http://forum.100percentile.com/iitjee_aieee_forums/posts/list/12.htm JForum - http://www.jforum.net http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/98795.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/98795.htm GMT
 [Eqn]Take\; \theta=\frac{\pi}{8}\\
 sin(2\;.\;\frac{\pi}{8})\\
 =sin(\frac{\pi}{4})\\
 =sin(45\;degrees)\\
 =\frac{1}{\sqrt{2}}\\
 \\
 tan(\frac{\pi}{8})
 =\sqrt{2}-1\\
 So:\\
 \frac{2(\sqrt{2}-1)}{1-{(\sqrt{2}-1)}^{2}}\\
 =\frac{2(\sqrt{2}-1)}{(1%2B\sqrt{2}-1)(1-\sqrt{2}%2B1})\\
 =1\\
 \ne\frac{1}{\sqrt{2}}[/Eqn]]]> http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/98877.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/98877.htm GMT ]]> http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99159.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99159.htm GMT http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99160.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99160.htm GMT http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99168.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99168.htm GMT
 [Eqn]sin(2\theta)=\frac{tan\theta}{1-{tan}^{2}\theta}\\
 \Rightarrow sin(2\theta)=\frac{2tan\theta}{2(1-{tan}^{2}\theta)}\\
 \Rightarrow sin(2\theta)=\frac{tan(2\theta)}{2}\\
 \Rightarrow sin(2\theta)=\frac{sin(2\theta)}{2cos(2\theta)}\\
 \Rightarrow sin(2\theta)=0,\;which\;means\;\theta=0\;or\\
 \Rightarrow cos(2\theta)=\frac{1}{2}\\
 So\;2\theta=\frac{\pi}{3}\;(1st\;Quadrant)\;or\;2\theta=\frac{5\pi}{3}\;(4th\;Quadrant)\\
 \Rightarrow \theta=\frac{\pi}{6},\;\theta=\frac{5\pi}{6}\\
 (and\;also\;their\;respective\;periodically\;occuring\;multiples)\\
 So:\\
 \theta= 0, \frac{\pi}{6} %2B 2k\pi, \frac{5\pi}{6} %2B 2k\pi\;(where\;k\;\in\;Z)

 [/Eqn]]]> http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99231.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99231.htm GMT http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99245.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99245.htm GMT http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99247.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99247.htm GMT

 [Eqn]\theta=30\;degrees\\
 LHS\\
 =sin(2\theta)\\
 =sin(60\;degrees)\\
 =\frac{\sqrt{3}}{2}\\
 \\
 RHS\\
 =\frac{tan(30\;degrees)}{1-{tan}^{2}30\;degrees}\\
 =\frac{\frac{1}{\sqrt{3}}}{1-\frac{1}{3}}\\
 =\frac{\frac{1}{\sqrt{3}}}{\frac{2}{3}}\\
 =\frac{3}{2\sqrt{3}}\\
 =\frac{\sqrt{3}}{2}\\
 \\
 So\;LHS=RHS.\;Hence,\;verified.\\[/Eqn]]]> http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99403.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99403.htm GMT http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99451.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99451.htm GMT  Then I think it's:

 [Eqn]\theta = k\pi (k \in Z)[/Eqn]
 as now we get
 sin(2[eqn]\theta[/eqn])=tan(2[eqn]\theta[/eqn])

 => sin(2[eqn]\theta[/eqn])=[eqn]\frac{sin(2\theta)}{cos(2\theta)}[/eqn]

 and sin(2[eqn]\theta[/eqn])=0 for k[eqn]\pi[/eqn] values of [eqn]\theta[/eqn], where k [eqn]\in[/eqn] Z...

 (values of [eqn]\theta[/eqn] obtained on equating cos(2[eqn]\theta[/eqn]) to 1 are included in this)

 tell me if I'm wrong again...

]]> http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99493.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99493.htm GMT  [/Eqn]]]> http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99986.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/99986.htm GMT  u will read in 11
 tan2theta=2tantheta/ (1-tan square theta)
 sin2theta=2 tan theta/(1+tan square theta)
 cos2theta=(1-tan square theta)/(1+tan square theta)]]> http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/100172.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/100172.htm GMT  no need to worry about this, its really very easy]]> http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/100173.htm http://forum.100percentile.com/iitjee_aieee_forums/posts/preList/25370/100173.htm GMT