trig identities

Hello, enea54!

Two identities: . $\sin\tfrac{x}{2} \;=\;\sqrt{\tfrac{1-\cos x}{2}} \qquad \cos\tfrac{x}{2} \;=\;\sqrt{\tfrac{1+\cos x}{2}}$

Quote:

$\text{Prove: }\:\frac{1+\cos x}{\sin x} \:=\:\cot\tfrac{x}{2}$

The right side is: . $\cot\tfrac{x}{2} \;\;=\;\;\dfrac{\cos\frac{x}{2}}{\sin\frac{x}{2}} \;\;=\;\; \dfrac{\sqrt{\frac{1+\cos x}{2}}}{\sqrt{\frac{1-\cos x}{2}}}$

. . $=\;\;\sqrt{\dfrac{\frac{1+\cos x}{2}}{\frac{1-\cos x}{2}}} \;\;=\;\;\sqrt{\dfrac{1+\cos x}{1-\cos x}} \;\;=\;\; \sqrt{\frac{1+\cos x}{1-\cos x}\cdot\color{blue}{\frac{1+\cos x}{1+\cos x}}}$

. . $=\;\;\sqrt{\frac{(1+\cos x)^2}{1-\cos^2\!x}} \;\;=\;\; \sqrt{\frac{(1+\cos x)^2}{\sin^2\!x}} \;\;=\;\;\frac{1+\cos x}{\sin x}$