Inverse Trigonometric Functions

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

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 sin1(x)\sin^{-1}(x) is the inverse of sine
  • arccos(x)\arccos(x) or cos1(x)\cos^{-1}(x) is the inverse of cosine
  • arctan(x)\arctan(x) or tan1(x)\tan^{-1}(x) is the inverse of tangent

Important: sin1(x)1sin(x)\sin^{-1}(x) \neq \frac{1}{\sin(x)}. The 1-1 is NOT an exponent!

  • sin1(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 π2yπ2-\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 0yπ0 \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:

  • limxarctan(x)=π2\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}
  • limxarctan(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 0x<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=π0.7752.366x = \pi - 0.775 \approx 2.366

Solutions: x0.775,2.366x \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 x1|x| \geq 1
  • y=\arcsec(x)=arccos(1x)y = \arcsec(x) = \arccos(\frac{1}{x}) for x1|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)=12andπ2yπ2\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 0θπ0 \leq \theta \leq \pi

We need to find: sin(θ)\sin(\theta)

Method: Use Pythagorean identity

sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

Substitute cos(θ)=35\cos(\theta) = \frac{3}{5}: sin2(θ)+(35)2=1\sin^2(\theta) + (\frac{3}{5})^2 = 1 sin2(θ)+925=1\sin^2(\theta) + \frac{9}{25} = 1 sin2(θ)=1925=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 = 5232=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)=π4arctan(2x)\arctan(x) = \frac{\pi}{4} - \arctan(2x)

Step 2: Take tangent of both sides x=tan(π4arctan(2x))x = \tan(\frac{\pi}{4} - \arctan(2x))

Step 3: Use tangent difference formula

tan(AB)=tanAtanB1+tanAtanB\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=12x1+(1)(2x)=12x1+2xx = \frac{1 - 2x}{1 + (1)(2x)} = \frac{1 - 2x}{1 + 2x}

Step 5: Solve for xx

x(1+2x)=12xx(1 + 2x) = 1 - 2x x+2x2=12xx + 2x^2 = 1 - 2x 2x2+x+2x1=02x^2 + x + 2x - 1 = 0 2x2+3x1=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+1740.281x_1 = \frac{-3 + \sqrt{17}}{4} \approx 0.281

x2=31741.781x_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 π40.785\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)