# Question #1fcac

Nov 8, 2016

graph{(abs(x^2-1)-y)(abs(x^2-3)-y)((x-sqrt2)^2+(y-1)^2-0.002)=0 [-1, 4, -0.5, 2]}

$\alpha = \pi - 2 \arctan 2 \sqrt{2}$

#### Explanation:

The curves can intersect only when the arguments signs are opposite so

${x}^{2} - 1 = 3 - {x}^{2} \setminus \setminus \setminus \setminus \implies \setminus \setminus \setminus \setminus x = \pm \sqrt{2}$

${x}_{0} = \sqrt{2}$

${f}_{1 , 2}^{'} \left({x}_{0}\right) = \pm 2 \sqrt{2}$

So the obtuse angle is

$\beta = \arctan 2 \sqrt{2} - \arctan \left(- 2 \sqrt{2}\right) = 2 \arctan 2 \sqrt{2}$

then

$\alpha = \pi - 2 \arctan 2 \sqrt{2}$

Now try to convert it into a single arctan value.

$\tan \left(\alpha\right) = - \tan \left(2 \arctan 2 \sqrt{2}\right)$

Now remember duplication formula for tan.

$\tan \left(2 x\right) = \frac{2 \tan x}{1 - {\tan}^{2} x}$

$\tan \left(\alpha\right) = - \frac{2 \tan \left(\arctan 2 \sqrt{2}\right)}{1 - {\tan}^{2} \left(\arctan 2 \sqrt{2}\right)} = - \frac{2 \cdot 2 \sqrt{2}}{1 - 4 \cdot 2} = \frac{4 \sqrt{2}}{7}$

So

$\alpha = \arctan \left(\frac{4 \sqrt{2}}{7}\right)$