Solving for $\tan \theta$ given $\sin \theta/2$ QUESTION:
I'm having a hard time figuring this problem out. I've looked through my lectures and cannot find a problem that relates to this one. I do have my identities pulled up in front of me. I'm unsure where to start though. Can someone give me a kick in the right direction. Also could someone give me some rep points so I can format my questions nicer next time. Thanks
PROBLEM:

Find $\tan \theta$ if $\sin(\theta/2) = 3/5$

 A: $$
\sin^2\frac\theta2+\cos^2\frac\theta2=1.
$$
So
$$
\left(\frac35\right)^2+\cos^2\frac\theta2=1.
$$
Given that, you can find that $\cos\frac\theta2=\pm\text{something}$.  Once you've got $\sin\frac\theta2$ and $\cos\frac\theta2$, use the fact that $\sin\theta=2\sin\frac\theta2\cos\frac\theta2$ and $\cos\theta=\cos^2\frac\theta2-\sin^2\frac\theta2$ (double-angle formulas).
Then there's still a "$\pm$" question.
A: Since $\sin(\theta)=2\sin(\theta/2)\cos(\theta/2)$ and $\cos(\theta)=1-2\sin^2(\theta/2)$, we get
$$
\begin{align}
\tan(\theta)
&=\frac{\sin(\theta)}{\cos(\theta)}\\
&=\frac{2\sin(\theta/2)\cos(\theta/2)}{1-2\sin^2(\theta/2)}\\
&=\pm\frac{2\sin(\theta/2)\sqrt{1-\sin^2(\theta/2)}}{1-2\sin^2(\theta/2)}
\end{align}
$$
In your case, $\sqrt{1-(3/5)^2}=4/5$, but the sign of $\cos(\theta/2)$ could be positive or negative, so you might need $\pm4/5$.
A: If it is ok with you, $user 91539$, i will use $x$ instead of $\theta$ and assume that $x=\theta$
Since we know:
$$1. tan(x)=\frac{sin(x)}{cos(x)}$$
$$2. cos^2(x)=1-sin^2(x)$$
We can derive the following by solving Equation [2] for $cos(x)$:
$$3. cos(x)=\pm\sqrt{1-sin^2(x)}$$
And by substituting Equation[3] into Equation[1], we may arrive to the conclusion that:
$$4. tan(x)=\frac{sin(x)}{\pm\sqrt{1-sin^2(x)}}$$
As well, since if $\left[f(x)=g(x)\right]$, then $\left[f(a)=g(a)\right]$
Therefore:
$$5. tan\left(\frac{x}{2}\right)=\frac{sin\left(\frac{x}{2}\right)}{\pm\sqrt{1-sin^2\left(\frac{x}{2}\right)}}$$
So we can begin substituting $sin(x/2)=\frac{3}{5}$ into Equation[5]:
$$tan\left(\frac{x}{2}\right)=\frac{\frac{3}{5}}{\pm\sqrt{1-\frac{9}{25}}}$$
$$tan\left(\frac{x}{2}\right)=\pm\frac{3}{5\sqrt{\frac{16}{25}}}$$
$$tan\left(\frac{x}{2}\right)=\pm\frac{3}{5(4/5)}$$
$$tan\left(\frac{x}{2}\right)=\pm\frac{3}{4}$$
Therefore, let us say, $\alpha=\frac{x}{2}$ and rewrite the equation as
$$tan(\alpha)=\pm\frac{3}{4}$$
We can evaluate this to
$$\alpha=\pm tan^{-1}(3/4)+\pi n , n\in Z$$
Therefore, since $\alpha=\frac{x}{2}$,
$$\frac{x}{2}=\pm tan^{-1}(3/4)+\pi n , n\in Z$$
$$x=\pm 2tan^{-1}(3/4)+2\pi n, n\in Z$$
$$tan(x)=tan(\pm 2tan^{-1}(3/4))+2\pi n, n\in Z$$
So, in general, $tan(x)$ is about
$$ tan(x)\dot{=}\pm 3.4286+6.2831n, n\in Z$$
From 0 to $2 \pi$, 
$$tan(x)\dot{=}3.4286$$
A: Using only geometry:


*

*The triangle $\Delta ABC$ is Pythagorean with sides of length $3,4,5$. 

*The triangles  $\Delta AB'C$ and $\Delta ABC$ are congruent.

*The angle $\angle BCA$ is $\theta/2$.

*The angles $\angle BCB'$ and $\angle DAB'$ are equal to $\theta$.


Can you take it from here?

A: Since we're given $\sin\frac{\theta}{2}=\frac{3}{5}$, we can use the Half-Angle Formula for Sine to write $\tan\theta$ as a function of $\sin\frac{\theta}{2}$.
Now normally the Half-Angle Formula for Sine is $\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$, with the $\pm$ being dependent on which quadrant the angle $\frac{\theta}{2}$ lies within.  However, for sake of simplicity, we'll just skip to starting with the equation $\left|\sin\frac{\theta}{2}\right|=\sqrt{\frac{1-\cos\theta}{2}}$ instead.

$\left|\sin\frac{\theta}{2}\right|=\sqrt{\frac{1-\cos\theta}{2}}$
$\left(\left|\sin\frac{\theta}{2}\right|\right)^2=\left(\sqrt{\frac{1-\cos\theta}{2}}\right)^2$
$\sin^2\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{2}$
$2\sin^2\left(\frac{\theta}{2}\right)=1-\cos\theta$
$\cos\theta=1-2\sin^2\left(\frac{\theta}{2}\right)$
$\sec\theta=\frac{1}{1-2\sin^2\left(\frac{\theta}{2}\right)}$
$\sec^2\theta=\frac{1}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}$
$\sec^2\theta-1=\frac{1}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}-1$
$\tan^2\theta=\frac{1}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}-\frac{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}$
$\tan^2\theta=\frac{1-\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}$
And in the end we get
$\tan\theta=\pm\frac{\sqrt{1-\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}}{1-2\sin^2\left(\frac{\theta}{2}\right)}$
Note that the $\pm$ is still dependent on quadrant.
From here, we can simply substitute in for $\sin\frac{\theta}{2}$ with $\frac{3}{5}$.
$\tan\theta=\pm\frac{\sqrt{1-\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}}{1-2\sin^2\left(\frac{\theta}{2}\right)}$
$\tan\theta=\pm\frac{\sqrt{1-\left[1-2\left(\frac{3}{5}\right)^2\right]^2}}{1-2\left(\frac{3}{5}\right)^2}$
$\tan\theta=\pm\frac{\sqrt{674}}{7}$
