Prove that the half-angle identity for tangent can be expressed as \( \tan \left(\frac{x}{2}\right) = \frac{1 - \cos x}{\sin x} \) starting from the definitions of sine and cosine.

To prove the half-angle identity for tangent: \[ \tan\left(\frac{x}{2}\right) = \frac{1 - \cos x}{\sin x}, \] we'll start from the fundamental definitions of sine and cosine and use some basic trigonometric identities. Here's a step-by-step derivation: ### Step 1: Express \(\tan\left(\frac{x}{2}\right)\) in Terms of Sine and Cosine By definition, the tangent of an angle is the ratio of its sine to its cosine. Therefore: \[ \tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} \] ### Step 2: Use the Double-Angle Formulas for Sine and Cosine Recall the double-angle formulas for sine and cosine: \[ \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \] \[ \cos x = \cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right) = 1 - 2 \sin^2\left(\frac{x}{2}\right) \] From the second identity, we can solve for \(\sin^2\left(\frac{x}{2}\right)\): \[ \sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos x}{2} \] Similarly, solve for \(\cos^2\left(\frac{x}{2}\right)\): \[ \cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos x}{2} \] ### Step 3: Substitute Back into the Expression for \(\tan\left(\frac{x}{2}\right)\) Using the expressions for \(\sin\left(\frac{x}{2}\right)\) and \(\cos\left(\frac{x}{2}\right)\), we have: \[ \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{2}} \] \[ \cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \cos x}{2}} \] Thus, \[ \tan\left(\frac{x}{2}\right) = \frac{\sqrt{\frac{1 - \cos x}{2}}}{\sqrt{\frac{1 + \cos x}{2}}} = \sqrt{\frac{1 - \cos x}{1 + \cos x}} \] ### Step 4: Rationalize the Expression To eliminate the square root, multiply the numerator and denominator by the conjugate \(\sqrt{1 - \cos x}\): \[ \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{1 - \cos x}{\sin x} \] Here’s why: 1. Note that \(1 - \cos^2 x = \sin^2 x\), which implies \(\sin x = \sqrt{1 - \cos^2 x}\). 2. Therefore, \[ \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{1 - \cos x}{\sqrt{(1 + \cos x)(1 - \cos x)}} = \frac{1 - \cos x}{\sin x} \] ### Conclusion Putting it all together, we have shown that: \[ \tan\left(\frac{x}{2}\right) = \frac{1 - \cos x}{\sin x} \] This completes the proof of the half-angle identity for tangent.