How do you find the value of sin(theta/2) given tantheta=2 and 0<theta<pi/2?

Jul 24, 2016

$\sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{\sqrt{5} - 1}{2 \sqrt{5}}}$

Explanation:

It is observed that in range $0 < \theta < \frac{\pi}{2}$ all trigonometric functions are positive. Further as $\theta$ is in first quadrant, so is $\frac{\theta}{2}$ hence trigonometric ratios of $t \frac{\eta}{2}$ too are all positive.

Now as $\tan \theta = 2$, $\sec \theta = \sqrt{1 + {\tan}^{2} \theta}$

= $\sqrt{1 + {2}^{2}} = \sqrt{5}$

and hence $\cos \theta = \frac{1}{\sqrt{5}}$

Now, $\sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}}$

= $\sqrt{\frac{1 - \frac{1}{\sqrt{5}}}{2}} = \sqrt{\frac{\sqrt{5} - 1}{2 \sqrt{5}}}$

Jul 24, 2016

$\therefore \sin \left(\frac{\theta}{2}\right) = \frac{\left(\sqrt{5} - 1\right) \sqrt{50 + 10 \sqrt{5}}}{20}$

Explanation:

Let us use the Identity : tantheta=(2tan(theta/2))/(1-tan^2(theta/2)

Letting, $\tan \left(\frac{\theta}{2}\right) = t , \tan \theta = \frac{2 t}{1 - {t}^{2}} \ldots \ldots \ldots \ldots . \left(1\right)$.

Since, $\tan \theta = 2$, we have, by $\left(1\right)$,

$2 = \frac{2 t}{1 - {t}^{2}} \Rightarrow 1 - {t}^{2} = t , \mathmr{and} , {t}^{2} + t - 1 + 0$

$t = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 1 \cdot \left(- 1\right)}}{2} = \frac{- 1 \pm \sqrt{5}}{2}$

Let us note that, by data, $0 < , \theta < , \frac{\pi}{2} \Rightarrow 0 < , \frac{\theta}{2} < , \frac{\pi}{4}$

$\therefore \frac{\theta}{2}$ lies in the First Quadrant, in which, $\tan \left(\frac{\theta}{2}\right) > 0 ,$

and, also, $\sin \left(\frac{\theta}{2}\right) > 0$

This means that $\tan \left(\frac{\theta}{2}\right) \ne \frac{- 1 - \sqrt{5}}{2}$

$\therefore \tan \left(\frac{\theta}{2}\right) = \frac{\sqrt{5} - 1}{2} \Rightarrow \cot \left(\frac{\theta}{2}\right) = \frac{2}{\sqrt{5} - 1}$

Now, ${\csc}^{2} \left(\frac{\theta}{2}\right) = 1 + {\cot}^{2} \left(\frac{\theta}{2}\right) = 1 + \frac{4}{\sqrt{5} - 1} ^ 2$

$\therefore {\csc}^{2} \left(\frac{\theta}{2}\right) = \frac{\left(5 - 2 \sqrt{5} + 1\right) + 4}{\sqrt{5} - 1} ^ 2 = \frac{10 - 2 \sqrt{5}}{\sqrt{5} - 1} ^ 2$

$\therefore \sin \left(\frac{\theta}{2}\right) = + \frac{\sqrt{5} - 1}{\sqrt{10 - 2 \sqrt{5}}} = \frac{\left(\sqrt{5} - 1\right) \sqrt{10 + 2 \sqrt{5}}}{\sqrt{10 - 2 \sqrt{5}} \sqrt{10 + 2 \sqrt{5}}}$

$= \frac{\left(\sqrt{5} - 1\right) \sqrt{10 + 2 \sqrt{5}}}{\sqrt{100 - 20}} = \frac{\left(\sqrt{5} - 1\right) \sqrt{10 + 2 \sqrt{5}}}{4 \sqrt{5}}$

$= \frac{\left(\sqrt{5} - 1\right) \left(\sqrt{5}\right) \sqrt{10 + 2 \sqrt{5}}}{4 \cdot 5} = \frac{\left(\sqrt{5} - 1\right) \sqrt{50 + 10 \sqrt{5}}}{20}$