Cot 2x Formula

What is the formula for Cot 2x ?

Cot 2x is equal to (Cot² x - 1)/2.Cot x.

Proof of Cot 2x's Formula

In the right triangle ABC above;

$AC=1$

$AB=\sin x$

$BC=\cos x$

$AD=DC=a$

$BD=\cos x-a$'dır.

For triangle ABC;

$(AB{)}^2+(BC{)}^2=(AC{)}^2$

${\sin }^{2}x+{\cos }^{2}x=1^2$

${\sin }^{2}x+{\cos }^{2}x=1$

For triangle ABD;

$(AB{)}^2+(BD{)}^2=(AD{)}^2$

${\sin }^{2}x+(\cos x-a{)}^2=a^2$

${\sin }^{2}x+{\cos }^{2}x-2.\cos x.a+a^2=a^2$

${\sin }^{2}x+{\cos }^{2}x-2.\cos x.a+\cancel{a^2}-\cancel{a^2}=0$

$1-2.\cos x.a=0$

$-2.\cos x.a=-1$

$\frac{\cancel{-2.\cos x}.a}{\cancel{-2.\cos x}}=\frac{-1}{-2.\cos x}$

$a=\frac{1}{2.\cos x}$

In the ABD triangle;

$\cot 2x=\frac{BD}{AB}=\frac{\cos x-a}{\sin x}$

$\cot 2x=\frac{\cos x-\frac{1}{2.\cos x}}{\sin x}$

$\cot 2x=\frac{\frac{2.\cos x.\cos x-1}{2.\cos x}}{\sin x}$

$\cot 2x=\frac{\frac{2.{\cos }^{2}x-1}{2.\cos x}}{\sin x}$

$\cot 2x=\frac{2.{\cos }^{2}x-1}{2.\cos x.\sin x}$

$\cot 2x=\frac{2.{\cos }^{2}x-({\cos }^{2}x+{\sin }^{2}x)}{2.\sin x.\cos x}$

$\cot 2x=\frac{2.{\cos }^{2}x-{\cos }^{2}x-{\sin }^{2}x}{2.\sin x.\cos x}$

$\cot 2x=\frac{{\cos }^{2}x-{\sin }^{2}x}{2.\sin x.\cos x}$

$\cot 2x=\frac{{\cos }^{2}x}{2.\sin x.\cos x}-\frac{{\sin }^{2}x}{2.\sin x.\cos x}$

$\cot 2x=\frac{\cancel{\cos x}.\cos x}{2.\sin x.\cancel{\cos x}}-\frac{\cancel{\sin x}.\sin x}{2.\cancel{\sin x}.\cos x}$

$\cot 2x=\frac{1}{2}.\frac{\cos x}{\sin x}-\frac{1}{2}.\frac{\sin x}{\cos x}$

$\cot 2x=\frac{1}{2}.(\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x})$

$\boxed{\cot x=\frac{\cos x}{\sin x}}$

$\cot 2x=\frac{1}{2}.(\cot x-\frac{1}{\cot x})$

$\cot 2x=\frac{1}{2}.\frac{\cot x.\cot x-1}{\cot x}$

$\cot 2x=\frac{1}{2}.\frac{{\cot }^{2}x-1}{\cot x}$

$\cot 2x=\frac{{\cot }^{2}x-1}{2.\cot x}$