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 $\sin(\theta/2) = 3/5$
$$ \sin^2\frac\theta2+\cos^2\frac\theta2=1. $$ 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.
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 $\pm4/5$.
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 $\frac{3}{5}$.
$\tan\theta=\pm\frac{\sqrt{1-\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}}{1-2\sin^2\left(\frac{\theta}{2}\right)}$
$\tan\theta=\pm\frac{\sqrt{1-\left[1-2\left(\frac{3}{5}\right)^2\right]^2}}{1-2\left(\frac{3}{5}\right)^2}$
$\tan\theta=\pm\frac{\sqrt{674}}{7}$
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 $2 \pi$, $$tan(x)\dot{=}3.4286$$
Using only geometry:
Can you take it from here?
