Inverse Trigonometric Functions Explained

Ordinary trigonometric functions take an angle and give a ratio. Inverse trigonometric functions do the reverse: they take a ratio and give back the angle. This topic is a core part of FSc Part 1, A Level Pure Maths and first-year university calculus, and it becomes easy once you understand why each function has a restricted range.

What Is an Inverse Trig Function?

If \( \sin \theta = \frac{1}{2} \), we want to know which angle \( \theta \) produces that value. The inverse sine function answers this:

$$ \theta = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} $$

The notation \( \sin^{-1} x \) is also written \( \arcsin x \). Importantly, \( \sin^{-1} x \) does not mean \( \frac{1}{\sin x} \). That reciprocal is \( \csc x \).

Why the Range Must Be Restricted

A function can only have an inverse if it is one-to-one: each output comes from exactly one input. Sine is not one-to-one, because \( \sin \frac{\pi}{6} = \sin \frac{5\pi}{6} = \frac{1}{2} \), and the values repeat every \( 2\pi \).

To fix this, we restrict each trig function to an interval where it takes every value exactly once. The answers the inverse function gives are called principal values.

Domains and principal ranges

• \( y = \sin^{-1} x \): domain \( -1 \le x \le 1 \), range \( -\frac{\pi}{2} \le y \le \frac{\pi}{2} \)
• \( y = \cos^{-1} x \): domain \( -1 \le x \le 1 \), range \( 0 \le y \le \pi \)
• \( y = \tan^{-1} x \): domain all real \( x \), range \( -\frac{\pi}{2} < y < \frac{\pi}{2} \)

A quick way to remember: \( \sin^{-1} \) and \( \tan^{-1} \) give angles in the 1st and 4th quadrants, while \( \cos^{-1} \) gives angles in the 1st and 2nd quadrants.

Key Identities

These identities appear again and again in exams:

1. Cancellation: \( \sin(\sin^{-1} x) = x \) for \( -1 \le x \le 1 \). But \( \sin^{-1}(\sin x) = x \) only when \( x \) lies in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
2. Negative arguments:

$$ \sin^{-1}(-x) = -\sin^{-1} x, \quad \tan^{-1}(-x) = -\tan^{-1} x, \quad \cos^{-1}(-x) = \pi - \cos^{-1} x $$

3. Complementary relation:

$$ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}, \qquad -1 \le x \le 1 $$

4. Sum of inverse tangents (valid when \( xy < 1 \)):

$$ \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) $$

Worked Example 1: Evaluating Principal Values

Question: Find (a) \( \cos^{-1}\left(-\frac{1}{2}\right) \) and (b) \( \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) \).

Solution (a):

• Step 1: We know \( \cos \frac{\pi}{3} = \frac{1}{2} \), so \( \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \).
• Step 2: Use the negative-argument identity for cosine:

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

• Check: \( \frac{2\pi}{3} \) lies in \( [0, \pi] \), and \( \cos \frac{2\pi}{3} = -\frac{1}{2} \). Correct.

Solution (b):

• Step 1: \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \).
• Step 2: Sine inverse is an odd function, so

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

• Check: \( -\frac{\pi}{3} \) lies in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Correct.

Worked Example 2: Using the Tangent Sum Formula

Question: Show that \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4} \).

Solution:

• Step 1: Check the condition. \( xy = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} < 1 \), so the formula applies.
• Step 2: Apply the formula.

$$ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}}\right) $$

• Step 3: Simplify the fraction.

$$ \frac{\frac{5}{6}}{\frac{5}{6}} = 1 $$

• Step 4: Evaluate.

$$ \tan^{-1}(1) = \frac{\pi}{4} $$

Hence proved.

Worked Example 3: Composite Expressions

Question: Find the exact value of \( \cos\left(\sin^{-1}\frac{3}{5}\right) \).

Solution:

• Step 1: Let \( \theta = \sin^{-1}\frac{3}{5} \). Then \( \sin \theta = \frac{3}{5} \), and \( \theta \) lies in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Since \( \sin\theta > 0 \), \( \theta \) is in the first quadrant, so \( \cos \theta > 0 \).
• Step 2: Use \( \sin^2\theta + \cos^2\theta = 1 \):

$$ \cos \theta = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} $$

So \( \cos\left(\sin^{-1}\frac{3}{5}\right) = \frac{4}{5} \). You can picture this as a 3-4-5 right triangle.

Common Mistakes

• Confusing \( \sin^{-1} x \) with \( \frac{1}{\sin x} \). The \(-1\) means inverse, not reciprocal.
• Giving answers outside the principal range. For example, \( \cos^{-1}\left(-\frac{1}{2}\right) \) is not \( -\frac{\pi}{3} \) or \( \frac{4\pi}{3} \).
• Assuming \( \sin^{-1}(\sin x) = x \) always. For \( x = \frac{5\pi}{6} \), \( \sin^{-1}\left(\sin\frac{5\pi}{6}\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \), not \( \frac{5\pi}{6} \).
• Calculator in degree mode when the question expects radians, or vice versa.

Tips

• Draw a right triangle for any expression like \( \tan\left(\cos^{-1} x\right) \). It turns the problem into Pythagoras.
• Memorise the standard values at \( 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} \).
• Always check that your final angle lies in the correct principal range.

Key Takeaways

• Inverse trig functions return an angle from a given ratio.
• Ranges are restricted so that each function is one-to-one: \( \sin^{-1} \) and \( \tan^{-1} \) give \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \( \cos^{-1} \) gives \( [0, \pi] \).
• \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) and \( \cos^{-1}(-x) = \pi - \cos^{-1} x \).
• The tangent sum formula combines two inverse tangents when \( xy < 1 \).
• A right-triangle sketch solves most composite expressions quickly.

If inverse trig proofs or exercise questions are slowing you down, Amathify's subject experts and PhD scholars can provide step-by-step solutions and assignment help. Contact us on WhatsApp or send your question through the "Ask a Question" form.