I'm always forgetting these even though
I need them from time to time. Fortunately, I have ben able to re-derive
them as needed.

We start both derivations with a given angle `a`. Let
`b = a/2`. Thus, `2b=a`

**Cosine**
From that,

`cos a = cos 2b`

` = cos`^{2}b - sin^{2}b

` = cos`^{2}b - sin^{2}b
+ 1 - 1

` = cos`^{2}b - sin^{2}b
+ (cos^{2}b + sin^{2}b) - 1

` = 2cos`^{2}b - 1

so

`2cos`^{2}b = cos a + 1

`cos`^{2}b = (cos a + 1) / 2

`cos b = sqrt ((cos a + 1) / 2)`

that is,

`cos (a/2) = sqrt ((cos a + 1) / 2)`

**Sine**
From above, we note that

`cos`^{2}b = (cos a + 1) / 2

so

`cos`^{2}b - 1 = (cos a + 1) / 2 - 1

`cos`^{2}b - cos^{2}b - sin^{2}b = (cos
a + 1) / 2 - 2 / 2

`-sin`^{2}b = (cos a + 1 - 2) / 2

` = (cos a - 1) / 2`

`sin`^{2}b = (1 - cos a) / 2

`sin b = sqrt ((1 - cos a) / 2)`

that is,

`sin (a/2) = sqrt ((1 - cos a) / 2)`

The other formulas are easily derived from the sine and cosine formulas.

`tan (a/2) = sin (a/2)/cos(a/2) = sqrt ((1 - cos a)/(1+cos a))`

`cot (a/2) = cos (a/2)/sin(a/2) = sqrt ((1+ cos a)/(1-cos a))`

`sec (a/2) = 1/cos(a/2) = sqrt (2/(1+cos a))`

`csc (a/2) = 1/sin(a/2) = sqrt (2/(1-cos a))`