Why Inverse Trigonometric Functions Restrict a Domain
A formula involving an inverse trigonometric function is defined only when the input to that function is allowed. For real-valued inverse sine and inverse cosine, this requirement places a specific bound on their arguments, whether the argument is a single variable or an expression in several variables.
For example, in \(f(x,y)=\arcsin(x+2y)\), it is not enough to know that \(x\) and \(y\) are real. The expression \(x+2y\) must also be an allowable input to real inverse sine. Finding the domain means identifying exactly which ordered pairs satisfy that requirement.
The Input and Output of Inverse Sine and Cosine
The real inverse sine function, written \(\arcsin(u)\), takes an input \(u\) between \(-1\) and \(1\), inclusive. Its output is an angle in \([-\pi/2,\pi/2]\). The real inverse cosine function, written \(\arccos(u)\), has the same input restriction, but its output is an angle in \([0,\pi]\).
The intervals for the outputs explain which inverse functions are being used: sine is one-to-one on \([-\pi/2,\pi/2]\), and cosine is one-to-one on \([0,\pi]\). For finding the domain of a formula, however, the relevant fact is the common input restriction.
Both endpoints count. In particular, \(\arcsin(-1)=-\pi/2\), \(\arcsin(1)=\pi/2\), \(\arccos(-1)=\pi\), and \(\arccos(1)=0\). Thus a valid boundary point must not be discarded just because its argument equals \(1\) or \(-1\).
Why the Argument Must Lie Between Negative One and One
The input restriction follows from the definition of real inverse trigonometric functions. The sine function takes values only in \([-1,1]\); after restricting its angle to \([-\pi/2,\pi/2]\), every value in that interval occurs exactly once. Thus its inverse on that interval can accept every input from \(-1\) through \(1\), and no real input outside that range.
Similarly, cosine takes values only in \([-1,1]\). On \([0,\pi]\), it takes every such value exactly once, so its inverse on that interval has the same input domain. Applying this fact to an expression \(g(\mathbf{x})\) gives a direct test for each tuple: calculate the argument, then check whether it lies in the allowed interval.
This condition applies to the value of the whole argument, not to each variable separately. Solving it may produce a band, a disk, a spherical shell, or another region, depending on the expression inside the inverse trigonometric function.
Worked Example: A Linear Argument in Two Variables
Worked Example: A Linear Argument in Two Variables
Find the domain of \(f(x,y)=\arcsin(x+2y)\).
The argument is \(x+2y\). It must lie in \([-1,1]\), so the domain condition is
To describe this region, solve the two inequalities for \(y\). Subtract \(x\) throughout and divide by \(2\), which is positive, so the inequality directions do not change:
Therefore the domain is the closed strip between the two lines \(y=(-1-x)/2\) and \(y=(1-x)/2\), including both boundary lines. The set can also be written \(\{(x,y)\in\mathbb{R}^2:-1\le x+2y\le 1\}\).
Check three points. At \((0,0)\), the argument is \(0+2(0)=0\), so it is allowed. At \((1,0)\), the argument is \(1+2(0)=1\), so the point is allowed and lies on the boundary. At \((0,1)\), the argument is \(0+2(1)=2\), which is greater than \(1\), so that point is not in the domain.
The calculation checks out: the boundary value \(1\) gives \(\arcsin(1)=\pi/2\), a real output, while the argument \(2\) has no real inverse sine.
Worked Example: A Curved Domain in Two Variables
Worked Example: A Curved Domain in Two Variables
Find the domain of \(g(x,y)=\arccos(1-x^2-4y^2)\).
Require the argument to be between \(-1\) and \(1\), inclusive:
Separate the two inequalities. The left inequality gives
The right inequality gives
The last inequality is true for all real \(x\) and \(y\), because squares are nonnegative. Thus only \(x^2+4y^2\le 2\) restricts the variables. The domain is the filled ellipse \(x^2+4y^2\le 2\), including its boundary.
For a check, at \((0,0)\) the argument is \(1-0-0=1\), which is allowed, and \(g(0,0)=\arccos(1)=0\). At \((\sqrt{2},0)\), the argument is \(1-2-0=-1\), also allowed, and the point is on the boundary. At \((2,0)\), the argument is \(1-4=-3\), outside \([-1,1]\); indeed \(2^2+4(0)^2=4>2\), so the point is excluded.
Worked Example: A Shell in Three Variables
Worked Example: A Shell in Three Variables
Find the domain of \(h(x,y,z)=\arcsin(x^2+y^2+z^2-2)\).
Set the argument between \(-1\) and \(1\):
Add \(2\) to each part of the inequality:
The domain consists of points in \(\mathbb{R}^3\) whose squared distance from the origin is between \(1\) and \(3\), inclusive. Equivalently, their distance from the origin lies between \(1\) and \(\sqrt{3}\), inclusive. Geometrically, this is a closed spherical shell: the region between two concentric spheres, with both spherical boundaries included.
Check the origin: at \((0,0,0)\), the argument is \(0+0+0-2=-2\), which is not allowed, and the squared distance \(0\) fails the lower bound. At \((1,0,0)\), the argument is \(1-2=-1\), which is allowed; the point lies on the inner boundary. At \((1,1,1)\), the argument is \(1+1+1-2=1\), also allowed; its squared distance is \(3\), so it lies on the outer boundary.
A Common Mistake: Excluding the Endpoints
A frequent error is to replace the required closed interval with an open interval, as if the argument had to be strictly between \(-1\) and \(1\). For example, an incorrect calculation for \(\arcsin(x)\) might say
and conclude that neither endpoint is in the domain. That conclusion is false: \(\arcsin(-1)=-\pi/2\) and \(\arcsin(1)=\pi/2\) are both real. The correct condition is
The same correction applies when the argument depends on multiple variables. Keep equality in both bounds unless another feature of the formula independently makes a boundary point invalid.
A Reliable Domain Check
For an inverse-sine or inverse-cosine expression, focus first on the entire argument inside the function. Write down the two-sided bound, solve it for the variables, and retain equality where it is allowed. For a multivariable argument, describe the set of tuples satisfying the resulting inequality; interpreting that set geometrically can make the answer easier to understand and check.
In \(\arcsin(g(x,y))\) or \(\arccos(g(x,y))\), the argument is the complete expression \(g(x,y)\).
Write \(-1\le g(x,y)\le 1\), including both endpoints.
Describe the set of variable values that satisfy the condition, then test representative points, including boundary points when useful.
The test is necessary and sufficient for the inverse-trigonometric part of the formula: if the argument falls in \([-1,1]\), the real inverse sine or cosine is defined; if it falls outside that interval, it is not. This makes the argument range, rather than the number of variables, the central issue.
Check Your Understanding
For each expression, write the inverse-trigonometric input condition and describe or simplify the resulting domain.
- Find the domain of \(p(x,y)=\arccos(x-y)\).
- Find the domain of \(q(x,y)=\arcsin(2x+y)\).
- For \(r(x,y)=\arccos(x^2+y^2)\), simplify the two-sided input condition.
- Is \((0,1)\) in the domain of \(\arcsin(x+2y)\)? Show the argument calculation.
- Are boundary points with argument \(-1\) or \(1\) included for real inverse sine and inverse cosine? Explain using one function value.