What is the formula for 1+sin theta?

There will be several formulas of expressing [math]1+sin\theta[/math]

Firstly [math]sin\theta \implies cos\theta[/math]

[math]sin^2\theta=1-cos^2\theta[/math]

[math]\implies sin\theta=\sqrt{1-cos^2\theta}[/math]

Adding 1 to both sides:-

[math]\implies 1+sin\theta=\sqrt{1-cos^2\theta} +1[/math]

EDIT (Added few more formulas) :

[math]1+sin\theta=sin^2(\theta/2)+cos^2(\theta/2)+sin\theta[/math]

[math]\implies sin^2(\theta/2)+cos^2(\theta/2)+2sin(\theta/2)cos(\theta/2)[/math]

[math]\implies (sin(\theta/2)+cos(\theta/2))^2[/math]

Here is another approach :

[math]1+sin\theta=sin(\pi/2)+sin\theta[/math]

Using [math]sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2][/math]

[math]\implies 2sin(\pi/4+\theta/2)cos(\pi/4-\theta/2)[/math]

Want more?

[math]1+sin\theta=1+\dfrac{2tan(\theta/2)}{1+tan^2(\theta/2)}[/math]

[math]\implies 1+sin\theta=\dfrac{(1+tan(\theta/2))^2}{1+tan^2(\theta/2)}[/math]

[math]\implies (\dfrac{1+tan(\theta/2)}{sec(\theta/2)})^2[/math]

OR

[math]\implies (cos(\theta/2)(1+tan(\theta/2))^2[/math]

Hope it helps.