IT"S A CHALLENGE

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Replies : 13	Last Replied :08/02/2010

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01/02/2010 20:41:16

Title: IT"S A CHALLENGE

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Deepak Kumar

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PROVE THAT SIN(2THETA)=2TAN(THETA)/1-TAN SQUARE(THETA)

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02/02/2010 00:32:26

Title: IT"S A CHALLENGE

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Abhishek Kapoor

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Questions seems to be wrong.

$Take\; \theta=\frac{\pi}{8}\\ sin(2\;.\;\frac{\pi}{8})\\ =sin(\frac{\pi}{4})\\ =sin(45\;degrees)\\ =\frac{1}{\sqrt{2}}\\ \\ tan(\frac{\pi}{8}) =\sqrt{2}-1\\ So:\\ \frac{2(\sqrt{2}-1)}{1-{(\sqrt{2}-1)}^{2}}\\ =\frac{2(\sqrt{2}-1)}{(1+\sqrt{2}-1)(1-\sqrt{2}+1})\\ =1\\ \ne\frac{1}{\sqrt{2}}$

This message was edited 1 time. Last update was at 04/02/2010 23:35:37

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03/02/2010 11:34:32

Title: Re:IT"S A CHALLENGE

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VANSH PAHWA

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$\theta$ =0or multiple of 2 pie

This message was edited 1 time. Last update was at 03/02/2010 12:44:32

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03/02/2010 11:40:53

Title: Re:IT"S A CHALLENGE

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VANSH PAHWA

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if we solve it conventionally value comes out to be odd multiple of 45 degree

This message was edited 1 time. Last update was at 03/02/2010 12:48:30

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03/02/2010 12:46:48

Title: Re:IT"S A CHALLENGE

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VANSH PAHWA

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if theta =0 then sin 2 theta =0 tan theta is also equal to zero and tan sq theta is equal to 0

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03/02/2010 20:20:38

Title: Re:IT"S A CHALLENGE

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Abhishek Kapoor

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If we solve this as an eqn in $\theta$ instead of trying to prove it,...

$sin(2\theta)=\frac{tan\theta}{1-{tan}^{2}\theta}\\ \Rightarrow sin(2\theta)=\frac{2tan\theta}{2(1-{tan}^{2}\theta)}\\ \Rightarrow sin(2\theta)=\frac{tan(2\theta)}{2}\\ \Rightarrow sin(2\theta)=\frac{sin(2\theta)}{2cos(2\theta)}\\ \Rightarrow sin(2\theta)=0,\;which\;means\;\theta=0\;or\\ \Rightarrow cos(2\theta)=\frac{1}{2}\\ So\;2\theta=\frac{\pi}{3}\;(1st\;Quadrant)\;or\;2\theta=\frac{5\pi}{3}\;(4th\;Quadrant)\\ \Rightarrow \theta=\frac{\pi}{6},\;\theta=\frac{5\pi}{6}\\ (and\;also\;their\;respective\;periodically\;occuring\;multiples)\\ So:\\ \theta= 0, \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi\;(where\;k\;\in\;Z)$

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03/02/2010 20:47:53

Title: Re:IT"S A CHALLENGE

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VANSH PAHWA

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abhishek:can u pls verify ur solution for $\theta$ =30 degree

This message was edited 1 time. Last update was at 03/02/2010 20:48:20

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03/02/2010 20:48:35

Title: Re:IT"S A CHALLENGE

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VANSH PAHWA

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can u pls verify ur solution for $\theta$ =30 degree

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04/02/2010 20:03:53

Title: Re:IT"S A CHALLENGE

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Abhishek Kapoor

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ok

$\theta=30\;degrees\\ LHS\\ =sin(2\theta)\\ =sin(60\;degrees)\\ =\frac{\sqrt{3}}{2}\\ \\ RHS\\ =\frac{tan(30\;degrees)}{1-{tan}^{2}30\;degrees}\\ =\frac{\frac{1}{\sqrt{3}}}{1-\frac{1}{3}}\\ =\frac{\frac{1}{\sqrt{3}}}{\frac{2}{3}}\\ =\frac{3}{2\sqrt{3}}\\ =\frac{\sqrt{3}}{2}\\ \\ So\;LHS=RHS.\;Hence,\;verified.\\$

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04/02/2010 22:05:01

Title: Re:IT"S A CHALLENGE

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VANSH PAHWA

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it is 2tan $\theta$ NOT TAN $\theta$

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04/02/2010 23:34:28

Title: Re:IT"S A CHALLENGE

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Abhishek Kapoor

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oh.. my mistake..
Then I think it's:

$\theta = k\pi (k \in Z)$
as now we get
sin(2 $\theta$ )=tan(2 $\theta$ )

=> sin(2 $\theta$ )= $\frac{sin(2\theta)}{cos(2\theta)}$

and sin(2 $\theta$ )=0 for k $\pi$ values of $\theta$ , where k $\in$ Z...

(values of $\theta$ obtained on equating cos(2 $\theta$ ) to 1 are included in this)

tell me if I'm wrong again...

This message was edited 7 times. Last update was at 04/02/2010 23:51:18

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07/02/2010 20:23:08

Title: Re:IT"S A CHALLENGE

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Abhishek Sharan

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$Sin(2\theta) = 2sin\thetacos\theta\\ = 2tan\theta{cos\theta}^{2}\\ = 2\frac{tan\theta}{{sec\theta}^{2}}\\ now {sec\theta}^{2} = 1+{tan\theta}^{2}$

This message was edited 1 time. Last update was at 07/02/2010 20:23:36

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08/02/2010 23:13:10

Title: IT"S A CHALLENGE

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AKASH RUPELA

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yaar ye to identity hai, ques is WRONG
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)

passion does it all

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08/02/2010 23:14:07

Title: IT"S A CHALLENGE

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AKASH RUPELA

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you guys can derive this in 11th yourself
no need to worry about this, its really very easy

passion does it all

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