# trig identity

• June 21st 2010, 03:12 PM
bigwave
trig identity
prove that:

$
\tan^{-1}{\left(\frac{1}{5}\right)}
+\tan^{-1}{\left(\frac{2}{3}\right)}
=\frac{\pi}{4}
$

first, the coordinates of $\frac{\pi}{4} \Rightarrow \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)
$

and the coordinates of
$
\tan^{-1}\left(\frac{1}{5}\right)
\Rightarrow \left(\frac{5}{\sqrt{26}},\frac{1}{\sqrt{26}}\righ t)
$

and
$
\tan^{-1}\left(\frac{2}{3}\right)
\Rightarrow \left(\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}\righ t)
$

so thot from these that using $\cos{\left(\alpha + \beta\right)}$ I could get the coordinates of $\frac{\pi}{4}$

but did not unless I made an error somewhere....
• June 21st 2010, 04:55 PM
undefined
Quote:

Originally Posted by bigwave
prove that:

$
\tan^{-1}{\left(\frac{1}{5}\right)}
+\tan^{-1}{\left(\frac{2}{3}\right)}
=\frac{\pi}{4}
$

first, the coordinates of $\frac{\pi}{4} \Rightarrow \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)
$

and the coordinates of
$
\tan^{-1}\left(\frac{1}{5}\right)
\Rightarrow \left(\frac{5}{\sqrt{26}},\frac{1}{\sqrt{26}}\righ t)
$

and
$
\tan^{-1}\left(\frac{2}{3}\right)
\Rightarrow \left(\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}\righ t)
$

so thot from these that using $\cos{\left(\alpha + \beta\right)}$ I could get the coordinates of $\frac{\pi}{4}$

but did not unless I made an error somewhere....

Hmm, the way I'm coming up with is pretty long. :(

Looking at the unit circle, you can define point P at $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ and Q at $\left(\frac{5\sqrt{26}}{26},\frac{\sqrt{26}}{26}\r ight)$. O is the origin. Then find the line perpendicular to line OQ going through P, call it line m. Find the intersection of m and OQ, call it point Z. Then verify that the length of line segement PZ divided by the length of line segment OZ is 2/3.

Someone care to share a better way?
• June 21st 2010, 06:18 PM
undefined
Okay, I should have focused more on the angle sum identity than on the unit circle aspect. We have this identity for tangent (we could also use the ones for sine and cosine, but this seems faster):

$\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}$

Here, $\tan\alpha = \frac{1}{5}$ and $\tan\beta = \frac{2}{3}$.

So $\tan(\alpha + \beta) = \frac{\frac{1}{5} + \frac{2}{3}}{1 - \frac{1}{5} \cdot \frac{2}{3}}=1$

$\alpha+\beta=\tan^{-1}(1)=\frac{\pi}{4}$

Much simpler than my first post.
• June 21st 2010, 10:42 PM
bigwave
thanks, I see how you got it...

I did go back to the $\cos\left(\alpha + \beta\right)$

we have:

$
\Rightarrow
\frac{5}{\sqrt{26}}\frac{2}{\sqrt{13}}
+\frac{1}{\sqrt{26}}\frac{3}{\sqrt{13}}
\Rightarrow
\frac{10}{13\sqrt{2}}+\frac{3}{13\sqrt{2}}
\Rightarrow
\frac{13}{13\sqrt{2}}
\Rightarrow
\frac{1}{\sqrt{2}} \Rightarrow \frac{\sqrt{2}}{2} = \frac{\pi}{4}
$

looks like just needed to be more carefull