We start by deriving the angle sum formulas and from there we can derive the other formulas.
Typically the angle sum formulas are proved with geometry but I will do a derivation using Euler's formula:
$${e^{i\theta} = \cos(\theta) + i \ \sin(\theta)}$$
But we want this in terms of the sum of 2 angles that we will call a and b:
$${e^{i(a+b)} = \cos(a+b) + i \ \sin(a+b)}$$
Using the rules of exponents this can be written differently:
$${e^{i(a+b)} = e^{ia} e^{ib} = (\cos(a) + i \ \sin(a))(\cos(b) + i \ \sin(b))}$$
Then we can multiply out the right side by distributing terms:
$${(\cos(a) + i \ \sin(a))(\cos(b) + i \ \sin(b)) = \cos(a)\cos(b) + i \ \sin(a)\cos(b) + i \ \sin(b)\cos(a) + i^2 \sin(a)\sin(b)}$$
Simplify and combine like terms:
$${\cos(a)\cos(b) - \sin(a)\sin(b) + i (\sin(a)\cos(b) + \sin(b)\cos(a))}$$
Now compare this with the formula we found originally:
$${\cos(a+b) + i \ \sin(a+b)}$$
Equate the real part and imaginary parts and we get:
$${\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)}$$
$${\sin(a+b) = \sin(a)\cos(b) + \sin(b)\cos(a)}$$
And those are the sum of angle formulas for sine and cosine!
Now finding the double angle formulas for sine and cosine is easy! Let's use the sum of angles formula but set \({b = a}\) and substitute.
$${\cos(a+a) = \cos(2a) = \cos(a)\cos(a) - \sin(a)\sin(a) = \cos^2(a) - \sin^2(a)}$$
$${\sin(a+a) = \sin(2a) = \sin(a)\cos(a) + \sin(a)\cos(a) = 2 \ \sin(a)\cos(a)}$$
We can then find a few variations on the cosine formula using the Pythagorean theorem:
$${\sin^2(a) + \cos^2(a) = 1}$$
Which can be also written as follows:
$${\sin^2(a) = 1 - \cos^2(a)}$$
$${\cos^2(a) = 1 - \sin^2(a)}$$
Then substituting into our double angle formula for cosine we end up with 3 formulas:
$${\cos(2a) = \cos^2(a) - \sin^2(a)}$$
$${\cos(2a) = 1 - 2 \ \sin^2(a)}$$
$${\cos(2a) = 2 \ \cos^2(a) - 1}$$
Finally we need a double angle formula for tangent:
$${\tan(2a) = \frac{\sin(2a)}{\cos(2a)} = \frac{2 \ \sin(a)\cos(a)}{\cos^2(a) - \sin^2(a)}}$$
To get this strictly in terms of tangent multiply numerator and denominator by \({\large{\frac{1}{\cos^2(a)}}}\):
$${\tan(2a) = \frac{2 \ \tan(a)}{1 - \tan^2(a)}}$$
The power reduction formulas can all be found by using the double angle formulas and rearranging the equations.
Use this formula but solve for \({\cos^2(a)}\)
$${\cos(2a) = 2 \ \cos^2(a) - 1}$$
The result will be:
$${\cos^2(a) = \frac{1}{2}(1 + \cos(2a))}$$
Use this formula but solve for \({\sin^2(a)}\)
$${\cos(2a) = 1 - 2 \ \sin^2(a)}$$
The result will be:
$${\sin^2(a) = \frac{1}{2}(1 - \cos(2a))}$$
Power reduction for tangent can be derived from the formulas for sine and cosine:
$${\tan^2(a) = \frac{\sin^2(a)}{\cos^2(a)} = \frac{\frac{1}{2}(1 - \cos(2a))}{ \frac{1}{2}(1 + \cos(2a))} = \frac{1 - \cos(2a)}{1 + \cos(2a)}}$$
The half angle formulas can be derived from the following Power reduction formulas:
$${\cos^2(a) = \frac{1}{2}(1 + \cos(2a))}$$
$${\sin^2(a) = \frac{1}{2}(1 - \cos(2a))}$$
Taking the square root on both sides you have:
$${\cos(a) = \pm \sqrt{\frac{1}{2}(1 + \cos(2a))}}$$
$${\sin(a) = \pm \sqrt{\frac{1}{2}(1 - \cos(2a))}}$$
Then make a simple substitution for half of the angles on both sides:
$${\cos\left(\frac{a}{2}\right) = \pm \sqrt{\frac{1}{2}(1 + \cos(a))}}$$
$${\sin\left(\frac{a}{2}\right) = \pm \sqrt{\frac{1}{2}(1 - \cos(a))}}$$
For tangent we can substitute the values above and simplify:
$${\tan\left(\frac{a}{2}\right) = \frac{\sin(\frac{a}{2})}{\cos(\frac{a}{2})} = \pm \sqrt{\frac{1 - \cos(a)}{1 + \cos(a)} } }$$
Then multiply numerator and denominator by the conjugate \({1 - \cos(a)}\)
$${\pm \sqrt{\frac{(1 - \cos(a))(1 - \cos(a))}{(1 + \cos(a))(1 - \cos(a))} } = \pm \sqrt{\frac{(1 - \cos(a))^2}{\sin^2(a)} }}$$
$${\pm \sqrt{\frac{(1 - \cos(a))^2}{\sin^2(a)} } = \pm \frac{|1 - \cos(a)|}{|\sin(a)|} }$$
But notice that in every case:
$${1 - \cos(a) \ge 0}$$
So we can drop the absolute value in the numerator which yields:
$${\pm \frac{1 - \cos(a)}{|\sin(a)|}}$$
But what do we do with this plus-minus and the absolute value??? We can evaluate this by Quadrant.
When the half angle of tangent is in Quadrants 1 & 3 we must use the plus sign. In quadrants 2 & 4 use the negative sign.
Then for the absolute value, whenever the value of sine is positive (Quadrants 1 & 2) we can drop the absolute value. When sine is negative (Quadrants 3 & 4) we need to add a negative sign.
So what we can do is evaluate this case by case. It turns out that when the half angle is in Quadrants 1 & 3 then the full angle is always in Quadrants 1 & 2.
Therefore in those cases we drop the absolute value and use the plus sign.
When the half angle is in Quadrants 2 & 4 we will use the negative sign and the full angle will be in Quadrants 3 & 4.
But in that case sine is negative so we will also need a negative sign on the sine.
But the 2 negative signs will cancel and that is why we end up with one formula to cover all cases without need for plus-minus or absolute value signs.
$${\tan\left(\frac{a}{2}\right) = \frac{1 - \cos(a)}{\sin(a)} = \frac{\sin(a)}{1 + \cos(a)} }$$