Question

What is the formula of $\frac{1-\cos x}{\sin x}$?

Answer 1

Determine the formula of $\frac{1-\cos x}{\sin x}$.

It is known that $\mathbf{1}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\left(2\theta \right)\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }$ and $\mathit{s}\mathit{i}\mathit{n}\left(2\theta \right)\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$.

So, $1-\cos x=2{\sin }^2\left(\frac{x}{2}\right)$ and $\sin x=2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)$.

Substitute the values into the expression $\frac{1-\cos x}{\sin x}$and simplify:

$\begin{array}{rcl}\frac{1-\cos x}{\sin x}& =& \frac{2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)}{2\sin \left({\displaystyle \frac{x}{2}}\right)\cos \left({\displaystyle \frac{x}{2}}\right)}\\ & \Rightarrow & \frac{1-\cos x}{\sin x}=\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}\\ & \Rightarrow & \frac{1-\cos x}{\sin x}=\tan \left(\frac{x}{2}\right)\end{array}$

Hence, the formula for $\frac{1-\cos x}{\sin x}\mathrm{is}\tan \left(\frac{x}{2}\right)$.