Question

In Exercises 59–68, verify each identity. cot⁡x2=sin⁡x1−cos⁡x\cot\frac{x}{2}=\frac{\sin x}{1-\cos x}

Accepted Answer

Start by writing down the given identity clearly: $$ \cot \left( \frac{x}{2} ight) = \frac{\sin x}{1 - \cos x} . $$Our goal is to verify this identity, meaning we want to show that both sides are equal for all valid values of $$ x . $$Recall the definition of cotangent in terms of sine and cosine: $$ \cot 	heta = \frac{\cos 	heta}{\sin 	heta} . $$Apply this to the left side: $$ \cot \left( \frac{x}{2} ight) = \frac{\cos \left( \frac{x}{2} ight)}{\sin \left( \frac{x}{2} ight)} . $$Use the half-angle formulas for sine and cosine: 
$$ \sin x = 2 \sin \left( \frac{x}{2} ight) \cos \left( \frac{x}{2} ight) \quad 	ext{and} \quad 1 - \cos x = 2 \sin^2 \left( \frac{x}{2} ight) . $$These will help rewrite the right side in terms of $$ \sin \left( \frac{x}{2} ight) $$ and $$ \cos \left( \frac{x}{2} ight) . $$Substitute these half-angle expressions into the right side: 
$$ \frac{\sin x}{1 - \cos x} = \frac{2 \sin \left( \frac{x}{2} ight) \cos \left( \frac{x}{2} ight)}{2 \sin^2 \left( \frac{x}{2} ight)} . $$Simplify the fraction by canceling common factors. After simplification, you should get $$ \frac{\cos \left( \frac{x}{2} ight)}{\sin \left( \frac{x}{2} ight)} $$, which matches the left side expression for $$ \cot \left( \frac{x}{2} ight) . $$This confirms the identity is true.

In Exercises 59–68, verify each identity. $\cot \frac{x}{2} = \frac{\sin x}{1 - \cos x}$

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Key Concepts

Trigonometric Identities

Cotangent and Its Relationship to Sine and Cosine

Algebraic Manipulation of Trigonometric Expressions

In Exercises 59–68, verify each identity. cotx2=sinx1−cosx\(\cot\]\frac{x}{2}\)=\(\frac{\sin x}{1-\cos x}\)