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Inverse Trigonometric Functions

Understand inverse trigonometric functions, their domains, ranges, and how to evaluate and use them to solve equations.

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Inverse Trigonometric Functions

Introduction

Inverse trigonometric functions reverse the action of the regular trig functions. They answer the question: "What angle has this trig value?"

Notation:

  • arcsin⁡(x)\arcsin(x) or sin⁡−1(x)\sin^{-1}(x) is the inverse of sine
  • arccos⁡(x)\arccos(x) or cos⁡−1(x)\cos^{-1}(x) is the inverse of cosine
  • arctan⁡(x)\arctan(x) or tan⁡−1(x)\tan^{-1}(x) is the inverse of tangent

Important: sin⁡−1(x)≠1sin⁡(x)\sin^{-1}(x) \neq \frac{1}{\sin(x)}. The −1-1 is NOT an exponent!

  • sin⁡−1(x)\sin^{-1}(x) means inverse sine (arcsin)
  • 1sin⁡(x)=csc⁡(x)\frac{1}{\sin(x)} = \csc(x) is the reciprocal

Why We Need Restrictions

Trig functions are periodic (repeat), so they're not one-to-one. To have an inverse function, we must restrict the domain.

Restricted Domains (Principal Values)

For y=sin⁡(x)y = \sin(x):

  • Restrict to [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] (Quadrants I and IV)
  • This gives all y-values from −1-1 to 11 exactly once

For y=cos⁡(x)y = \cos(x):

  • Restrict to [0,π][0, \pi] (Quadrants I and II)
  • This gives all y-values from −1-1 to 11 exactly once

For y=tan⁡(x)y = \tan(x):

  • Restrict to (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) (Quadrants I and IV)
  • This gives all real y-values exactly once

Inverse Sine: y=arcsin⁡(x)y = \arcsin(x)

Definition: y=arcsin⁡(x)y = \arcsin(x) means sin⁡(y)=x\sin(y) = x where −π2≤y≤π2-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

Domain: [−1,1][-1, 1] (input must be a valid sine value)

Range: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] (output is an angle)

Key values:

  • arcsin⁡(0)=0\arcsin(0) = 0
  • arcsin⁡(12)=π6\arcsin(\frac{1}{2}) = \frac{\pi}{6}
  • arcsin⁡(22)=π4\arcsin(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}
  • arcsin⁡(32)=π3\arcsin(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}
  • arcsin⁡(1)=π2\arcsin(1) = \frac{\pi}{2}
  • arcsin⁡(−1)=−π2\arcsin(-1) = -\frac{\pi}{2}

Inverse Cosine: y=arccos⁡(x)y = \arccos(x)

Definition: y=arccos⁡(x)y = \arccos(x) means cos⁡(y)=x\cos(y) = x where 0≤y≤π0 \leq y \leq \pi

Domain: [−1,1][-1, 1]

Range: [0,π][0, \pi]

Key values:

  • arccos⁡(1)=0\arccos(1) = 0
  • arccos⁡(32)=π6\arccos(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}
  • arccos⁡(22)=π4\arccos(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}
  • arccos⁡(12)=π3\arccos(\frac{1}{2}) = \frac{\pi}{3}
  • arccos⁡(0)=π2\arccos(0) = \frac{\pi}{2}
  • arccos⁡(−1)=π\arccos(-1) = \pi

Inverse Tangent: y=arctan⁡(x)y = \arctan(x)

Definition: y=arctan⁡(x)y = \arctan(x) means tan⁡(y)=x\tan(y) = x where −π2<y<π2-\frac{\pi}{2} < y < \frac{\pi}{2}

Domain: All real numbers (−∞,∞)(-\infty, \infty)

Range: (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})

Key values:

  • arctan⁡(0)=0\arctan(0) = 0
  • arctan⁡(1)=π4\arctan(1) = \frac{\pi}{4}
  • arctan⁡(3)=π3\arctan(\sqrt{3}) = \frac{\pi}{3}
  • arctan⁡(−1)=−π4\arctan(-1) = -\frac{\pi}{4}

Asymptotes:

  • lim⁡x→∞arctan⁡(x)=π2\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}
  • lim⁡x→−∞arctan⁡(x)=−π2\lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}

Composition Properties

Inverse function property:

For values in the appropriate domains:

  • sin⁡(arcsin⁡(x))=x\sin(\arcsin(x)) = x for x∈[−1,1]x \in [-1, 1]
  • arcsin⁡(sin⁡(x))=x\arcsin(\sin(x)) = x for x∈[−π2,π2]x \in [-\frac{\pi}{2}, \frac{\pi}{2}]

Similarly for cosine and tangent.

Warning: arcsin⁡(sin⁡(x))\arcsin(\sin(x)) does NOT always equal xx!

Example: arcsin⁡(sin⁡(3π4))=π4\arcsin(\sin(\frac{3\pi}{4})) = \frac{\pi}{4} (not 3π4\frac{3\pi}{4})

Why? Because 3π4\frac{3\pi}{4} is outside the range of arcsin.

Using Inverse Trig to Solve Equations

Example: Solve sin⁡(x)=0.7\sin(x) = 0.7 for 0≤x<2π0 \leq x < 2\pi

Step 1: Find the reference angle xref=arcsin⁡(0.7)≈0.775 radiansx_{ref} = \arcsin(0.7) \approx 0.775 \text{ radians}

Step 2: Determine quadrants (sine is positive in I and II)

  • Quadrant I: x=0.775x = 0.775
  • Quadrant II: x=π−0.775≈2.366x = \pi - 0.775 \approx 2.366

Solutions: x≈0.775,2.366x \approx 0.775, 2.366

Graphs of Inverse Trig Functions

All inverse trig functions are reflections of the restricted trig functions over the line y=xy = x.

Graph characteristics:

y=arcsin⁡(x)y = \arcsin(x):

  • Domain: [−1,1][-1, 1], Range: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]
  • Increasing function
  • Passes through origin

y=arccos⁡(x)y = \arccos(x):

  • Domain: [−1,1][-1, 1], Range: [0,π][0, \pi]
  • Decreasing function
  • yy-intercept at (0,π2)(0, \frac{\pi}{2})

y=arctan⁡(x)y = \arctan(x):

  • Domain: (−∞,∞)(-\infty, \infty), Range: (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})
  • Increasing function
  • Horizontal asymptotes at y=±π2y = \pm\frac{\pi}{2}
  • Passes through origin

Other Inverse Trig Functions

  • y=\arccsc(x)=arcsin⁡(1x)y = \arccsc(x) = \arcsin(\frac{1}{x}) for ∣x∣≥1|x| \geq 1
  • y=\arcsec(x)=arccos⁡(1x)y = \arcsec(x) = \arccos(\frac{1}{x}) for ∣x∣≥1|x| \geq 1
  • y=\arccot(x)=arctan⁡(1x)y = \arccot(x) = \arctan(\frac{1}{x}) (with adjustments)

These are less commonly used but follow similar principles.

📚 Practice Problems

1Problem 1easy

❓ Question:

Evaluate arcsin⁡(12)\arcsin(\frac{1}{2}) exactly in radians.

💡 Show Solution

Solution:

We need to find the angle yy such that: sin⁡(y)=12and−π2≤y≤π2\sin(y) = \frac{1}{2} \quad \text{and} \quad -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

From the unit circle:

sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2}

And π6\frac{\pi}{6} is in the range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] ✓

Answer: arcsin⁡(12)=π6\arcsin(\frac{1}{2}) = \frac{\pi}{6}

Verification: sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2} ✓

Note: Even though sin⁡(5π6)=12\sin(\frac{5\pi}{6}) = \frac{1}{2} as well, we don't choose 5π6\frac{5\pi}{6} because it's outside the range of arcsin.

2Problem 2medium

❓ Question:

Find the exact value of sin⁡(arccos⁡(35))\sin(\arccos(\frac{3}{5})).

💡 Show Solution

Solution:

Let θ=arccos⁡(35)\theta = \arccos(\frac{3}{5})

This means: cos⁡(θ)=35\cos(\theta) = \frac{3}{5} where 0≤θ≤π0 \leq \theta \leq \pi

We need to find: sin⁡(θ)\sin(\theta)

Method: Use Pythagorean identity

sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

Substitute cos⁡(θ)=35\cos(\theta) = \frac{3}{5}: sin⁡2(θ)+(35)2=1\sin^2(\theta) + (\frac{3}{5})^2 = 1 sin⁡2(θ)+925=1\sin^2(\theta) + \frac{9}{25} = 1 sin⁡2(θ)=1−925=1625\sin^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25} sin⁡(θ)=±45\sin(\theta) = \pm\frac{4}{5}

Determine the sign:

Since θ=arccos⁡(35)\theta = \arccos(\frac{3}{5}) and the range of arccos is [0,π][0, \pi], θ\theta is in Quadrant I or II.

In both quadrants, sine is positive.

Answer: sin⁡(arccos⁡(35))=45\sin(\arccos(\frac{3}{5})) = \frac{4}{5}

Alternative method (right triangle):

If cos⁡(θ)=35=adjacenthypotenuse\cos(\theta) = \frac{3}{5} = \frac{\text{adjacent}}{\text{hypotenuse}}:

  • Adjacent = 3
  • Hypotenuse = 5
  • Opposite = 52−32=16=4\sqrt{5^2 - 3^2} = \sqrt{16} = 4

Therefore: sin⁡(θ)=45\sin(\theta) = \frac{4}{5} ✓

3Problem 3hard

❓ Question:

Solve for xx: arctan⁡(x)+arctan⁡(2x)=π4\arctan(x) + \arctan(2x) = \frac{\pi}{4}

💡 Show Solution

Solution:

Given: arctan⁡(x)+arctan⁡(2x)=π4\arctan(x) + \arctan(2x) = \frac{\pi}{4}

Step 1: Isolate one arctan arctan⁡(x)=π4−arctan⁡(2x)\arctan(x) = \frac{\pi}{4} - \arctan(2x)

Step 2: Take tangent of both sides x=tan⁡(π4−arctan⁡(2x))x = \tan(\frac{\pi}{4} - \arctan(2x))

Step 3: Use tangent difference formula

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

Here: A=π4A = \frac{\pi}{4}, B=arctan⁡(2x)B = \arctan(2x)

x=tan⁡(π4)−tan⁡(arctan⁡(2x))1+tan⁡(π4)tan⁡(arctan⁡(2x))x = \frac{\tan(\frac{\pi}{4}) - \tan(\arctan(2x))}{1 + \tan(\frac{\pi}{4})\tan(\arctan(2x))}

Step 4: Simplify

Since tan⁡(π4)=1\tan(\frac{\pi}{4}) = 1 and tan⁡(arctan⁡(2x))=2x\tan(\arctan(2x)) = 2x:

x=1−2x1+(1)(2x)=1−2x1+2xx = \frac{1 - 2x}{1 + (1)(2x)} = \frac{1 - 2x}{1 + 2x}

Step 5: Solve for xx

x(1+2x)=1−2xx(1 + 2x) = 1 - 2x x+2x2=1−2xx + 2x^2 = 1 - 2x 2x2+x+2x−1=02x^2 + x + 2x - 1 = 0 2x2+3x−1=02x^2 + 3x - 1 = 0

Step 6: Use quadratic formula

x=−3±9+84=−3±174x = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}

Step 7: Check which solution is valid

x1=−3+174≈0.281x_1 = \frac{-3 + \sqrt{17}}{4} \approx 0.281

x2=−3−174≈−1.781x_2 = \frac{-3 - \sqrt{17}}{4} \approx -1.781

Both are in the domain of arctan, so check by substitution:

For x=−3+174x = \frac{-3 + \sqrt{17}}{4}: Calculate arctan⁡(x)+arctan⁡(2x)\arctan(x) + \arctan(2x) numerically to verify it equals π4≈0.785\frac{\pi}{4} \approx 0.785

Answer: x=−3+174x = \frac{-3 + \sqrt{17}}{4}

(The negative solution may or may not work depending on domain restrictions)

Explain using:

⚠️ Common Mistakes: Inverse Trigonometric Functions

Avoid these 4 frequent errors

🌍 Real-World Applications: Inverse Trigonometric Functions

See how this math is used in the real world

📝 Worked Example: Related Rates — Expanding Circle

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 22 cm/s. How fast is the area of the circle increasing when the radius is 1010 cm?

2Write the relationship between variables
3Differentiate both sides with respect to time
4Substitute known values

📌 Related Topics in Trigonometric Functions

❓ Frequently Asked Questions

What is Inverse Trigonometric Functions?▾
Understand inverse trigonometric functions, their domains, ranges, and how to evaluate and use them to solve equations.
How can I study Inverse Trigonometric Functions effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Inverse Trigonometric Functions?▾
Inverse Trigonometric Functions is part of the AP Precalculus course on Study Mondo, specifically in the Trigonometric Functions section. You can explore the full course for more related topics and practice resources.
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.