$6.2.10.$ Two infinite planes intersecting at an angle $\alpha$ divide space into four regions. What is the electric field strength in regions 1 and 2 if the surface charge density of the planes is $\pm \sigma$?
Consider the following figure...
It's known that the electric field intensity for one of the faces of a infinite plane with surface charge density is $E = \frac{\sigma}{2\varepsilon_0}$.
For region 1 (see part $a$ of above figure):
$$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})$$
and for $x$-direction, $$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}.$$ So, as $$E_{R1} = \sqrt{E_{R1x}^2+E_{R1y}^2},$$ and taking in account that $$\sin{\alpha} = \sqrt{\frac{1-\cos{\alpha}}{2}},$$ it is obtained
$$E_{R1} = \frac{\sigma}{\varepsilon_0}\sin{\frac{\alpha}{2}}$$
For region 2, for $y$-direction: $$E_{R2y} = E(1+\cos{\alpha})=\frac{\sigma}{2\varepsilon_0}(1+\cos{\alpha})$$ and for $x$-axis, $$E_{R2x} = E\sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}.$$
Again, $$E_{R2} = \sqrt{E_{R2x}^2+E_{R2y}^2}$$ and taking in account that $$\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$$
$$E_{R2} = \frac{\sigma}{\varepsilon_0}\cos{\frac{\alpha}{2}}$$