Inverse Trigonometric Functions - Comprehensive Lecture Notes

Introduction and Importance of Trigonometry

  • Trigonometry serves as a foundation for mathematics in engineering and university entrance examinations. It is described as "crucial" and "important" for the following exams:

    • NUST (NET)

    • FAST

    • IBA

    • LUMS

    • NED University of Engineering & Technology

    • UET

    • COMSATS

    • GIKI

  • Without trigonometry, completing a university-level math paper is considered impossible.

  • Inverse Trigonometric Functions represent a new level of complexity within trigonometry that builds upon basic definitions to find angular values from numerical ratios.

Fundamental Distinction: Inverse vs. Reciprocal

  • Conceptual Error: A common misconception among students is that the inverse of a function is the same as its reciprocal. For example, many mistakenly believe that f−1f^{-1} is equal to 1f\frac{1}{f}.

  • Inverse Specifics: In the context of trigonometry, sin⁡−1\sin^{-1} does not mean 1sin⁡\frac{1}{\sin}. The negative exponent indicates the inverse operation, not a reciprocal power.

  • Reciprocal Functions (Normal Trig):

    • sin⁡(θ)=1csc⁡(θ)\sin(\theta) = \frac{1}{\csc(\theta)}

    • cos⁡(θ)=1sec⁡(θ)\cos(\theta) = \frac{1}{\sec(\theta)}

    • tan⁡(θ)=1cot⁡(θ)\tan(\theta) = \frac{1}{\cot(\theta)}

  • The Reversible Process Metaphor:

    • Normal Process: Applying a trigonometric function to an angle yields a numeric value. Example: tan⁡(45∘)=1\tan(45^{\circ}) = 1 or sin⁡(90∘)=1\sin(90^{\circ}) = 1.

    • Inverse Process: Applying an inverse trigonometric function to a numeric value yields an angle. Example: tan⁡−1(1)=45∘\tan^{-1}(1) = 45^{\circ}.

    • This is likened to reversible chemical reactions; the inverse function acts as the reverse chain returning a value back to its angular origin.

Functional Properties and Cancellation Laws

  • Standard Identity Rule: Every function is cancelled out by its own inverse.

  • Formula 1: f(f−1(x))=xf(f^{-1}(x)) = x

  • Formula 2: f−1(f(y))=yf^{-1}(f(y)) = y

  • Application in Solving Equations:

    • If a simple trigonometric function is moved to the other side of an equation without its angle, it converts into an inverse function.

    • Example Case: sin⁡(sin⁡−1(x))=x\sin(\sin^{-1}(x)) = x.

    • Example Case: sin⁡−1(sin⁡(π4+x2))=π4+x2\sin^{-1}(\sin(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}.

  • This property allows for complex simplification in calculus (derivatives) where trigonometric parts might otherwise seem difficult to differentiate.

Handling Negatives in Trigonometric Functions

  • Sine and Tangent (Odd Functions): These functions and their inverses "throw out" the negative sign.

    • sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta)

    • sin⁡−1(−x)=−sin⁡−1(x)\sin^{-1}(-x) = -\sin^{-1}(x)

    • tan⁡(−θ)=−tan⁡(θ)\tan(-\theta) = -\tan(\theta)

    • tan⁡−1(−x)=−tan⁡−1(x)\tan^{-1}(-x) = -\tan^{-1}(x)

  • Example: tan⁡−1(−1)=−tan⁡−1(1)=−45∘\tan^{-1}(-1) = -\tan^{-1}(1) = -45^{\circ}.

  • Cosine (Even Function Property): Normal cosine absorbs the negative sign.

    • cos⁡(−θ)=cos⁡(θ)\cos(-\theta) = \cos(\theta)

  • Inverse Cosine Exception (Crucial Property): The negative sign in an inverse cosine function cannot be ignored or simply thrown out. It requires a specific formula involving a phase shift.

    • Identity: cos⁡−1(−x)=π−cos⁡−1(x)\cos^{-1}(-x) = \pi - \cos^{-1}(x)

    • Note: π\pi is equivalent to 180∘180^{\circ}.

  • Step-by-step Example: Solving cos⁡−1(−12)\cos^{-1}(-\frac{1}{2}):

    1. Use the formula: 180∘−cos⁡−1(12)180^{\circ} - \cos^{-1}(\frac{1}{2}).

    2. We know cos⁡−1(12)=60∘\cos^{-1}(\frac{1}{2}) = 60^{\circ}.

    3. 180∘−60∘=120∘180^{\circ} - 60^{\circ} = 120^{\circ}.

  • Step-by-step Example: Solving cos⁡−1(−32)\cos^{-1}(-\frac{\sqrt{3}}{2}):

    1. Formula: 180∘−cos⁡−1(32)180^{\circ} - \cos^{-1}(\frac{\sqrt{3}}{2}).

    2. We know cos⁡−1(32)=30∘\cos^{-1}(\frac{\sqrt{3}}{2}) = 30^{\circ}.

    3. 180∘−30∘=150∘180^{\circ} - 30^{\circ} = 150^{\circ}.

Interdependence and Conversion of Ratios (PBH Method)

  • Inverse functions can be converted into one another (e.g., converting Sine Inverse to Tangent Inverse) using the Pythagorean Theorem.

  • The Pythagorean Mnemonic:

    • Some People Have (Sin): sin⁡(θ)=PerpendicularHypotenuse\sin(\theta) = \frac{\text{Perpendicular}}{\text{Hypotenuse}}

    • Curly Brown Hair (Cos): cos⁡(θ)=BaseHypotenuse\cos(\theta) = \frac{\text{Base}}{\text{Hypotenuse}}

    • They Painted Black (Tan): tan⁡(θ)=PerpendicularBase\tan(\theta) = \frac{\text{Perpendicular}}{\text{Base}}

  • Pythagoras Theorem: h2=p2+b2h^2 = p^2 + b^2.

  • Case Study: Solving tan⁡(sin⁡−1(x2))\tan(\sin^{-1}(\frac{x}{2})):

    1. Identify Sin components: p=xp = x, h=2h = 2.

    2. Find Base (bb): 22=x2+b2→4−x2=b2→b=4−x22^2 = x^2 + b^2 \rightarrow 4 - x^2 = b^2 \rightarrow b = \sqrt{4 - x^2}.

    3. Convert inside to Tan inverse: tan⁡−1(pb)=tan⁡−1(x4−x2)\tan^{-1}(\frac{p}{b}) = \tan^{-1}(\frac{x}{\sqrt{4 - x^2}}).

    4. Result: tan⁡(tan⁡−1(x4−x2))=x4−x2\tan(\tan^{-1}(\frac{x}{\sqrt{4 - x^2}})) = \frac{x}{\sqrt{4 - x^2}}.

Reciprocal Relationships in Inverse Trigonometry

  • In inverse trigonometry, functions are reciprocals of their angles, not the function itself.

  • Formula Set:

    • sin⁡−1(x)=csc⁡−1(1x)\sin^{-1}(x) = \csc^{-1}(\frac{1}{x})

    • cos⁡−1(x)=sec⁡−1(1x)\cos^{-1}(x) = \sec^{-1}(\frac{1}{x})

    • tan⁡−1(x)=cot⁡−1(1x)\tan^{-1}(x) = \cot^{-1}(\frac{1}{x})

  • Defining "Arc": The term "Arc" is synonymous with inverse. ArcSin(x)\text{ArcSin}(x) is sin⁡−1(x)\sin^{-1}(x), ArcCos(x)\text{ArcCos}(x) is cos⁡−1(x)\cos^{-1}(x), and ArcTan(x)\text{ArcTan}(x) is tan⁡−1(x)\tan^{-1}(x).

Fundamental Addition and Subtraction Identities

  • Addition Formula for Tangent Inverse:

    • tan⁡−1(a)+tan⁡−1(b)=tan⁡−1(a+b1−(a×b))\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}(\frac{a + b}{1 - (a \times b)})

  • Subtraction Formula for Tangent Inverse:

    • tan⁡−1(a)−tan⁡−1(b)=tan⁡−1(a−b1+(a×b))\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}(\frac{a - b}{1 + (a \times b)})

  • Mnemonic for signs: In addition, the numerator matches the positive sign while the denominator is opposite. In subtraction, the numerator matches the negative sign while the denominator is opposite.

  • Sum of Complementary Ratios:

    • sin⁡−1(x)+cos⁡−1(x)=π2\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}

    • tan⁡−1(x)+cot⁡−1(x)=π2\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}

    • sec⁡−1(x)+csc⁡−1(x)=π2\sec^{-1}(x) + \csc^{-1}(x) = \frac{\pi}{2}

Advanced Example: The Double Angle Scenario

  • Problem: Solve sin⁡(2×ArcTan(3))\sin(2 \times \text{ArcTan}(3)).

  • Strategy: Recognize this as sin⁡(2θ)\sin(2\theta) where θ=tan⁡−1(3)\theta = \tan^{-1}(3).

  • Formula: sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta).

  • Execution:

    1. Let tan⁡−1(3)\tan^{-1}(3) imply p=3,b=1p = 3, b = 1.

    2. Calculate Hypotenuse: h=32+12=10h = \sqrt{3^2 + 1^2} = \sqrt{10}.

    3. Convert components: sin⁡(θ)=310\sin(\theta) = \frac{3}{\sqrt{10}} and cos⁡(θ)=110\cos(\theta) = \frac{1}{\sqrt{10}}.

    4. Apply formula: 2×(310)×(110)2 \times (\frac{3}{\sqrt{10}}) \times (\frac{1}{\sqrt{10}}).

    5. Simplify: 610=35\frac{6}{10} = \frac{3}{5}.

    6. Rule of Roots: 10×10\sqrt{10} \times \sqrt{10} results in the number being written once without the root (e.g., 1010).

Domain and Range of Inverse Functions

  • Definitions:

    • Domain: The input value (likened to fuel in a bike).

    • Range: The output value (likened to the distance/speed the bike travels).

  • Properties on Axes (The Table):

    • Sine: Always 00 on the x-axis (0∘,180∘,360∘0^{\circ}, 180^{\circ}, 360^{\circ}). Value is 11 or −1-1 on the y-axis (90∘,270∘90^{\circ}, 270^{\circ}).

    • Cosine: Always 00 on the y-axis. Value is 11 or −1-1 on the x-axis.

    • Tangent: Always 00 on the x-axis. Value is ∞\infty (Infinity) on the y-axis.

  • Inverse Domain and Range Summary:

    • sin⁡−1(x)\sin^{-1}(x): Domain: [−1,1][-1, 1]. Range: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

    • cos⁡−1(x)\cos^{-1}(x): Domain: [−1,1][-1, 1]. Range: [0,π][0, \pi].

    • tan⁡−1(x)\tan^{-1}(x): Domain: All Real Numbers (R)\text{All Real Numbers } (R). Range: (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) (exclusive of limits due to infinity).

    • cot⁡−1(x)\cot^{-1}(x): Domain: RR. Range: (0,π)(0, \pi).

Questions & Discussion

  • Question from student: Can you explain the double angle formula again?

  • Response: sin⁡(2θ)\sin(2\theta) means you take the value 22 as a coefficient outside according to the formula, then multiply by the sine of the angle and the cosine of the angle. For example, sin⁡(2×30∘)=2sin⁡(30∘)cos⁡(30∘)\sin(2 \times 30^{\circ}) = 2\sin(30^{\circ})\cos(30^{\circ}).

  • Challenge Problem: Solve for xx in ArcCos(2x2−2x)=2π3\text{ArcCos}(2x^2 - 2x) = \frac{2\pi}{3}.

    1. Move ArcCos to other side: 2x2−2x=cos⁡(2π3)2x^2 - 2x = \cos(\frac{2\pi}{3}).

    2. 2π3=120∘\frac{2\pi}{3} = 120^{\circ}.

    3. cos⁡(120∘)\cos(120^{\circ}) is calculated via cos⁡(90∘+30∘)=−sin⁡(30∘)=−12\cos(90^{\circ} + 30^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2}. (Reference: vertical line/co-function rules).

    4. Equation becomes: 2x2−2x=−122x^2 - 2x = -\frac{1}{2}.

    5. Multiply by 22: 4x2−4x+1=04x^2 - 4x + 1 = 0.

    6. Recognise perfect square: (2x−1)2=0(2x - 1)^2 = 0.

    7. Final answer: x=12x = \frac{1}{2}.