Inverse Trigonometric Functions

sources:

• text: Standard textbook reference

Inverse trigonometric functions reverse the action of trigonometric functions. They are essential for solving equations involving angles and for integration.

• $\sin^{-1}: [-1, 1] \to [-\pi/2, \pi/2]$
• $\cos^{-1}: [-1, 1] \to [0, \pi]$
• $\tan^{-1}: \mathbb{R} \to (-\pi/2, \pi/2)$
• $\sin^{-1}x + \cos^{-1}x = \pi/2$
• $\tan^{-1}x + \cot^{-1}x = \pi/2$
• $\sin^{-1}(-x) = -\sin^{-1}x$
• $\cos^{-1}(-x) = \pi - \cos^{-1}x$
• $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy < 1$

Problem: Simplify $\sin^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right)$.

Solution:

Let $\alpha = \sin^{-1}(3/5)$, so $\sin\alpha = 3/5$, $\cos\alpha = 4/5$.

Let $\beta = \sin^{-1}(5/13)$, so $\sin\beta = 5/13$, $\cos\beta = 12/13$.

$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$

$= \frac{3}{5} \cdot \frac{12}{13} + \frac{4}{5} \cdot \frac{5}{13} = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}$

Therefore: $\sin^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \sin^{-1}\left(\frac{56}{65}\right)$

Common mistake: Forgetting to check that the sum is within the range $[-\pi/2, \pi/2]$.

Problem: Find the value of \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1)(3).

Solution:

First, use the addition formula for $\tan^{-1}(1) + \tan^{-1}(2)$: Since $1 \times 2 = 2 > 1$, use the identity $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ when $xy > 1$ and $x > 0$:

$\tan^{-1}(1) + \tan^{-1}(2) = \pi + \tan^{-1}\left(\frac{1+2}{1-2}\right) = \pi + \tan^{-1}(-3) = \pi - \tan^{-1}(3)$

Now add $\tan^{-1}(3)$: $\pi - \tan^{-1}(3) + \tan^{-1}(3) = \pi$

Common mistake: Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ without checking whether $xy < 1$.

Problem: Solve $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{3}$.

Solution:

Using the identity $\sin^{-1}x + \cos^{-1}x = \pi/2$:

$\frac{\pi}{2} = \frac{\pi}{3}$

This is a contradiction! The equation has no solution.

Common mistake: Not recognizing standard identities. If the equation were $\sin^{-1}x + \cos^{-1}x = \pi/2$, every $x \in [-1, 1]$ would be a solution.

1. Simplify $\cos^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right)$.
2. Find the value of $\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right)$.
3. Solve $\tan^{-1}(2x) + \tan^{-1}(3x) = \pi/4$.

Inverse trigonometric functions are essential for integration (they appear as antiderivatives), solving equations in physics and engineering, and in any context where angles need to be computed from ratios.

Asking “which angle gives this ratio?”: Inverse trigonometric functions reverse the usual trig functions — instead of asking “what’s the sine of 30°?”, you ask “what angle has a sine of 0.5?” Think of it as a lookup table: you give the function a ratio (like 0.5), and it tells you the angle. But there’s a catch — trig functions repeat (sine of 30° = sine of 150°), so we restrict the output to a specific range to make the inverse a proper function. The complementary identities (sin⁻¹x + cos⁻¹x = π/2) are like two sides of the same coin — they always add up to a right angle.

Why it matters: Inverse trig functions appear everywhere — in integration (they’re antiderivatives), in physics (calculating angles from components), in engineering (signal processing), and in any situation where you need to recover an angle from a ratio. They’re essential tools for solving equations that involve angles.

The key insight: The range restrictions on inverse trig functions aren’t arbitrary — they ensure each input has exactly one output, making the function well-defined. The domain of sin⁻¹ is [-1,1] because sine never exceeds 1, and its range is [-π/2, π/2] because that’s where sine is one-to-one.

• Always check the range of the inverse function
• Use the addition formulas carefully, checking conditions on $xy$
• Standard values: $\sin^{-1}(1/2) = \pi/6$, $\tan^{-1}(1) = \pi/4$, $\cos^{-1}(1/2) = \pi/3$
• Practice converting between different inverse trig forms using complementary angle identities

• Domain and range:
  • $\sin^{-1} x$: domain $[-1, 1]$, range $[-\pi/2, \pi/2]$
  • $\cos^{-1} x$: domain $[-1, 1]$, range $[0, \pi]$
  • $\tan^{-1} x$: domain $\mathbb{R}$, range $(-\pi/2, \pi/2)$
• Complementary identities:
  • $\sin^{-1} x + \cos^{-1} x = \pi/2$
  • $\tan^{-1} x + \cot^{-1} x = \pi/2$
  • $\sec^{-1} x + \csc^{-1} x = \pi/2$
• Negative argument:
  • $\sin^{-1}(-x) = -\sin^{-1} x$
  • $\cos^{-1}(-x) = \pi - \cos^{-1} x$
  • $\tan^{-1}(-x) = -\tan^{-1} x$
• Addition formulas:
  • $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ when $xy < 1$
  • $\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ when $xy > 1, x > 0$

Problem: Express $\cos^{-1}\left(\frac{3}{5}\right)$ in terms of $\sin^{-1}$.

Solution:

Let $\theta = \cos^{-1}(3/5)$, so $\cos\theta = 3/5$ and $\theta \in [0, \pi]$.

Since $\cos\theta = 3/5 > 0$, $\theta$ is in the first quadrant, so $\sin\theta > 0$.

$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

Therefore: $\cos^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{4}{5}\right)$

Common mistake: Forgetting that $\sin^{-1} x + \cos^{-1} x = \pi/2$, so $\cos^{-1} x = \pi/2 - \sin^{-1} x$. This is another valid answer.

Problem: Find the value of $\sin\left(2\tan^{-1}\frac{3}{4}\right)$.

Solution:

Let $\theta = \tan^{-1}(3/4)$, so $\tan\theta = 3/4$ and $\theta \in (-\pi/2, \pi/2)$.

Since $\tan\theta = 3/4 > 0$, $\theta$ is in the first quadrant.

$\sin\theta = \frac{3}{5}, \quad \cos\theta = \frac{4}{5}$

$\sin(2\theta) = 2\sin\theta\cos\theta = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$

Common mistake: Forgetting that $\tan^{-1}(3/4)$ gives an angle, not a ratio. After finding the angle, use it in the double-angle formula.

Problem: Solve $\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\left(\frac{8}{31}\right)$.

Solution:

Using the addition formula (check $xy = (x+1)(x-1) = x^2 - 1$):

Case 1: $x^2 - 1 < 1$, i.e., $x^2 < 2$: $\tan^{-1}\left(\frac{(x+1)+(x-1)}{1-(x+1)(x-1)}\right) = \tan^{-1}\left(\frac{8}{31}\right)$

$\frac{2x}{1-(x^2-1)} = \frac{8}{31}$

$\frac{2x}{2-x^2} = \frac{8}{31}$

$62x = 16 - 8x^2$

$8x^2 + 62x - 16 = 0$

$4x^2 + 31x - 8 = 0$

$(4x - 1)(x + 8) = 0$

$x = 1/4$ or $x = -8$

Check: For $x = 1/4$: $(1/4)^2 = 1/16 < 2$ (valid). For $x = -8$: $(-8)^2 = 64 > 2$ (not valid for Case 1).

For Case 2 ($x^2 > 1$), we need to check separately. Testing $x = -8$ in the original equation shows it doesn’t satisfy.

Answer: $x = 1/4$

Common mistake: Not checking the condition $xy < 1$ when using the addition formula. Always verify the domain condition after solving.

1. Memorize the standard values: $\sin^{-1}(0) = 0$, $\sin^{-1}(1/2) = \pi/6$, $\sin^{-1}(1/\sqrt{2}) = \pi/4$, $\sin^{-1}(\sqrt{3}/2) = \pi/3$, $\sin^{-1}(1) = \pi/2$
2. When simplifying inverse trig expressions, always check that the result is within the correct range
3. For addition formulas, the condition on $xy$ determines which form to use
4. Converting between inverse trig functions: use complementary identities or construct a right triangle
5. Practice with both numerical and algebraic arguments

The formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ is only valid when $xy < 1$. When $xy > 1$ and $x > 0$, the correct formula is $\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$. Students frequently apply the first formula blindly and get answers that are off by $\pi$. Always check the product $xy$ before choosing which form to use.

Each inverse trigonometric function has a specific range: $\sin^{-1}$ maps to $[-\pi/2, \pi/2]$, $\cos^{-1}$ maps to $[0, \pi]$, and $\tan^{-1}$ maps to $(-\pi/2, \pi/2)$. Students often forget that $\cos^{-1}(-x) = \pi - \cos^{-1}x$ (not $-\cos^{-1}x$) because the range of $\cos^{-1}$ is $[0, \pi]$. Always verify that your answer falls within the correct range before finalising.

The identity $\sin^{-1}x + \cos^{-1}x = \pi/2$ is extremely useful for converting between inverse trig functions, but students often overlook it. For example, $\cos^{-1}(3/5) = \pi/2 - \sin^{-1}(3/5) = \sin^{-1}(4/5)$. When a problem gives you one inverse trig function and asks for another, check whether the complementary identity simplifies the calculation.

• Trigonometry — Inverse trigonometric functions are defined as inverses of restricted trigonometric functions, requiring understanding of domain restrictions.
• Matrices — Rotation matrices use trigonometric functions, and their inverses connect to inverse trigonometric representations.
• Calculus — Differentiation and integration of inverse trigonometric functions are key applications in calculus.