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📐 Key Concepts: Inverse Trigonometric Functions

Definitions and Ranges:
$y = \arcsin x$: domain $[-1,1]$, range $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$
$y = \arccos x$: domain $[-1,1]$, range $[0,\pi]$
$y = \arctan x$: domain $(-\infty,\infty)$, range $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

Exact Values to Know:
$\arcsin 0 = 0°$, $\arcsin\tfrac{1}{2} = 30°$, $\arcsin\tfrac{\sqrt{2}}{2} = 45°$, $\arcsin\tfrac{\sqrt{3}}{2} = 60°$, $\arcsin 1 = 90°$
$\arccos 1 = 0°$, $\arccos\tfrac{\sqrt{3}}{2} = 30°$, $\arccos\tfrac{\sqrt{2}}{2} = 45°$, $\arccos\tfrac{1}{2} = 60°$, $\arccos 0 = 90°$
$\arctan 0 = 0°$, $\arctan\tfrac{1}{\sqrt{3}} = 30°$, $\arctan 1 = 45°$, $\arctan\sqrt{3} = 60°$

Key Identities:
$\sin(\arcsin x) = x$ for $x \in [-1,1]$
$\cos(\arccos x) = x$ for $x \in [-1,1]$
$\tan(\arctan x) = x$ for all $x$
$\arcsin x + \arccos x = \dfrac{\pi}{2}$ for $x \in [-1,1]$

Composition Technique: To evaluate $\sin(\arctan(a/b))$, let $\theta = \arctan(a/b)$, draw a right triangle with opposite $= a$ and adjacent $= b$, find hypotenuse $= \sqrt{a^2+b^2}$, then read off the required function directly.

Problem 1: Exact Values of Inverse Trig Functions

Find the exact value of each expression in both radians and degrees:

  1. $\arcsin\dfrac{1}{2}$
  2. $\arccos\dfrac{\sqrt{2}}{2}$
  3. $\arctan\sqrt{3}$

(a) We need $\theta \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ such that $\sin\theta = \dfrac{1}{2}$.

From the unit circle, $\sin 30° = \dfrac{1}{2}$ and $30° \in [-90°,90°]$:

$$\arcsin\frac{1}{2} = \frac{\pi}{6} = 30°$$

(b) We need $\theta \in [0,\pi]$ such that $\cos\theta = \dfrac{\sqrt{2}}{2}$.

From the unit circle, $\cos 45° = \dfrac{\sqrt{2}}{2}$ and $45° \in [0°,180°]$:

$$\arccos\frac{\sqrt{2}}{2} = \frac{\pi}{4} = 45°$$

(c) We need $\theta \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ such that $\tan\theta = \sqrt{3}$.

From the unit circle, $\tan 60° = \sqrt{3}$ and $60° \in (-90°,90°)$:

$$\arctan\sqrt{3} = \frac{\pi}{3} = 60°$$

Problem 2: Composition of Inverse and Direct Trig

Find the exact value of $\tan\!\left(\arcsin\dfrac{3}{5}\right)$.

Let $\theta = \arcsin\dfrac{3}{5}$, so $\sin\theta = \dfrac{3}{5}$ and $\theta \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.

Since $\dfrac{3}{5} > 0$, angle $\theta$ is in the first quadrant and $\cos\theta > 0$.

Draw a right triangle: opposite $= 3$, hypotenuse $= 5$, so

$$\text{adjacent} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$$

Therefore:

$$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}$$
$$\tan\!\left(\arcsin\frac{3}{5}\right) = \boxed{\frac{3}{4}}$$

Problem 3: Engineering Application — Roof Inclination

A roof has a standard 8:12 pitch, meaning it rises 8 inches vertically for every 12 inches of horizontal run. Find the angle of inclination of the roof surface to the nearest minute.

The rise-to-run ratio gives the tangent of the inclination angle $\theta$:

$$\tan\theta = \frac{\text{rise}}{\text{run}} = \frac{8}{12} = \frac{2}{3} \approx 0.6\overline{6}$$

Apply the inverse tangent:

$$\theta = \arctan\!\left(\frac{2}{3}\right) \approx 33.690°$$

Convert the decimal portion to minutes: $0.690° \times 60'/° \approx 41'$

$$\theta \approx 33°41'$$

Problem 4: Resultant Force Direction

A horizontal force $F_1 = 100\text{ N}$ and a vertical force $F_2 = 75\text{ N}$ act simultaneously on a bolt. Find the angle (in degrees and minutes) that the resultant force makes with the horizontal direction, and compute the magnitude of the resultant.

The resultant $R$ makes angle $\theta$ with the horizontal, where:

$$\tan\theta = \frac{F_2}{F_1} = \frac{75}{100} = \frac{3}{4} = 0.75$$

Apply the inverse tangent:

$$\theta = \arctan(0.75) \approx 36.870°$$

Convert the decimal to minutes: $0.870° \times 60'/° \approx 52'$

$$\theta \approx 36°52'$$

Magnitude of the resultant:

$$R = \sqrt{F_1^2 + F_2^2} = \sqrt{100^2 + 75^2} = \sqrt{10000 + 5625} = \sqrt{15625} = 125\text{ N}$$

Problem 5: Solving an Inverse Trig Equation

Solve for $x$: $\quad \arctan(2x + 1) = \dfrac{\pi}{4}$

Apply the tangent function to both sides to undo the arctangent (valid since $\dfrac{\pi}{4} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$):

$$\tan\!\bigl(\arctan(2x+1)\bigr) = \tan\!\left(\frac{\pi}{4}\right)$$
$$2x + 1 = 1$$

Solve the resulting linear equation:

$$2x = 0 \implies \boxed{x = 0}$$

Verification: $\arctan\bigl(2(0)+1\bigr) = \arctan(1) = \dfrac{\pi}{4}$ ✓