7.5 Inverse Trigonometric Functions

7.5 Inverse Trigonometric Functions

Inverse trigonometric functions are named twice in the official content outline: once among the eleven elementary families you must recognize symbolically, graphically, or in tabular form (the 30% area), and again inside Trigonometry and Its Applications as "the basic trigonometric functions and their inverses." They are also the family students handle least reliably, because their whole behavior is governed by an artificial-looking domain restriction. This section makes that restriction make sense.

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1. Why a Restriction Is Necessary

A function has an inverse only if it is one-to-one — it must pass the horizontal line test. Every trigonometric function fails that test spectacularly: the line $y = \frac{1}{2}$ crosses $y = \sin x$ infinitely many times.

The fix is to shrink the domain to a stretch on which the function is one-to-one and still hits every output value. The conventional choices:

Commit the last two columns to memory. Almost every inverse-trigonometry error on the exam is a range violation.

Note the asymmetry: $\arcsin$ and $\arctan$ return negative angles for negative inputs, while $\arccos$ returns angles in the second quadrant for negative inputs. They do not behave alike.

Note also that the domains of $\arcsin$ and $\arccos$ are $[-1, 1]$. Expressions such as $\arcsin(2)$ are undefined — no angle has a sine of $2$. Inverse tangent, by contrast, accepts every real number, because $\tan x$ is unbounded.

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2. Graphs of the Inverse Functions

Each inverse graph is the reflection of the restricted original across the line $y = x$.

$\arccos$ is the only decreasing one, and $\arctan$ is the only one with asymptotes. On a graph-identification item, those two facts alone distinguish all three.

   y = arctan x
        y
    π/2 +- - - - - - - - -  (horizontal asymptote)
        |          ______
        |     ____/
    ----+----/---------------- x
        |   /
        |__/
   -π/2 +- - - - - - - - -  (horizontal asymptote)

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3. Evaluating Inverse Expressions

Step 1: identify the reference angle from the exact-value table. Step 2: place it inside the correct output range.

Worked Example 7.5A: A negative input

Find the exact value of $\arcsin\left(-\frac{1}{2}\right)$ and of $\arccos\left(-\frac{1}{2}\right)$.

Solution:

• Reference angle for $\frac{1}{2}$ is $\frac{\pi}{6}$ in both cases.
• $\arcsin$ must return a value in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Sine is negative in Quadrant IV, so the answer is $-\frac{\pi}{6}$.
• $\arccos$ must return a value in $[0, \pi]$. Cosine is negative in Quadrant II, so the answer is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

Same input, same reference angle, completely different outputs. That is the restriction at work.

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4. Compositions

Cancellation — and its limits

$\sin(\arcsin x) = x \quad \text{for } x \in [-1, 1]$

$\arcsin(\sin \theta) = \theta \quad \textbf{only if } \theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$

The outer-then-inner direction always cancels; the inner-then-outer direction cancels only when the angle already lies in the output range.

Worked Example 7.5B: When cancellation fails

Evaluate $\arccos\left(\cos\frac{7\pi}{6}\right)$.

Solution: The answer is not $\frac{7\pi}{6}$, because $\frac{7\pi}{6}$ lies outside $[0, \pi]$.

1. Evaluate inside first: $\cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$.
2. Now find the angle in $[0, \pi]$ whose cosine is $-\frac{\sqrt{3}}{2}$: the reference angle is $\frac{\pi}{6}$, and Quadrant II gives $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
3. Therefore $\arccos\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$.

Mixed compositions — build a triangle

Worked Example 7.5C

Evaluate $\sin\left(\arctan\frac{3}{4}\right)$.

Solution:

1. Let $\theta = \arctan\frac{3}{4}$, so $\tan\theta = \frac{3}{4}$ with $\theta$ in Quadrant I (positive input, and $\arctan$ returns $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$).
2. Build a right triangle: opposite $= 3$, adjacent $= 4$, hypotenuse $= \sqrt{3^2 + 4^2} = 5$.
3. Read off the sine: $\sin\theta = \frac{3}{5}$.

Worked Example 7.5D: An algebraic composition

Write $\cos(\arcsin x)$ as an algebraic expression in $x$, for $x \in [-1, 1]$.

Solution: Let $\theta = \arcsin x$, so $\sin\theta = x$ and $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. From $\sin^2\theta + \cos^2\theta = 1$:

$\cos\theta = \pm\sqrt{1 - x^2}$

Cosine is non-negative throughout $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so the sign is positive:

$\cos(\arcsin x) = \sqrt{1 - x^2}$

The range restriction is exactly what resolves the $\pm$.

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5. CLEP Exam Traps & Common Errors

• Trap 1: Reading $\sin^{-1} x$ as $\frac{1}{\sin x}$. The reciprocal is $\csc x$. The $-1$ superscript on a function name means inverse function, not a negative exponent. (Confusingly, $\sin^2 x$ does mean $(\sin x)^2$ — the $-1$ is the sole exception.)
• Trap 2: Returning a coterminal angle. $\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$, not $\frac{11\pi}{6}$. Both have sine $-\frac{1}{2}$, but only one lies in the output range.
• Trap 3: Assuming $\arccos$ behaves like $\arcsin$. For a negative input, $\arcsin$ goes to Quadrant IV and $\arccos$ goes to Quadrant II.
• Trap 4: Cancelling in the wrong direction. $\sin(\arcsin 0.4) = 0.4$ always, but $\arcsin(\sin 3) \ne 3$ because $3 > \frac{\pi}{2} \approx 1.571$.
• Trap 5: Evaluating an out-of-domain expression. $\arccos(1.5)$ and $\arcsin(-2)$ are undefined; only $\arctan$ takes every real input.