# How do you find the exact value of arctan(1/2)+arctan(1/3)?

May 14, 2015

Suppose $\alpha = \arctan \left(\frac{1}{2}\right)$ and $\beta = \arctan \left(\frac{1}{3}\right)$

What is the value of $\tan \left(\alpha + \beta\right)$ ?

It's probably easiest to calculate $\sin \alpha$, $\cos \alpha$, $\sin \beta$ and $\cos \beta$ first:

If $\alpha = \arctan \left(\frac{1}{2}\right)$, then $\frac{1}{2} = \tan \alpha = \sin \frac{\alpha}{\cos} \alpha$.

Multiplying through by $2 \cos \alpha$ we get

$\cos \alpha = 2 \sin \alpha$

Squaring both sides and using ${\sin}^{2} \alpha + {\cos}^{2} \alpha = 1$ we get

$4 {\sin}^{2} \alpha = {\cos}^{2} \alpha = 1 - {\sin}^{2} \alpha$

Adding ${\sin}^{2} \alpha$ to both sides and dividing by 5 we get

${\sin}^{2} \alpha = \frac{1}{5}$

So $\sin \alpha = \frac{1}{\sqrt{5}}$ and $\cos \alpha = 2 \sin \alpha = \frac{2}{\sqrt{5}}$.

Similarly, we can find $\sin \beta = \frac{1}{\sqrt{10}}$ and $\cos \beta = \frac{3}{\sqrt{10}}$.

Now we can calculate

$\tan \left(\alpha + \beta\right) = \sin \frac{\alpha + \beta}{\cos} \left(\alpha + \beta\right)$

$= \frac{\sin \alpha \cos \beta + \sin \beta \cos \alpha}{\cos \alpha \cos \beta - \sin \alpha \sin \beta}$

The numerator:

$\sin \alpha \cos \beta + \sin \beta \cos \alpha$

$= \left(\frac{1}{\sqrt{5}}\right) \left(\frac{3}{\sqrt{10}}\right) + \left(\frac{1}{\sqrt{10}}\right) \left(\frac{2}{\sqrt{5}}\right)$

$= \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}}$

The denominator:

$\cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\left(\frac{2}{\sqrt{5}}\right) \left(\frac{3}{\sqrt{10}}\right) - \left(\frac{1}{\sqrt{5}}\right) \left(\frac{1}{\sqrt{10}}\right)$

$= \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}}$

So putting these together:

$\tan \left(\alpha + \beta\right) = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1$

So $\alpha + \beta = \frac{\pi}{4}$