MHB Trig Functions: When Plugging in x Returns x

[RidiculousName]

I have the statement $$\sin[\sin^{-1}(x)] = x \hspace{7pt} if -1 \leq x \leq 1$$. How can I tell if plugging in x will return x for $$\cos[\cos^{-1}(x)] $$ and $$\tan[\tan^{-1}(x)] $$? What if the positions of the regular and inverse functions were reversed? For example, $$\cos^{-1}[\cos(x)]$$.

I am only looking for answers regarding cosine, sine, and tangent here.

 

[I like Serena]

I have the statement $$\sin[\sin^{-1}(x)] = x \hspace{7pt} if -1 \leq x \leq 1$$. How can I tell if plugging in x will return x for $$\cos[\cos^{-1}(x)] $$ and $$\tan[\tan^{-1}(x)] $$? What if the positions of the regular and inverse functions were reversed? For example, $$\cos^{-1}[\cos(x)]$$.

I am only looking for answers regarding cosine, sine, and tangent here.

Hi RidiculousName,

The functions $\sin$, $\cos$, and $\tan$ are all functions that are not actually considered invertible functions.
That's because there are multiple values for $x$ for which they give the same result.
Consider for instance $y=\frac 12\sqrt 2$.
The values of $x$ for which $\cos(x)=y=\frac 12\sqrt 2$ are $x=\pm \frac\pi 4, \pm 2\pi\pm \frac\pi 4, ...$.
This shows that we cannot uniquely find a value $x$ for $y=\frac 12\sqrt 2$ where the $\cos$ is concerned.
Not even when $-1\le x\le 1$.
More specifically: $\cos^{-1}(\cos(-\frac\pi 4))=\cos^{-1}(\frac 12\sqrt 2)=\frac\pi 4\ne-\frac \pi 4$.

That's why $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ have special definitions that define in which interval they will return their result.
Consequently these special inverses are invertible themselves.
For instance $\sin^{-1}$ is defined to return its result in the interval $[-\frac\pi 2,+\frac\pi 2]$.
If we apply the $\sin$ again on a value in this interval, we will get the original value back.
For instance, we have $\sin(\sin^{-1}(-\frac 12\sqrt 2))=\sin(-\frac\pi 4)=-\frac 12\sqrt 2$.

Do you perchance already know what the defined intervals are for $\cos^{-1}$ and $\tan^{-1}$?
Can you tell now which combinations will return the same result, and which ones won't?

 

[RidiculousName]

Hi RidiculousName,
Do you perchance already know what the defined intervals are for $\cos^{-1}$ and $\tan^{-1}$?
Can you tell now which combinations will return the same result, and which ones won't?

Well, I know the definition of $\tan^{-1}$ is, $\tan\theta = y$, and $\frac{-\pi}{2} < \theta < \frac{\pi}{2}$. I forget exactly what the definition of inverse sine and cosine were, but I think the second half of sine was $\frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}$, and for cosine it was the same as sine, but with 0 and $\pi$.

I don't understand why x would have to be $-1 \leq x \leq 1$ to work unless it is referring to the unit-circle. But I am not sure this is correct, or if it would apply to cosine and tangent too.

I am still confused about which combinations will return the same result, but if I had to guess, if the regular trig function comes first x must be $-1 \leq x \leq 1$, except possibly for tangent because it is undefined at $\frac{\pi}{2}$ radians or 90 degrees. So, I'm not sure what values of x would return x.

If the inverse trig function comes first, it seems like x would have to be $0 \leq x \leq \pi$, but again, I'm not sure about tangent.

 

[I like Serena]

Well, I know the definition of $\tan^{-1}$ is, $\tan\theta = y$, and $\frac{-\pi}{2} < \theta < \frac{\pi}{2}$. I forget exactly what the definition of inverse sine and cosine were, but I think the second half of sine was $\frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}$, and for cosine it was the same as sine, but with 0 and $\pi$.

That is correct.

I don't understand why x would have to be $-1 \leq x \leq 1$ to work unless it is referring to the unit-circle. But I am not sure this is correct, or if it would apply to cosine and tangent too.

As we can see in the unit circle, we are indeed limited to $-1 \leq x \leq 1$ for $\cos^{-1}$ and $\sin^{-1}$.
This limitation does not apply to $\tan^{-1}$ though, which is defined for every $x$.
Either way, for this problem we would still limit $x$ to the given range.

I am still confused about which combinations will return the same result, but if I had to guess, if the regular trig function comes first x must be $-1 \leq x \leq 1$, except possibly for tangent because it is undefined at $\frac{\pi}{2}$ radians or 90 degrees. So, I'm not sure what values of x would return x.

If the regular trig function is left most, it means that we evaluate the inverse function first.
Since the inverse function is actually invertible, that means that applying the regular trig functions afterwards, will always return the same result.

Let's take a look at $\tan$.
The question then is if $\tan(\tan^{-1}(x))\overset ?= x$ if $-1\le x\le 1$.
The range of $\tan^{-1}$ is defined to be $\left(-\frac\pi 2,\frac\pi 2\right)$.
We won't use the entire range though, since $\tan^{-1}(-1)=-\frac\pi 4$ and $\tan^{-1}(1)=\frac\pi 4$.
Check the unit circle to see why that is.
Either way, if we apply $\tan$ afterwards, we get exactly the original $x$ back don't we?

If the inverse trig function comes first, it seems like x would have to be $0 \leq x \leq \pi$, but again, I'm not sure about tangent.

The important part is whether there are multiple values of $x$ with $-1\le x\le 1$ that map to the same value.
In this interval $\cos$ first goes up to $1$ and then goes down again, so that one is no good.
$\sin$ is an increasing function in this interval, so that one works.
And $\tan$ is also an increasing function in this interval, so that one works as well.