Multiple Angle formulas, alternate forms Relatively simple question, that might not be simple to answer:
I have noticed that there are ways of expressing every double angle formula of a given trigonometric function using only that function except for $\sin$ and $\csc$. That is,
$\sin2\theta=2\sin\theta\cos\theta=?$
$\cos2\theta=2\cos^2\theta-1$
$\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$
$\csc2\theta=\dfrac{1}{2}\csc\theta\sec\theta=?$
$\sec2\theta=\dfrac{\sec^2\theta}{2-\sec^2\theta}$
$\cot2\theta=\dfrac{\cot^2\theta-1}{2\cot\theta}$
For triple angle formulas, all 6 trig functions have expressions using only the given trig function. They are
$\sin3\theta=3\sin\theta-4\sin^3\theta$
$\cos3\theta=4\cos^{3}\theta-3\cos\theta$
$\tan3\theta=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$
$\csc3\theta=\dfrac{\csc^3\theta}{3\csc^2\theta-4}$
$\sec3\theta=\dfrac{\sec^3\theta}{4-3\sec^2\theta}$
$\cot3\theta=\dfrac{3\cot\theta-\cot^3\theta}{1-3\cot^2\theta}$
My question is: What are the the formulas, provided they exist, for $\sin2\theta$ and $\csc2\theta$ in terms of $\sin\theta$ and $\csc\theta$ respectively. If they do not exist, then some explanation as to why it is not possible would be most insightful.
Edit: I appreciate the answers so far, but what I really wanting to know is: Is there a known closed form (no piecewise-defined function) expression for the given expressions. Or if not, how to show that there is no such expression?
 A: A formula using no trig functions other than the sine, and
without the ambiguity of symbols such as $\pm$, is
$$
\sin(2\theta) = \begin{cases}
2\sin\theta \sqrt{1-\sin^2\theta} &
 \text{if $2n\pi-\frac\pi2 \leq \theta \leq 2n\pi+\frac\pi2$
for some integer $n$,} \\
-2\sin\theta \sqrt{1-\sin^2\theta} & \text{otherwise}.
\end{cases}
$$
The reason this works is that 
$\sqrt{1-\sin^2\theta} = \lvert \cos\theta\rvert$,
which is $\cos\theta$ when $\cos\theta \geq 0$, which occurs
whenever $2n\pi-\frac\pi2 \leq \theta \leq 2n\pi+\frac\pi2$
for some integer $n$;
but for any other $\theta$, $\cos\theta < 0$ and therefore
$\lvert\cos\theta\rvert = -\cos\theta$.
In fact, this formula is really just a combination of the identity
$\sin(2\theta) = 2\sin\theta\cos\theta$ with the identity
$$
\cos\theta = \begin{cases}
\sqrt{1-\sin^2\theta} &
 \text{if $2n\pi-\frac\pi2 \leq \theta \leq 2n\pi+\frac\pi2$
for some integer $n$,} \\
-\sqrt{1-\sin^2\theta} & \text{otherwise}.
\end{cases}
$$
Not surprisingly, few people choose to write out such a complicated
formula merely to have a formula for $\sin(2\theta)$ that involves
no trig functions other than $\sin\theta$.
A: You could express $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ as $\pm2\sin(\theta)(1-\sin^2(\theta))^{1/2}$, and furthermore, the same should be possible for $\csc$.
Note: I used the relation $\sin^2(\theta)+\cos^2(\theta)=1$ to derive the expression $\cos(\theta)=\pm(1-\sin^2(\theta))^{1/2}$ where the $\pm$ is a result of taking the square root.
