| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Mar 2007 14:01:45 IST
|
|
|
sin2x+cos2x=? 2cosx/2cosecx=?
|
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 05:24:13 IST
|
|
|
sin2x + cos2x = 1
2cosx/2cosecx = sinxcosx
|
Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 09:06:41 IST
|
|
|
which is 1/2 sin2x
|
Life Ka fundaa hai
Jiyo aur jino do |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 09:54:58 IST
|
|
|
taruntanuj, 2nd one is 1/2 sin2x bcoz 1/2 . 2 sinx .cos x = 1/2 sin2x
|
DO NOT FOLLOW MY WAY.
IT'S TOO DANGER. |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 09:55:21 IST
|
|
|
rate me quickly
|
DO NOT FOLLOW MY WAY.
IT'S TOO DANGER. |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 16:40:53 IST
|
|
|
cosec (x/2)
|
sachin |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 17:02:51 IST
|
|
|
woah!!! those were really tough ones!!
|
--
 |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 17:45:35 IST
|
|
|
YES SUBS YOU ARE RIGHT THEY ARE SILLY QUESTIONS
|
DO NOT FOLLOW MY WAY.
IT'S TOO DANGER. |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2007 18:46:56 IST
|
|
|
sin^2x+cos^2x=1. your question is not clear. there are two possibilities a)[2cosx/2].[cosecx] b)(2cosx)/(2cosecx)
solutions a)(2cosx/2).(1/'sinx) =(2cosx/2).[1/2(sinx/2.)(cosx/2)] =1/(sinx/2) =cosec(x/2) b)(2cosx)/(2cosecx) =cosx.sinx
hope you have understood properly.
|
vision for iit never ends........ |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|