# Implied domain of composite trig function

I know that to find the implied domain of y=arcsin(cos(x)) one must first find the domain of arcsin and then set up an inequation and solve for x. For this particular example: cos(x) is less than or equal to 1 and greater than or equal to -1. However, when you have an equation such as y=arctan(cos(x)) and you take the domain of arctan you end up with all real numbers.

My question then is this: how does one find the implied domain of an equation of this kind using this method, given that you can't take the inverse cos of R to solve for x?

Or, alternatively, is there a more effective way of approaching these kinds of problems?

The domain of $\arcsin(\cos x)$ is $\mathbb R$, because $\cos$ is defined on all real numbers and for all real numbers, the value of $\cos x$ is between $-1$ and $1$ so $\arcsin$ is defined on it.

• Hmm, my textbook gives the implied domain as 0 to pi. Could it be possible that my book wants me restrict the domain 0 to pi? I'm not sure why it would, but I'm at a lost.
– Jack
Feb 23, 2015 at 8:29
• @Jack Are you sure you are talking about $\arcsin(\cos(x))$ and not $\cos(\arcsin(x))$?
– 5xum
Feb 23, 2015 at 8:43
• Positive. Another example from my textbook gives the implied domain of arccos(sin(2x)) as -pi/4 to pi/4. Perhaps my textbook means something entirely different by 'implied domain', because my intuition tells me that you are correct.
– Jack
Feb 23, 2015 at 8:51
• Actually, it appears that my textbook is assuming a restricted domain of -pi/2 to pi/2 and 0 to pi for sin and cos, respectively.
– Jack
Feb 23, 2015 at 9:13
• @Jack That's a very limited view of both of these functions...
– 5xum
Feb 23, 2015 at 9:19
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