$3 / \frac{{c o s (5 x) + i s i n (5 x) + 1}^{2}}{{c o s (5 x) + 1}^{2} - {i s i n (5 x)}^{2}}$ $= \frac{{c o s}^{2} (5 x) - {s i n}^{2} (5 x) + 1 + 2 c o s (5 x) + 2 i s i n (5 x) + 2 i s i n (5 x) c o s (5 x)}{{c o s}^{2} (5 x) + {s i n}^{2} (5 x) + 1 + 2 c o s (5 x)}$ $[∵ i^{2} = - 1]$ $= \frac{2 {c o s}^{2} (5 x) + 2 c o s (5 x) + 2 i s i n (5 x) {1 + c o s (5 x)}}{2 {1 + c o s (5 x)}}$ $= \frac{2 c o s (5 x) {1 + c o s (5 x)} + 2 i s i n (5 x) {1 + c o s (5 x)}}{2 {1 + c o s (5 x)}}$ $= \frac{2 {1 + c o s (5 x)} {c o s (5 x) + i s i n (5 x)}}{2 {1 + c o s (5 x)}}$ $= c o s (5 x) + i s i n (5 x)$
	Commented by M±th+et£s last updated on 31/Mar/20

	$t h a n k y o u s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum