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I re-arranged: $$3\tan^{2} x -1=0$$ to get $\tan^{2}\theta = \frac{1}3$. I noticed the inverse of the $cos, sin$ and $tan$ functions are written as $\cos^{-1}\theta, \sin^{-1}\theta$ and $\tan^{-1}\theta$ respectively, does this mean the inverse of $\tan^{2}\theta$ would equal $\tan^{(2-1=1)}\theta = \tan\theta$ ? Also is it referred to Arc-$function$ or the inverse of the function, I've heard they're two different things but the distinction is ambiguous to me.

Tobi
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  • Maybe this answer can help to address some notation issues, particularly those involving trig functions: http://math.stackexchange.com/questions/728476/is-sin2x-the-same-as-sin-x2/728591#728591 – Xoque55 May 13 '16 at 01:11

2 Answers2

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No, this is not correct. Perhaps the notation is misleading, but

$$\cos^{-1} \theta \neq \frac{1}{\cos \theta}.$$

That is why I prefer to use the arc notation as in $\arccos \theta$.

The notations $\cos^{-1} \theta$ and $\arccos \theta$ represent the same thing, which is, roughly speaking, the inverse of $\cos \theta$ (although it is not a true inverse since $\cos$ is not injective).

Back to your question:

We can simplify $3\tan^{2} \theta -1=0$ to get $|\tan \theta|= \frac{1}{\sqrt 3}$. You may then use the fact that $\tan \theta$ is odd and that it is $\pi$-periodic to find all of the solutions.

MathMajor
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  • is $arccos\theta = \frac{1}{\cos\theta}$ ? – Tobi May 13 '16 at 01:05
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    @James Unfortunately not. It really is just a matter of bad notation. – MathMajor May 13 '16 at 01:05
  • So... 1, Is arccos the same as $\cos^{-1}$ . 2, how is the inverse of a function squared found? Square root and then inverse? – Tobi May 13 '16 at 01:06
  • Yes, they are the same. To find the inverse is a bit tricky. Take the square root and account for the fact that same values of $\tan$ are negative. – MathMajor May 13 '16 at 01:07
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    Not so much "bad" notation, as the unfortunate confluence of notation from two different usages in mathematics (somewhat like the over-use of the word "normal"). – colormegone May 13 '16 at 01:07
  • does $cos\theta^{-1} $ not equal $\frac{1}{cos\theta}$ – Tobi Jun 12 '16 at 06:04
  • @Tobi That is quite ambiguous notation. It could mean $\arccos \theta$, $\cos \frac{1}{\theta}$, or $\frac{1}{\cos \theta}$. – MathMajor Jun 12 '16 at 20:26
  • When I posted the question I asked if arccos$\theta$ was equivalent to $\frac{1}{cos\theta}$ and I was told no, but then I asked if arccos was equal to $cos^{-1}$ and was told yes, so its quite confusing... – Tobi Jun 13 '16 at 07:34
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REMEMBER: $tan^2\;x$ is a simplification of $(tan(x))^2$.

It's easier than it seems, root both sides so $tan(x) = \frac{\pm 1}{\sqrt3}$

Now inverse tan $\frac{1}{\sqrt{3}} $ ... $tan^{-1}(\frac{1}{\sqrt{3}})$

and you get: $\theta = 30$ this is the principal value (closest to the origin); you can find the limitless other solutions by $\pm 180$

Tobi
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