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I'm making no headway in trying to prove that $$\frac{\sin \theta}{1+ \cos \theta} = \frac{\sin (\theta / 2)}{\cos (\theta /2)}.$$ I know that $$\frac{\sin (\theta / 2)}{\cos (\theta /2)} = \tan (\theta /2) = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}},$$ but I'm pretty stuck. Any help would be appreciated.

  • From Wikipedia article inscribed angle: "The inscribed angle theorem (Also known as Star Trek Lemma) relates the measure of an inscribed angle to that of the central angle subtending the same arc. ". Apply this to an inscribed angle with one chord being a diameter. Let $\theta$ be the central angle. Then $\theta/2$ is the inscribed angle. – Somos Aug 13 '21 at 19:30

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According to the identities

\begin{align*} \begin{cases} \sin(2x) = 2\sin(x)\cos(x)\\\\ \cos(2x) = 2\cos^{2}(x) - 1 \end{cases} \end{align*}

we arrive at the desired identity as follows: \begin{align*} \frac{\sin(\theta)}{1 + \cos(\theta)} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^{2}(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)} \end{align*}

Hopefully this helps!

user0102
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Note that, \begin{align} \frac{\sin\theta}{1+\cos \theta } =& \frac{2\cdot \sin(\theta /2)\cdot\cos(\theta/2)} {1+\cos(\theta/2)^2-\sin(\theta/2)^2} \\ =& \frac{2\cdot \sin(\theta /2)\cdot\cos(\theta/2)} {\cos(\theta/2)^2+\sin(\theta/2)^2+\cos(\theta/2)^2-\sin(\theta/2)^2} \\ =& \frac{2\cdot \sin(\theta /2)\cdot\cos(\theta/2)} {2\cdot \cos(\theta/2)^2} \\ =& \frac{\sin(\theta /2)} { \cos(\theta/2)} \\ \end{align}

Elias Costa
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HINT: $\sin\theta=2\sin(\theta/2)\cos(\theta/2)$, $\cos\theta=\cos^2(\theta/2)-\sin^2(\theta/2)$.

Przemysław Scherwentke
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