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,







For triple angle formulas, all 6 trig functions have expressions using only the given trig function. They are







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 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 $\sin\theta$.


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.

  • $\begingroup$ How exactly is that formula arrived at (and how is one to interpret the $\pm$ in it)? $\endgroup$ – Justin Benfield Apr 17 '16 at 23:28
  • 1
    $\begingroup$ Because $\sin^2t+\cos^2t=1$ so $|\cos t|=|(1-\sin^2t|^{1/2}.$ Unfortunately we cannot get rid of the "$\pm$" in the above answer. But we have $\cos 2 t=2\cos^t-1=\cos^2t-\sin^2t=1-2\sin^2 t $ from $\cos^2t+\sin^2t=1.$...... $A=\pm B$ means ($A=B $ or $A=-B$). $\endgroup$ – DanielWainfleet Apr 17 '16 at 23:35

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