Multiple Angle formulas, alternate forms

Question

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?

Answer 1

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.

Answer 2

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 $\sin\theta$.