# How do you use the half angle formula to solve Tan( θ/2) = Sin θ?

Sep 16, 2016

$\theta = 2 k \pi \mathmr{and} \theta = \left(2 k \pm \frac{1}{2}\right) \pi , k = 0 , 1 , 2 , 3 , \ldots$.

In $\left[0 , 2 \pi\right] , \theta = 0 , \frac{\pi}{2} , \frac{3}{2} \pi \mathmr{and} 2 \pi$.

#### Explanation:

Use $\sin \theta = \frac{2 \tan \left(\frac{\theta}{2}\right)}{1 + {\tan}^{2} \left(\frac{\theta}{2}\right)}$

Here,

tan(theta/2)=sin theta = (2 tan (theta/2))/(1+tan^2(theta/2)

So, tan( theta/2)((1+tan^2(theta/2)-2)=0

And so, $\tan \left(\frac{\theta}{2}\right) = 0$ gives $\frac{\theta}{2} = k \pi , k = 0 , 1 , 2 , 3 , . .$

the other factor = 0 gives tan^(theta/2)=1 to tan(theta/2)=+-1 to

theta/2=kpi+-pi/4.

Combining both for values of $\theta$,.

$\theta = 2 k \pi \mathmr{and} \theta = \left(2 k \pm \frac{1}{2}\right) \pi , k = 0 , 1 , 2 , 3 , \ldots$

Sep 16, 2016

pi/4; (3pi)/4; (5pi)/4; (7pi)/4

#### Explanation:

Use the trig identity
sin a = 2sin (a/2).cos (a/2)
Re-write the equation:
$\frac{\sin \left(\frac{t}{2}\right)}{\cos \left(\frac{t}{2}\right)} = 2 \sin \left(\frac{t}{2}\right) . \cos \left(\frac{t}{2}\right)$
Simplify both sides by sin (t/2(, we get:
$\frac{1}{\cos \left(\frac{t}{2}\right)} = 2 \cos \left(\frac{t}{2}\right)$
$2 {\cos}^{2} \left(\frac{t}{2}\right) = 1$
${\cos}^{2} \left(\frac{t}{2}\right) = \frac{1}{2}$
$\cos \left(\frac{t}{2}\right) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$
Trig table of special arcs, and trig unit circle give 4 solution arcs:
a. $\cos x = \frac{\sqrt{2}}{2}$ --> $x = \pm \frac{\pi}{4}$
Two solution arcs:
$x = \frac{\pi}{4}$ and $x = \frac{7 \pi}{4}$ (arc co-terminal to arc $\left(- \frac{\pi}{4}\right)$)
b. $\cos x = - \frac{\sqrt{2}}{2}$ --> $x = \pm \frac{3 \pi}{4}$
Two solution arcs:
$x = \frac{3 \pi}{4}$, and $x = \frac{5 \pi}{4}$ (arc co-terminal to arc $\frac{- 3 \pi}{4}$)
Answers for $\left(0 , 2 \pi\right)$
pi/4; (3pi)/4, (5pi)/4, (7pi)/4