Question

if t = tan(x/2), -π<x<π, sketch aright triangle or use trigonometric identities to show that
cos(x/2) = 1/sqrt(1+t^2) and sin(x/2) = t/sqrt(1+t^2)

Answer 1

Answer on Question #53984 – Math – Trigonometry

Question

If $t = \tan \left(\frac{x}{2}\right)$

then

$$\cos \left(\frac {x}{2}\right) = \frac {1}{\sqrt {1 + t ^ {2}}}, \qquad \sin \left(\frac {x}{2}\right) = \frac {t}{\sqrt {1 + t ^ {2}}}, \qquad 0 \leq x \leq \pi .$$

Solution

If $0 \leq x \leq \pi$, then $0 \leq \frac{x}{2} \leq \frac{\pi}{2}$ and

$$\tan \left(\frac {x}{2}\right) \geq 0, \quad \cos \left(\frac {x}{2}\right) \geq 0, \quad \sin \left(\frac {x}{2}\right) \geq 0.$$

Rewrite

$$\begin{array}{l} \frac {1}{\sqrt {1 + t ^ {2}}} = \frac {1}{\sqrt {1 + \left(\tan \left(\frac {x}{2}\right)\right) ^ {2}}} = \frac {1}{\sqrt {1 + \left(\frac {\sin \left(\frac {x}{2}\right)}{\cos \left(\frac {x}{2}\right)}\right) ^ {2}}} = \frac {1}{\frac {\sqrt {\left(\cos \left(\frac {x}{2}\right)\right) ^ {2} + \left(\sin \left(\frac {x}{2}\right)\right) ^ {2}}}{\sqrt {\left(\cos \left(\frac {x}{2}\right)\right) ^ {2}}}} = \\ = \frac {1}{\frac {\sqrt {1}}{\cos \left(\frac {x}{2}\right)}} = \cos \left(\frac {x}{2}\right). \\ \end{array}$$

Thus, we have $\frac{1}{\sqrt{1 + t^2}} = \cos \left(\frac{x}{2}\right)$.

Next,

$$\frac {t}{\sqrt {1 + t ^ {2}}} = t \cdot \frac {1}{\sqrt {1 + t ^ {2}}} = \tan \left(\frac {x}{2}\right) \cdot \cos \left(\frac {x}{2}\right) = \frac {\sin \left(\frac {x}{2}\right)}{\cos \left(\frac {x}{2}\right)} \cdot \cos \left(\frac {x}{2}\right) = \sin \left(\frac {x}{2}\right).$$

So

$$\cos \left(\frac {x}{2}\right) = \frac {1}{\sqrt {1 + t ^ {2}}}, \qquad \sin \left(\frac {x}{2}\right) = \frac {t}{\sqrt {1 + t ^ {2}}}, \qquad 0 \leq x \leq \pi$$

where

$$t = \tan \left(\frac {x}{2}\right).$$

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