Cotangent half-angle identity

 

Statement of the identity

$$\boxed{\displaystyle \cot\!\left(\tfrac{\theta}{2}\right)=\frac{1+\cos\theta}{\sin\theta}}$$

This relates the cotangent of half an angle to the sine and cosine of the whole angle (with $\theta\not\equiv 2k\pi$ so $\sin\theta\neq0$).

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Proof using double-angle formulas

1. Start from double-angle expressions

   $$\sin\theta \;=\; 2\sin\!\bigl(\tfrac{\theta}{2}\bigr)\cos\!\bigl(\tfrac{\theta}{2}\bigr), \quad 1+\cos\theta \;=\; 2\cos^{2}\!\bigl(\tfrac{\theta}{2}\bigr).$$

2. Form the ratio $\dfrac{1+\cos\theta}{\sin\theta}$

   $$\frac{1+\cos\theta}{\sin\theta} \;=\; \frac{2\cos^{2}\!\bigl(\tfrac{\theta}{2}\bigr)}{2\sin\!\bigl(\tfrac{\theta}{2}\bigr)\cos\!\bigl(\tfrac{\theta}{2}\bigr)} \;=\; \frac{\color{royalblue}{\cancel{2}}\cos^{\cancel{2}}\!\bigl(\tfrac{\theta}{2}\bigr)} {\color{royalblue}{\cancel{2}}\sin\!\bigl(\tfrac{\theta}{2}\bigr)\cos\!\bigl(\tfrac{\theta}{2}\bigr)} \;=\; \frac{\cos\!\bigl(\tfrac{\theta}{2}\bigr)}{\sin\!\bigl(\tfrac{\theta}{2}\bigr)}.$$

3. Recognize the cotangent

   $$\frac{\cos\!\bigl(\tfrac{\theta}{2}\bigr)}{\sin\!\bigl(\tfrac{\theta}{2}\bigr)} \;=\; \cot\!\bigl(\tfrac{\theta}{2}\bigr).$$

Therefore

$$\boxed{\displaystyle \cot\!\left(\tfrac{\theta}{2}\right)=\frac{1+\cos\theta}{\sin\theta}}.$$

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Notes & alternatives

• Domain caveat: The identity requires $\sin\theta\neq0$ (no division by zero), which excludes integer multiples of $\pi$.

• Alternate proof: Starting from the Pythagorean identity $\cot^{2}x+1=\csc^{2}x$, substitute $x=\theta/2$ and use $\csc(\theta/2)=\dfrac{1}{\sin(\theta/2)}=\dfrac{2\sin(\theta/2)\cos(\theta/2)}{\sin\theta}$ to reach the same result, though the double-angle route above is the most direct.

• Geometric viewpoint: On the unit circle, draw a diameter perpendicular to the initial side of angle $\theta$; similar right triangles give the same ratio $\cot(\theta/2)$.

This completes both the statement and its proof.