arctan ?

**thereddevils** · August 24th 2009, 12:24 AM

Find all the values of x which satisfy the equation

(a) $\tan^{-1}x+\tan^{-1}2x=\frac{1}{2}\pi$

(b) Show that the sum of the three angles whose tangent have the following values
$\frac{3}{8}$ , $\frac{1}{3}$ , $\frac{2}{19}$ is $\frac{1}{4}\pi$

**mr fantastic** · August 24th 2009, 01:28 AM

Originally Posted by thereddevils

Find all the values of x which satisfy the equation

(a) $\tan^{-1}x+\tan^{-1}2x=\frac{1}{2}\pi$

(b) Show that the sum of the three angles whose tangent have the following values
$\frac{3}{8}$ , $\frac{1}{3}$ , $\frac{2}{19}$ is $\frac{1}{4}\pi$

(a) Let $\tan^{-1} (x) = \alpha \Rightarrow \tan \alpha = x$ and $\tan^{-1} (2x) = \beta \Rightarrow \tan \beta = 2x$ .

Then you need the value(s) of x such that

$\alpha + \beta = \frac{\pi}{2}$

$\Rightarrow \sin ( \alpha + \beta ) = 1 \Rightarrow \sin \alpha \cos \beta + \sin \beta \cos \alpha = 1$

$\Rightarrow \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{4x^2 + 1}} + \frac{2x}{\sqrt{4x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = 1$

$\Rightarrow \frac{3x}{\sqrt{x^2 + 1} \sqrt{4x^2 + 1}} = 1$

$\Rightarrow 3x = \sqrt{x^2 + 1} \sqrt{4x^2 + 1} \Rightarrow 9x^2 = 4x^4 + 5x^2 + 1$

$\Rightarrow (2x^2 - 1)^2 = 0$ .

Solve this equation carefully - one of the solutions is extraneous (Google it).

I might have time for (b) later (although I'd prefer to see if you now have some ideas on how to do it ....)

**red_dog** · August 24th 2009, 01:33 AM

Use the identity: $\arctan a+\arctan b=\arctan\frac{a+b}{1-ab}$

a) $\arctan\frac{3x}{1-2x^2}=\frac{\pi}{2}\Rightarrow 1-2x^2=0\Rightarrow x=\pm\frac{\sqrt{2}}{2}$

The correct solution is $x=\frac{\sqrt{2}}{2}$

b) $arctan\frac{3}{8}+\arctan\frac{1}{3}+\arctan\frac{ 2}{19}=\arctan\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{1}{8}}+\arctan\frac{2}{19}=$

$=\arctan\frac{17}{21}+\arctan\frac{2}{19}=\arctan 1=\frac{\pi}{4}$

**thereddevils** · August 24th 2009, 03:10 AM

Originally Posted by mr fantastic

(a) Let $\tan^{-1} (x) = \alpha \Rightarrow \tan \alpha = x$ and $\tan^{-1} (2x) = \beta \Rightarrow \tan \beta = 2x$ .

Then you need the value(s) of x such that

$\alpha + \beta = \frac{\pi}{2}$

$\Rightarrow \sin ( \alpha + \beta ) = 1 \Rightarrow \sin \alpha \cos \beta + \sin \beta \cos \alpha = 1$

$\Rightarrow \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{4x^2 + 1}} + \frac{2x}{\sqrt{4x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = 1$

$\Rightarrow \frac{3x}{\sqrt{x^2 + 1} \sqrt{4x^2 + 1}} = 1$

$\Rightarrow 3x = \sqrt{x^2 + 1} \sqrt{4x^2 + 1} \Rightarrow 9x^2 = 4x^4 + 5x^2 + 1$

$\Rightarrow (2x^2 - 1)^2 = 0$ .

Solve this equation carefully - one of the solutions is extraneous (Google it).

I might have time for (b) later (although I'd prefer to see if you now have some ideas on how to do it ....)

thanks , i will give it a shot !

**pacman** · August 24th 2009, 06:31 AM

below is the graph of arctan x + arctan 2x = pi/2;

(sqrt of 2)/2 is indeed correct

**thereddevils** · August 24th 2009, 07:04 AM

Originally Posted by thereddevils

Find all the values of x which satisfy the equation

(b) Show that the sum of the three angles whose tangent have the following values
$\frac{3}{8}$ , $\frac{1}{3}$ , $\frac{2}{19}$ is $\frac{1}{4}\pi$

Lets call the 3 angles a , b, and c .

$\tan [(a+b)+c] = \frac{\frac{\tan a +\tan b}{1-\tan a\tan b}+\tan c}{1-\frac{\tan a+\tan b}{1-\tan a\tan b}(\tan c)}$

$<br /> =\frac{\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{3}{8}\frac{1}{3}}+\frac{2}{19}}{1-\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{3}{8}\frac{1}{3}}(\frac{2}{19})}=1 <br />$

and $arctan 1$ gives $\frac{\pi}{4}$

Am i correct ?

**mr fantastic** · August 24th 2009, 03:38 PM

Originally Posted by thereddevils

Lets call the 3 angles a , b, and c .

$\tan [(a+b)+c] = \frac{\frac{\tan a +\tan b}{1-\tan a\tan b}+\tan c}{1-\frac{\tan a+\tan b}{1-\tan a\tan b}(\tan c)}$

$<br /> =\frac{\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{3}{8}\frac{1}{3}}+\frac{2}{19}}{1-\frac{\frac{3}{8}+\frac{1}{3}}{1-\frac{3}{8}\frac{1}{3}}(\frac{2}{19})}=1 <br />$

and $arctan 1$ gives $\frac{\pi}{4}$

Am i correct ?

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