The Hook: The GPS Problem
Imagine you’re watching Mission: Impossible and Ethan Hunt needs to calculate the angle to throw his grappling hook. He knows:
- Distance to building: 20 meters (adjacent)
- Height of window: 15 meters (opposite)
He needs the angle θ. He can’t use sin θ = 15/hypotenuse because he doesn’t know hypotenuse yet. But he CAN use:
$$\theta = \tan^{-1}\left(\frac{15}{20}\right) = \tan^{-1}(0.75) \approx 36.87°$$That’s an inverse trig function! It answers: “What angle gives me this ratio?”
JEE Reality: Inverse trig shows up in calculus integration, limits, complex numbers, and physics. Master it, and you’ve unlocked 8-10 marks in JEE!
Question: Why is sin⁻¹(sin 200°) ≠ 200°? By the end of this, you’ll understand principal values and domain restrictions.
The Core Concept
What Are Inverse Trigonometric Functions?
Normal trig function: Given angle θ → Find ratio
- sin 30° = 1/2
Inverse trig function: Given ratio → Find angle
- sin⁻¹(1/2) = 30°
In simple terms: If y = sin x means “x is the angle whose sine is y,” then x = sin⁻¹ y means “x is the angle whose sine is y.”
Critical Difference from Algebra:
- In algebra: If y = x², then x = ±√y (two values)
- In trig: If y = sin x, then x = sin⁻¹ y (infinite values possible!)
- Solution: We restrict to principal value (one specific value in a defined range)
The Six Inverse Functions
Notation (Both are correct!)
| Function | Notation 1 | Notation 2 | Reads As |
|---|---|---|---|
| Inverse Sine | sin⁻¹ x | arcsin x | “Sine inverse x” or “Arc sine x” |
| Inverse Cosine | cos⁻¹ x | arccos x | “Cosine inverse x” |
| Inverse Tangent | tan⁻¹ x | arctan x | “Tangent inverse x” |
| Inverse Cotangent | cot⁻¹ x | arccot x | “Cotangent inverse x” |
| Inverse Secant | sec⁻¹ x | arcsec x | “Secant inverse x” |
| Inverse Cosecant | cosec⁻¹ x | arccosec x | “Cosecant inverse x” |
Warning: sin⁻¹ x ≠ 1/sin x ≠ cosec x
- sin⁻¹ x is the inverse function
- (sin x)⁻¹ = 1/sin x is the reciprocal
Memory Tricks & Patterns
Domain and Range: The “Mirror Rule”
For a function and its inverse, domain and range are swapped!
$$\boxed{ \begin{array}{|c|c|c|} \hline \textbf{Function} & \textbf{Domain} & \textbf{Range (Principal Value)} \\ \hline \sin^{-1} x & [-1, 1] & [-\pi/2, \pi/2] \text{ or } [-90°, 90°] \\ \cos^{-1} x & [-1, 1] & [0, \pi] \text{ or } [0°, 180°] \\ \tan^{-1} x & \mathbb{R} \text{ (all reals)} & (-\pi/2, \pi/2) \text{ or } (-90°, 90°) \\ \cot^{-1} x & \mathbb{R} & (0, \pi) \text{ or } (0°, 180°) \\ \sec^{-1} x & (-\infty, -1] \cup [1, \infty) & [0, \pi] - \{\pi/2\} \\ \csc^{-1} x & (-\infty, -1] \cup [1, \infty) & [-\pi/2, \pi/2] - \{0\} \\ \hline \end{array} }$$Mnemonic: “SCOOT at 90, REST at 180”
For ranges:
- Sin⁻¹ and Csc⁻¹: centered at 0, span ±90° → [-90°, 90°]
- Tan⁻¹: centered at 0, approaches ±90° → (-90°, 90°) (open interval)
- Cos⁻¹, Sec⁻¹, Cot⁻¹: start at 0, go to 180° → [0°, 180°]
Visual Memory:
sin⁻¹ range cos⁻¹ range
| |
90° ---- 180° ----
| |
0° ---- 90° ----
| |
-90° ---- 0° ----
Why These Specific Ranges?
Principle: Choose range where original function is one-to-one (passes horizontal line test).
Example for sin⁻¹:
- Sin x repeats every 360°
- We need a portion where it’s strictly increasing (one y for each x)
- [-90°, 90°] works! Sin increases from -1 to +1 here
- This becomes the range of sin⁻¹ x
Principal Value vs General Value
Principal Value (What Calculators Give)
Definition: The unique value of the inverse function in its defined range.
Example:
sin⁻¹(1/2) = 30° (principal value)
But sin 150° = 1/2 also! And sin 390°, sin 510°, ...
We choose 30° because it's in [-90°, 90°]
Why sin⁻¹(sin θ) ≠ θ Always
$$\boxed{\sin^{-1}(\sin\theta) = \theta \text{ ONLY if } \theta \in [-90°, 90°]}$$Example:
sin⁻¹(sin 200°) ≠ 200°
Step 1: Find sin 200°
sin 200° = sin(180° + 20°) = -sin 20°
Step 2: Find sin⁻¹(-sin 20°)
sin⁻¹(-sin 20°) = -sin⁻¹(sin 20°) = -20°
Answer: -20° (not 200°!)
JEE Trap: Questions often ask sin⁻¹(sin 5π/4). Answer is NOT 5π/4!
Important Properties and Identities
Complementary Angle Relations
$$\boxed{ \begin{align} \sin^{-1} x + \cos^{-1} x &= \frac{\pi}{2} \text{ (always!)} \\ \tan^{-1} x + \cot^{-1} x &= \frac{\pi}{2} \\ \sec^{-1} x + \csc^{-1} x &= \frac{\pi}{2} \end{align} }$$Memory Trick: “Sine and Cosine are 90° Complementary Partners”
Proof for sin⁻¹ x + cos⁻¹ x = π/2:
Let sin⁻¹ x = θ
Then sin θ = x
And cos(90° - θ) = x (co-function identity)
So 90° - θ = cos⁻¹ x
Therefore θ + cos⁻¹ x = 90°
sin⁻¹ x + cos⁻¹ x = π/2
JEE Application: If you forget cos⁻¹ formula, use cos⁻¹ x = π/2 - sin⁻¹ x!
Negative Argument Properties
$$\boxed{ \begin{align} \sin^{-1}(-x) &= -\sin^{-1} x \\ \tan^{-1}(-x) &= -\tan^{-1} x \\ \csc^{-1}(-x) &= -\csc^{-1} x \\ \cos^{-1}(-x) &= \pi - \cos^{-1} x \\ \sec^{-1}(-x) &= \pi - \sec^{-1} x \\ \cot^{-1}(-x) &= \pi - \cot^{-1} x \end{align} }$$Pattern Recognition:
- Sin, Tan, Csc are ODD functions: f(-x) = -f(x)
- Cos, Sec, Cot use the π shift: f(-x) = π - f(x)
Mnemonic: “STC are Odd, CSC use Pi” (Sine-Tan-Cosec vs Cos-Sec-Cot)
Addition Formulas
$$\boxed{ \begin{align} \tan^{-1} x + \tan^{-1} y &= \tan^{-1}\left(\frac{x+y}{1-xy}\right), \quad xy < 1 \\ \tan^{-1} x + \tan^{-1} y &= \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right), \quad xy > 1, x > 0 \\ \tan^{-1} x - \tan^{-1} y &= \tan^{-1}\left(\frac{x-y}{1+xy}\right) \\ \end{align} }$$Memory Trick: Same as tan(A+B) formula, but with tan⁻¹ on the outside!
JEE Shortcut: If xy < 1, simple formula. If xy > 1, add π.
Special Double Argument Formulas
$$\boxed{ \begin{align} 2\sin^{-1} x &= \sin^{-1}(2x\sqrt{1-x^2}) \\ 2\cos^{-1} x &= \cos^{-1}(2x^2 - 1) \\ 2\tan^{-1} x &= \tan^{-1}\left(\frac{2x}{1-x^2}\right), \quad |x| < 1 \\ 2\tan^{-1} x &= \pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right), \quad x > 1 \\ 2\tan^{-1} x &= -\pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right), \quad x < -1 \end{align} }$$Derivation Hint: Let θ = tan⁻¹ x, then tan θ = x. Use tan 2θ = 2 tan θ/(1 - tan²θ).
Conversion Between Inverse Functions
sin⁻¹ x in Terms of Others
$$\boxed{ \begin{align} \sin^{-1} x &= \cos^{-1}\sqrt{1-x^2} \\ &= \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \\ &= \cot^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \end{align} }$$Visual Method: Draw a right triangle!
If sin θ = x, then:
•
/|
1 / | x (opposite)
/ |
/θ |
•----•
√(1-x²) (adjacent)
From this triangle:
cos θ = √(1-x²)
tan θ = x/√(1-x²)
tan⁻¹ x in Terms of sin⁻¹ and cos⁻¹
$$\boxed{ \begin{align} \tan^{-1} x &= \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \\ &= \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \end{align} }$$Triangle for tan⁻¹ x:
If tan θ = x, then:
•
/|
√(1+x²) / | x (opposite)
/ |
/θ |
•----•
1 (adjacent)
sin θ = x/√(1+x²)
cos θ = 1/√(1+x²)
When to Use Inverse Trig Functions
Integration of 1/√(1-x²)? → Answer: sin⁻¹ x + C
Integration of 1/(1+x²)? → Answer: tan⁻¹ x + C
Simplifying sin⁻¹ x + cos⁻¹ x? → Use: sin⁻¹ x + cos⁻¹ x = π/2
Finding tan⁻¹ 1 + tan⁻¹ 2 + tan⁻¹ 3? → Use addition formula (watch for xy > 1 case!)
Solving equations like sin⁻¹ x = cos⁻¹ x? → Use: x = π/4 (since both must equal π/4 to sum to π/2)
Differentiation of inverse trig functions? → Use standard formulas (d/dx sin⁻¹ x = 1/√(1-x²), etc.)
Given right triangle sides, find angle? → Use inverse trig directly (arctan(opp/adj), arcsin(opp/hyp), etc.)
Common Mistakes to Avoid
Wrong: sin⁻¹ x = 1/sin x ❌
Right:
- sin⁻¹ x is the inverse function (arcsin x)
- 1/sin x is the reciprocal (cosec x)
Example:
sin⁻¹(1/2) = 30° ✓
1/sin(1/2) = 1/0.479 ≈ 2.09 (completely different!)
Fix: Use arcsin notation if confused, or think “sin to the power minus 1” means inverse operation, not reciprocal.
Wrong: sin⁻¹(2) exists ❌
Right: sin⁻¹ x is defined ONLY for x ∈ [-1, 1]
Why: Sine values can never exceed 1 or go below -1!
JEE Trap:
Find sin⁻¹(sin 7π/6)
Step 1: sin 7π/6 = sin(π + π/6) = -sin π/6 = -1/2
Step 2: sin⁻¹(-1/2) = -π/6 (in range [-π/2, π/2])
Answer: -π/6 (NOT 7π/6!)
Wrong: tan⁻¹ 2 + tan⁻¹ 3 = tan⁻¹[(2+3)/(1-6)] = tan⁻¹(-1) = -π/4
Right:
Here xy = 2·3 = 6 > 1 and x, y > 0
So tan⁻¹ 2 + tan⁻¹ 3 = π + tan⁻¹[(2+3)/(1-6)]
= π + tan⁻¹(-1)
= π - π/4
= 3π/4
Fix: Always check if xy < 1 or xy > 1 before applying tan⁻¹ addition formula!
Wrong: sin⁻¹(sin 250°) = 250° ❌
Right:
Step 1: Bring angle to standard position
sin 250° = sin(180° + 70°) = -sin 70°
Step 2: Apply sin⁻¹
sin⁻¹(-sin 70°) = -sin⁻¹(sin 70°) = -70°
General Rule:
$$\sin^{-1}(\sin\theta) = \begin{cases} \theta & \text{if } \theta \in [-90°, 90°] \\ 180° - \theta & \text{if } \theta \in [90°, 270°] \\ \theta - 360° & \text{if } \theta \in [270°, 450°] \end{cases}$$Fix: First reduce the argument to [-1, 1], then find principal value in range.
Practice Problems
Level 1: Foundation (NCERT)
Question: Find the exact value of sin⁻¹(√3/2)
Solution: We need angle θ such that sin θ = √3/2 and θ ∈ [-90°, 90°]
From standard values: sin 60° = √3/2
Since 60° ∈ [-90°, 90°]:
Answer: sin⁻¹(√3/2) = 60° = π/3
Question: Prove that sin⁻¹(3/5) + cos⁻¹(3/5) = π/2
Solution: Use the identity: sin⁻¹ x + cos⁻¹ x = π/2 for all x ∈ [-1, 1]
Since 3/5 ∈ [-1, 1], the identity holds.
Therefore: sin⁻¹(3/5) + cos⁻¹(3/5) = π/2 ✓
Alternatively (verification):
Let sin⁻¹(3/5) = α, so sin α = 3/5
Draw right triangle:
•
/|
5 / | 3
/ |
/α |
•----•
4
cos α = 4/5
So cos⁻¹(4/5) = α = sin⁻¹(3/5)
But we need cos⁻¹(3/5):
If sin α = 3/5, then cos(90° - α) = 3/5
So 90° - α = cos⁻¹(3/5)
α + cos⁻¹(3/5) = 90° ✓
Proved!
Question: Find tan⁻¹(-1)
Solution: Method 1: Use property tan⁻¹(-x) = -tan⁻¹(x)
tan⁻¹(-1) = -tan⁻¹(1) = -π/4 = -45°
Method 2: Direct
We need θ such that tan θ = -1 and θ ∈ (-90°, 90°)
tan(-45°) = -1 ✓
Answer: -π/4 or -45°
Level 2: JEE Main
Question: Simplify: sin⁻¹(3/5) + cos⁻¹(12/13)
Solution: Let α = sin⁻¹(3/5) and β = cos⁻¹(12/13)
Find sin α and cos α:
sin α = 3/5
cos α = √(1 - 9/25) = 4/5
Find sin β and cos β:
cos β = 12/13
sin β = √(1 - 144/169) = 5/13
Use sin(α + β) = sin α cos β + cos α sin β:
sin(α + β) = (3/5)(12/13) + (4/5)(5/13)
= 36/65 + 20/65
= 56/65
Therefore: α + β = sin⁻¹(56/65)
Answer: sin⁻¹(56/65)
Alternative: If you recognize 3-4-5 and 5-12-13 Pythagorean triples, use geometry!
Question: Find tan⁻¹(1/2) + tan⁻¹(1/3)
Solution: Use formula: tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x+y)/(1-xy)] when xy < 1
Here x = 1/2, y = 1/3:
xy = (1/2)(1/3) = 1/6 < 1 ✓
tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹[(1/2 + 1/3)/(1 - 1/6)]
= tan⁻¹[(5/6)/(5/6)]
= tan⁻¹(1)
= π/4
Answer: π/4 or 45°
Question: Find the value of sin⁻¹(sin 3π/4)
Solution: Wrong approach: sin⁻¹(sin 3π/4) = 3π/4 ❌ (3π/4 is NOT in [-π/2, π/2])
Correct approach:
Step 1: Find sin 3π/4
sin 3π/4 = sin(π - π/4) = sin π/4 = √2/2
Step 2: Find principal value
sin⁻¹(√2/2) = π/4 (since π/4 ∈ [-π/2, π/2])
Answer: π/4 (NOT 3π/4!)
General Pattern:
For θ ∈ [π/2, π]:
sin⁻¹(sin θ) = π - θ
Verify: sin⁻¹(sin 3π/4) = π - 3π/4 = π/4 ✓
Level 3: JEE Advanced
Question: Prove that tan⁻¹(1/2) + tan⁻¹(1/5) + tan⁻¹(1/8) = π/4
Solution: Step 1: Add first two terms using formula
tan⁻¹(1/2) + tan⁻¹(1/5)
= tan⁻¹[(1/2 + 1/5)/(1 - 1/10)]
= tan⁻¹[(7/10)/(9/10)]
= tan⁻¹(7/9)
Step 2: Add result with third term
tan⁻¹(7/9) + tan⁻¹(1/8)
= tan⁻¹[(7/9 + 1/8)/(1 - 7/72)]
= tan⁻¹[(56/72 + 9/72)/(65/72)]
= tan⁻¹[(65/72)/(65/72)]
= tan⁻¹(1)
= π/4
Proved!
JEE Pattern: Such sums are designed to collapse to π/4 or other standard values!
Question: Simplify: 2 tan⁻¹(1/3) - tan⁻¹(1/7)
Solution: Step 1: Use 2 tan⁻¹ x formula
2 tan⁻¹(1/3) = tan⁻¹[2(1/3)/(1 - 1/9)]
= tan⁻¹[(2/3)/(8/9)]
= tan⁻¹(2/3 × 9/8)
= tan⁻¹(3/4)
Step 2: Subtract using formula tan⁻¹ x - tan⁻¹ y = tan⁻¹[(x-y)/(1+xy)]
tan⁻¹(3/4) - tan⁻¹(1/7)
= tan⁻¹[(3/4 - 1/7)/(1 + 3/28)]
= tan⁻¹[(21/28 - 4/28)/(31/28)]
= tan⁻¹[(17/28)/(31/28)]
= tan⁻¹(17/31)
Answer: tan⁻¹(17/31)
Question: The value of cos⁻¹(cos 10) is: (Where 10 is in radians)
Solution: Key insight: 10 radians is NOT in [0, π] which is the range of cos⁻¹!
Step 1: Find equivalent angle in [0, 2π]
10 rad ≈ 10 rad
Since 2π ≈ 6.28, we have:
10 = 2π - (2π - 10) = 2π - (6.28 - 10)
Wait, 10 > 2π, so 10 = 3·2π + remainder
10 ÷ 6.28 ≈ 1.59
10 = 1·(2π) + (10 - 2π)
10 - 2π ≈ 10 - 6.28 = 3.72 rad
Step 2: Since cos is 2π periodic:
cos 10 = cos(10 - 2π) = cos(3.72...)
But 3.72 > π, so use:
cos θ = cos(2π - θ)
cos 3.72 = cos(2π - 3.72) = cos(6.28 - 3.72) = cos 2.56
Actually, simpler approach:
For cos⁻¹(cos θ):
- If θ ∈ [0, π]: answer is θ
- If θ ∈ (π, 2π]: answer is 2π - θ
- If θ > 2π: reduce modulo 2π first
10 rad = 10 - 2π ≈ 3.72 rad (after one cycle)
Since 3.72 > π:
cos⁻¹(cos 10) = cos⁻¹(cos 3.72) = 2π - 3.72 ≈ 2.56
But exact form:
10 = 3π + (10 - 3π)
cos 10 = cos(10 - 2π) since cos has period 2π
10 - 2π ≈ 3.717
Since 10 - 2π > π:
Answer = 2π - (10 - 2π) = 4π - 10
Answer: 4π - 10 rad
JEE Trick: For cos⁻¹(cos θ), reduce to [0, 2π] then to [0, π] using symmetry!
Integration and Differentiation (Quick Reference)
Derivatives of Inverse Trig Functions
$$\boxed{ \begin{align} \frac{d}{dx}\sin^{-1} x &= \frac{1}{\sqrt{1-x^2}}, \quad |x| < 1 \\ \frac{d}{dx}\cos^{-1} x &= \frac{-1}{\sqrt{1-x^2}}, \quad |x| < 1 \\ \frac{d}{dx}\tan^{-1} x &= \frac{1}{1+x^2} \\ \frac{d}{dx}\cot^{-1} x &= \frac{-1}{1+x^2} \\ \frac{d}{dx}\sec^{-1} x &= \frac{1}{|x|\sqrt{x^2-1}}, \quad |x| > 1 \\ \frac{d}{dx}\csc^{-1} x &= \frac{-1}{|x|\sqrt{x^2-1}}, \quad |x| > 1 \end{align} }$$Pattern: Co-functions (cos⁻¹, cot⁻¹, csc⁻¹) have negative derivatives!
Standard Integrals
$$\boxed{ \begin{align} \int \frac{1}{\sqrt{1-x^2}} dx &= \sin^{-1} x + C \\ \int \frac{1}{1+x^2} dx &= \tan^{-1} x + C \\ \int \frac{1}{x\sqrt{x^2-1}} dx &= \sec^{-1} x + C \end{align} }$$JEE High-Yield: The tan⁻¹ x integral appears in 50% of inverse trig integration problems!
Quick Revision Box
| Function | Domain | Range | Key Property |
|---|---|---|---|
| sin⁻¹ x | [-1, 1] | [-π/2, π/2] | sin⁻¹(-x) = -sin⁻¹ x |
| cos⁻¹ x | [-1, 1] | [0, π] | cos⁻¹(-x) = π - cos⁻¹ x |
| tan⁻¹ x | ℝ | (-π/2, π/2) | tan⁻¹(-x) = -tan⁻¹ x |
| sin⁻¹ + cos⁻¹ | — | — | = π/2 (always) |
| tan⁻¹ x + tan⁻¹ y | xy < 1 | — | = tan⁻¹[(x+y)/(1-xy)] |
| sin⁻¹(sin θ) | — | — | = θ only if θ ∈ range |
Links to Related Topics
Prerequisites:
- Trigonometric Identities - Essential for simplification
- Trigonometric Equations - Principal values connection
Connected Concepts:
- Limits and Continuity - Limits involving inverse trig
- Differentiation - Chain rule with inverse trig
- Integration Techniques - Standard integrals
- Complex Numbers - Euler’s Form - Argument of complex numbers
- Definite Integrals - Area under curves
What’s Next:
- Properties of Triangles - Apply inverse trig to find angles
- Differential Calculus - Advanced applications
Teacher’s Summary
Inverse ≠ Reciprocal: sin⁻¹ x is the inverse function (arcsin), NOT 1/sin x. Keep this crystal clear!
Domain is Sacred: sin⁻¹ and cos⁻¹ work ONLY for |x| ≤ 1. Going outside this gives no real solution.
Principal Value Range Matters: sin⁻¹(sin θ) = θ ONLY if θ is in [-π/2, π/2]. Otherwise, reduce to principal range first.
The Magic π/2 Formula: sin⁻¹ x + cos⁻¹ x = π/2 is your best friend. Use it to convert between functions instantly.
Addition Formulas Have Conditions: For tan⁻¹ x + tan⁻¹ y, check if xy < 1 or xy > 1. Formula changes with π addition!
Integration Shortcuts: Memorize the big 3: ∫1/√(1-x²) = sin⁻¹ x, ∫1/(1+x²) = tan⁻¹ x, ∫1/(x√(x²-1)) = sec⁻¹ x. These save minutes!
Exam Strategy: When stuck on inverse trig, draw a right triangle! Visual representation often reveals the answer faster than algebra.
“Inverse trig is not about formulas—it’s about understanding that every ratio has an angle, and we’re just finding it within the right neighborhood (principal range).”
Weightage:
- JEE Main: 1-2 direct questions + 3-4 in calculus (integration, differentiation)
- JEE Advanced: 1 question + heavily used in complex numbers and calculus
Time-Saving Tip: Create a “Domain-Range Cheat Card” with all 6 inverse functions. Laminate it. Look at it daily for 2 weeks. You’ll never forget!