2
$\begingroup$

I wonder how to solve this equation: $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ in an elegant/shorter way. My way: $$\frac{\sin(x+20^{\circ })-\sin(x+20^{\circ })\cdot cos(x+20^{\circ})}{\sin(x+20^{\circ })+\sin(x+20^{\circ })\cdot \cos(x+20^{\circ})}=2[1-(\cos(x+20^{\circ}))]$$ so $$1-\cos(x+20^{\circ})=2[1-\cos(x+20^{\circ})][1+\cos(x+20^{\circ})]$$ which gives $$\cos(x+20^{\circ})=1\Rightarrow x=-20^{\circ}+360^{\circ}k\\\cos(x+20^{\circ})=-0.5\Rightarrow x=140^{\circ}+360^{\circ}k,\, \, x=-100^{\circ}+360^{\circ}k$$ Thanks.

$\endgroup$
0
$\begingroup$

Setting $\displaystyle\frac x2+10^\circ=y\iff x+20^\circ=2y$

$$\frac{\tan2y-\sin2y}{\tan2y+\sin2y}=\frac{\sin2y\cdot\dfrac{1-\cos2y}{\cos2y}}{\sin2y\cdot\dfrac{1+\cos2y}{\cos2y}}=\frac{\cos2y\sin2y(1-\cos2y)}{\cos2y\sin2y(1+\cos2y)}=\tan^2y$$ if $\cos2y\sin2y\ne0$

Now $\cos2y\sin2y=0\iff\sin4y=2\sin2y\cos2y=0\implies4y=n\cdot180^\circ$ where $n$ is any integer

So, we need $\displaystyle y\ne n\cdot45^\circ\ \ \ \ (1)$ and we have $\displaystyle\tan^2y=4\sin^2y,$

$\displaystyle\implies\sin^2y=4\sin^2y\cos^2y\iff\sin^2y(1-4\cos^2y)=0$

If $\displaystyle\sin^2y=0,\sin y=0\implies y=m\cdot180^\circ$ which is inadmissible by $(1)$

So, $\displaystyle1-4\cos^2y=0\iff\cos^2y=\frac14\iff\cos2y=2\cos^2y-1=-\frac12=\cos120^\circ$

$\displaystyle\implies2y=360^\circ\cdot r\pm120^\circ\iff y=180^\circ\cdot r\pm60^\circ$ where $r$ is any integer

So, $x=2y-20=\cdots$

$\endgroup$
  • $\begingroup$ @aric, How about this? $\endgroup$ – lab bhattacharjee May 12 '14 at 6:54
  • $\begingroup$ can you please check my answer? $\endgroup$ – shaurya gupta May 12 '14 at 7:22
  • $\begingroup$ @shauryagupta, I think you proceed a little further to discard the inadmissible value $\endgroup$ – lab bhattacharjee May 12 '14 at 12:48
  • $\begingroup$ @labbhattacharjee - so $x=2y-20^\circ=-20^\circ+360^\circ\cdot r\pm120^\circ$? $\endgroup$ – aric May 12 '14 at 21:42
  • $\begingroup$ @aric, Any doubt? Please validate the values $\endgroup$ – lab bhattacharjee May 13 '14 at 2:47
1
$\begingroup$

$$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=\\\frac{\frac{\sin(x+20^{\circ })}{\cos(x+20^{\circ })}-\sin(x+20^{\circ })}{\frac{\sin(x+20^{\circ })}{\cos(x+20^{\circ })}+\sin(x+20^{\circ })}=\\\frac{\sin(x+20^{\circ })}{\sin(x+20^{\circ })}\frac{\frac{1}{\cos(x+20^{\circ })}-1}{\frac{1}{\cos(x+20^{\circ })}+1}=\\\frac{\sin(x+20^{\circ })}{\sin(x+20^{\circ })}\frac{\cos(x+20^{\circ })}{\cos(x+20^{\circ })}\frac{1-\cos(x+20^{\circ })}{1+\cos(x+20^{\circ })}=\\\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\frac{2\sin^2(\frac{x}{2}+10^{\circ })}{2\cos^2(\frac{x}{2}+10^{\circ })}=\\\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\tan^2\left(\frac{x}{2}+10^{\circ }\right)$$ Thus you can rewrite $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ as $$\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\tan^2\left(\frac{x}{2}+10^{\circ }\right)=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)\\\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\frac{1}{\cos^2(\frac{x}{2}+10^{\circ })}=\frac{4}{\cos^2(\frac{x}{2}+10^{\circ })}\\\frac{\tan(x+20^{\circ })}{\cos^2(\frac{x}{2}+10^{\circ })}=\frac{4\tan(x+20^{\circ })}{\cos^2(\frac{x}{2}+10^{\circ })}\\\frac{3\tan(x+20^{\circ })}{\cos^2(\frac{x}{2}+10^{\circ })}=0\Rightarrow\\\tan(x+20^{\circ })=0\\x+20^{\circ }=0^{\circ }\\x=-20^{\circ }+k\pi^{\circ }\qquad k\in\Bbb{Z}$$

$\endgroup$
  • $\begingroup$ why is $4sin^{2}(\frac{x}{2}+10^{\circ})=\frac{4}{cos^{2}(\frac{x}{2}+10^{\circ})}$ $\endgroup$ – aric May 11 '14 at 18:55
  • $\begingroup$ @Foga- also you got: $1=4$ so i think something is wrong in your solution. $\endgroup$ – aric May 11 '14 at 19:08
  • $\begingroup$ Nothing! You obtain $1=4$ by dividing by $0$... $\endgroup$ – sirfoga May 11 '14 at 19:39
  • $\begingroup$ @Foga, But $x=180^\circ n-20^\circ$ will make the LHS $\dfrac00$ $\endgroup$ – lab bhattacharjee May 12 '14 at 6:55
  • $\begingroup$ I've edited the answer to make as clear as possible... $\endgroup$ – sirfoga May 12 '14 at 11:09
0
$\begingroup$

$4sin^2(\frac{x + 20^\circ}2) = 4\cdot1/2\cdot(1-\cos(x+20^\circ)) = 2-2\cos(x+20^\circ)$

Applying componendo and dividendo on $\frac{tg(x+20^{\circ })-sin(x+20^{\circ })}{tg(x+20^{\circ })+sin(x+20^{\circ })}=4sin^{2}(\frac{x}{2}+10^{\circ })$ ,

$-\sec(x+20^\circ) = \frac{2-2\cos(x+20^\circ)+1}{2-2\cos(x+20^\circ) -1}$, $cos(x+20^\circ)\ne1/2$
$-\sec(x+20^\circ)\cdot(1-{2\cos(x+20^\circ)}) = {3-2\cos(x+20^\circ)}$
$-\sec(x+20^\circ)+2=3-2\cos(x+20^\circ)$
$2cos(x+20^\circ)-\sec(x+20^\circ)-1=0$
$2cos^2(x+20^\circ)-cos(x+20^\circ)-1=0$
Can you solve it further?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.