If all the bands are of equal width and intensity, then pattern is produced by double slit. Bands of the pattern produced by single slit have the width and intensity both varying.
(i) Derivation of expression for angular position of bright fringe produced by single slit diffraction:

The single slit is now divided in three equal parts.
If waves from two parts of the slit cancel each other, the wave from the third part will produce a maxima at a point between two minima.
So, $\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_1=\frac{3\mathit{\lambda }}{2\mathit{a}}$
Similarly, if the slit is divided in five equal parts, then another maxima will be produced at
$\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_2=\frac{5\mathit{\lambda }}{2\mathit{a}}$
Similarly for other fringes,
$\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathit{n}}=\frac{\left(2\mathit{n}+1\right)\mathit{\lambda }}{2\mathit{a}}$
Or, ${\mathit{\theta }}_{\mathit{n}}=\frac{\left(2\mathit{n}+1\right)\mathit{\lambda }}{2\mathit{a}}$
For central maxima, $\mathit{\theta }=0°$
(ii) Derivation of expression for angular position of dark fringe produced by single slit diffraction:

The single slit is divided into two equal halves. For every point in one half has a corresponding point in the other half. The path difference between two waves arriving at point $\mathit{P}$ is
$\frac{\mathit{a}}{2\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_1}=\frac{\mathit{\lambda }}{2}$
This means the contributions are in opposite phase, so they cancel each other and the intensity falls to zero.
So, far 1st dark fringe, $\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_1=\frac{\mathit{\lambda }}{\mathit{a}}$
Similarly for other dark fringes,
$\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathit{n}}=\frac{\mathit{n}\mathit{\lambda }}{\mathit{a}}$
${\mathit{\theta }}_{\mathit{n}}=\frac{\mathit{n}\mathit{\lambda }}{\mathit{a}}$