Calculus Lecture on Functions: Trigonometric and Inverse Properties

Fundamental Trigonometric Functions

  • Definitions and Right-Angled Triangle Relationships

    • sin⁡(θ)=oppositehypotenuse\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}

    • cos⁡(θ)=adjacenthypotenuse\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}

    • tan⁡(θ)=oppositeadjacent=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin(\theta)}{\cos(\theta)}

    • cosec(θ)=1sin⁡(θ)=hypotenuseopposite\text{cosec}(\theta) = \frac{1}{\sin(\theta)} = \frac{\text{hypotenuse}}{\text{opposite}}

    • sec⁡(θ)=1cos⁡(θ)=hypotenuseadjacent\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{\text{hypotenuse}}{\text{adjacent}}

    • cot⁡(θ)=1tan⁡(θ)=adjacentopposite\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{\text{adjacent}}{\text{opposite}}

  • Angle Measurements and Conversions

    • In calculus, angles θ\theta are primarily measured in radians.

    • Conversion from Degrees to Radians: Multiply by π180\frac{\pi}{180}.

    • Conversion from Radians to Degrees: Multiply by 180π\frac{180}{\pi}.

  • Common Angle Values Table

Degrees

Radians

30∘30^{\circ}

π6\frac{\pi}{6}

45∘45^{\circ}

π4\frac{\pi}{4}

60∘60^{\circ}

π3\frac{\pi}{3}

90∘90^{\circ}

π2\frac{\pi}{2}

180∘180^{\circ}

π\pi

Properties of Specific Trigonometric Functions

  • Function: y=sin⁡(x)y = \sin(x)

    • Type: Odd function.

    • Domain: R\mathbb{R}.

    • Range: [−1,1][-1, 1].

    • Period: 2π2\pi.

    • Frequency: 12π\frac{1}{2\pi}.

    • Amplitude: 11.

  • Function: y=sin⁡(2x)y = \sin(2x)

    • Type: Odd function.

    • Domain: R\mathbb{R}.

    • Range: [−1,1][-1, 1].

    • Period: 2π2=π\frac{2\pi}{2} = \pi.

    • Frequency: 1π\frac{1}{\pi}.

    • Amplitude: 11.

  • Function: y=sin⁡(12x)y = \sin(\frac{1}{2}x)

    • Type: Odd function.

    • Domain: R\mathbb{R}.

    • Range: [−1,1][-1, 1].

    • Period: 2π1/2=4π\frac{2\pi}{1/2} = 4\pi.

    • Frequency: 14π\frac{1}{4\pi}.

    • Amplitude: 11.

  • Function: y=sin⁡(x+π2)y = \sin(x + \frac{\pi}{2})

    • Note: This is equivalent to cos⁡(x)\cos(x).

    • Type: Even function.

    • Domain: R\mathbb{R}.

    • Range: [−1,1][-1, 1].

    • Period: 2π2\pi.

    • Frequency: 12π\frac{1}{2\pi}.

    • Amplitude: 11.

  • Function: y=3sin⁡(2x)y = 3\sin(2x)

    • Type: Odd function.

    • Domain: R\mathbb{R}.

    • Range: [−3,3][-3, 3].

    • Period: 2π2=π\frac{2\pi}{2} = \pi.

    • Frequency: 1π\frac{1}{\pi}.

    • Amplitude: 33.

  • Function: y=cos⁡(x)y = \cos(x)

    • Type: Even function.

    • Domain: R\mathbb{R}.

    • Range: [−1,1][-1, 1].

    • Period: 2π2\pi.

    • Frequency: 12π\frac{1}{2\pi}.

    • Amplitude: 11.

  • Function: y=tan⁡(x)y = \tan(x)

    • Type: Odd function.

    • Domain: R−{±nπ2:n is odd}\mathbb{R} - \{\pm \frac{n\pi}{2} : n \text{ is odd}\}.

    • Range: R\mathbb{R}.

    • Period: π\pi.

    • Frequency: 1π\frac{1}{\pi}.

    • Amplitude: ∞\infty.

Trigonometric Identities

  1. sin⁡2(□)+cos⁡2(□)=1\sin^2(\square) + \cos^2(\square) = 1

  2. 1+tan⁡2(□)=sec⁡2(□)1 + \tan^2(\square) = \sec^2(\square)

  3. cot⁡2(□)+1=cosec2(□)\cot^2(\square) + 1 = \text{cosec}^2(\square)

  4. sin⁡(x±y)=sin⁡(x)cos⁡(y)±cos⁡(x)sin⁡(y)\sin(x \pm y) = \sin(x)\cos(y) \pm \cos(x)\sin(y)

  5. cos⁡(x±y)=cos⁡(x)cos⁡(y)∓sin⁡(x)sin⁡(y)\cos(x \pm y) = \cos(x)\cos(y) \mp \sin(x)\sin(y)

  6. tan⁡(x±y)=tan⁡(x)±tan⁡(y)1∓tan⁡(x)tan⁡(y)\tan(x \pm y) = \frac{\tan(x) \pm \tan(y)}{1 \mp \tan(x)\tan(y)}

  7. sin⁡(2□)=2sin⁡(□)cos⁡(□)\sin(2\square) = 2\sin(\square)\cos(\square)

  8. cos⁡(2□)=cos⁡2(□)−sin⁡2(□)=2cos⁡2(□)−1=1−2sin⁡2(□)\cos(2\square) = \cos^2(\square) - \sin^2(\square) = 2\cos^2(\square) - 1 = 1 - 2\sin^2(\square)

  9. tan⁡(2□)=2tan⁡(□)1−tan⁡2(□)\tan(2\square) = \frac{2\tan(\square)}{1 - \tan^2(\square)}

  10. sin⁡2(□)=12(1−cos⁡(2□))\sin^2(\square) = \frac{1}{2}(1 - \cos(2\square))

  11. cos⁡2(□)=12(1+cos⁡(2□))\cos^2(\square) = \frac{1}{2}(1 + \cos(2\square))

  12. sin⁡(x)cos⁡(y)=12[sin⁡(x+y)+sin⁡(x−y)]\sin(x)\cos(y) = \frac{1}{2}[\sin(x + y) + \sin(x - y)]

  13. cos⁡(x)cos⁡(y)=12[cos⁡(x+y)+cos⁡(x−y)]\cos(x)\cos(y) = \frac{1}{2}[\cos(x + y) + \cos(x - y)]

  14. sin⁡(x)sin⁡(y)=12[cos⁡(x−y)−cos⁡(x+y)]\sin(x)\sin(y) = \frac{1}{2}[\cos(x - y) - \cos(x + y)]

Inverse Functions

  • Definition

    • An inverse function f−1f^{-1} undoes the operation performed by the original function ff.

    • Example: If f(x)=x3f(x) = x^3, then the inverse is f−1(x)=x3f^{-1}(x) = \sqrt[3]{x}.

    • Mapping Concept: If f(x)f(x) map an input of 22 to an output of 88, then f−1(x)f^{-1}(x) maps an input of 88 to an output of 22.

  • Procedure to Find f−1(x)f^{-1}(x)

    1. Interchange xx and yy in the equation.

    2. Solve the equation for yy to express it as a function of xx.

    • Example: Find the inverse for y=x3+2y = x^3 + 2.

      1. x=y3+2x = y^3 + 2

      2. x−2=y3x - 2 = y^3

      3. y=x−23y = \sqrt[3]{x - 2}

      4. Therefore, f−1(x)=x−23f^{-1}(x) = \sqrt[3]{x - 2}.

  • The One-to-One Condition

    • To have an inverse, a function must be one-to-one.

    • A function is one-to-one if every horizontal line intersects its graph at most once (Horizontal Line Test).

    • If a function is not one-to-one (e.g., y=x2y = x^2), its domain must be restricted to a suitable interval where it becomes one-to-one (e.g., x≥0x \geq 0 or x≤0x \leq 0).

  • Verification of Inverse Functions

    • A function g(x)g(x) is the inverse of f(x)f(x) if and only if:

      • f(g(x))=xf(g(x)) = x (Valid within the restricted domain of g(x)g(x))

      • g(f(x))=xg(f(x)) = x (Valid within the restricted domain of f(x)f(x))

  • Domain and Range Relationships

    • Domain of f(x)=Range of f−1(x)\text{Domain of } f(x) = \text{Range of } f^{-1}(x)

    • Range of f(x)=Domain of f−1(x)\text{Range of } f(x) = \text{Domain of } f^{-1}(x)

  • Graphical Representation

    • The graph of f−1(x)f^{-1}(x) is the reflection of the graph of f(x)f(x) across the line y=xy = x.

  • Inverse Calculation Examples

    • Example 1: f(x)=xx−2f(x) = \frac{x}{x - 2}

      • Interchange: x=yy−2x = \frac{y}{y - 2}

      • Solve: xy−2x=y→xy−y=2x→y(x−1)=2x→y=2xx−1xy - 2x = y \rightarrow xy - y = 2x \rightarrow y(x - 1) = 2x \rightarrow y = \frac{2x}{x - 1}

      • Answer: f−1(x)=2xx−1f^{-1}(x) = \frac{2x}{x - 1}

      • Df=R−{2}D_f = \mathbb{R} - \{2\}, Rf=Df−1=R−{1}R_f = D_{f^{-1}} = \mathbb{R} - \{1\}.

    • Example 2: f(x)=4x−3x+1f(x) = \frac{4x - 3}{x + 1}

      • Interchange: x=4y−3y+1x = \frac{4y - 3}{y + 1}

      • Solve: xy+x=4y−3→xy−4y=−x−3→y(x−4)=−(x+3)→y=−(x+3)x−4xy + x = 4y - 3 \rightarrow xy - 4y = -x - 3 \rightarrow y(x - 4) = -(x + 3) \rightarrow y = \frac{-(x + 3)}{x - 4}

      • Answer: f−1(x)=−x−3x−4f^{-1}(x) = \frac{-x - 3}{x - 4}

      • Df=R−{−1}D_f = \mathbb{R} - \{-1\}, Rf=Df−1=R−{4}R_f = D_{f^{-1}} = \mathbb{R} - \{4\}.

    • Example 3: f(x)=x−2f(x) = \sqrt{x - 2}

      • Interchange: x=y−2x = \sqrt{y - 2}

      • Solve: x2=y−2→y=x2+2x^2 = y - 2 \rightarrow y = x^2 + 2

      • Restriction: Df=[2,∞[D_f = [2, \infty[; therefore Rf=[0,∞[R_f = [0, \infty[.

      • Answer: f−1(x)=x2+2f^{-1}(x) = x^2 + 2 where Df−1=[0,∞[D_{f^{-1}} = [0, \infty[.

Inverse Trigonometric Functions

  • Core Concepts

    • If sin⁡(30∘)=12\sin(30^{\circ}) = \frac{1}{2}, then sin⁡−1(12)=30∘=π6\sin^{-1}(\frac{1}{2}) = 30^{\circ} = \frac{\pi}{6}.

    • To graph inverse trigonometric functions, the domain of the parent function is restricted to make it one-to-one, then reflected over y=xy = x.

  • Properties Summary

Function

Domain

Range (Principal Value)

Type

y=sin⁡−1(□)y = \sin^{-1}(\square)

[−1,1][-1, 1]

[−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

Odd

y=cos⁡−1(□)y = \cos^{-1}(\square)

[−1,1][-1, 1]

[0,π][0, \pi]

General

y=tan⁡−1(□)y = \tan^{-1}(\square)

R\mathbb{R}

]−π2,π2[]-\frac{\pi}{2}, \frac{\pi}{2}[

Odd

  • Reciprocal Relationship in Inverses

    • cosec−1(□)=sin⁡−1(1□)\text{cosec}^{-1}(\square) = \sin^{-1}(\frac{1}{\square})

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

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

  • Important Constraints and Notes

    • sin⁡−1(x)≠(sin⁡(x))−1\sin^{-1}(x) \neq (\sin(x))^{-1}. Note that (sin⁡(x))−1=cosec(x)(\sin(x))^{-1} = \text{cosec}(x).

    • sin⁡(sin⁡−1(x))=x\sin(\sin^{-1}(x)) = x is valid only for −1≤x≤1-1 \leq x \leq 1.

    • sin⁡−1(sin⁡(x))=x\sin^{-1}(\sin(x)) = x is valid only for −π2≤x≤π2-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}.

    • sin⁡(3sin⁡−1(x))≠3x\sin(3\sin^{-1}(x)) \neq 3x.

    • sin⁡2(sin⁡−1(x))=x2\sin^2(\sin^{-1}(x)) = x^2.

    • sin⁡−1(−x)=−sin⁡−1(x)\sin^{-1}(-x) = -\sin^{-1}(x) (due to the function being odd).

    • These properties apply similarly to cos⁡\cos and tan⁡\tan.

  • Examples of Domain and Range Calculation

    • Example 1: f(x)=3sin⁡−1(x−1)f(x) = 3\sin^{-1}(x - 1)

      • Domain: −1≤x−1≤1→0≤x≤2-1 \leq x - 1 \leq 1 \rightarrow 0 \leq x \leq 2. Domain: [0,2][0, 2].

      • Range: 3×[−π2,π2]=[−3π2,3π2]3 \times [-\frac{\pi}{2}, \frac{\pi}{2}] = [-\frac{3\pi}{2}, \frac{3\pi}{2}]. Range: [−3π2,3π2][-\frac{3\pi}{2}, \frac{3\pi}{2}].

    • Example 2: f(x)=5cos⁡−1(2x+1)f(x) = 5\cos^{-1}(2x + 1)

      • Domain: −1≤2x+1≤1→−2≤2x≤0→−1≤x≤0-1 \leq 2x + 1 \leq 1 \rightarrow -2 \leq 2x \leq 0 \rightarrow -1 \leq x \leq 0. Domain: [−1,0][-1, 0].

      • Range: 5×[0,π]=[0,5π]5 \times [0, \pi] = [0, 5\pi]. Range: [0,5π][0, 5\pi].

    • Example 3: f(x)=2tan⁡−1(x−3)f(x) = 2\tan^{-1}(x - 3)

      • Domain: R\mathbb{R} (horizontal shifts do not alter the domain of tan⁡−1(x)\tan^{-1}(x)).

      • Range: 2×]−π2,π2[=]−π,π[2 \times ]-\frac{\pi}{2}, \frac{\pi}{2}[ = ]-\pi, \pi[. Range: ]−π,π[]-\pi, \pi[.

Simplification Using the Triangle Method

  • This method is used to solve functions where an inverse trigonometric function is nested inside another trigonometric type.

  • Example 1: Same type nesting

    • f(x)=tan⁡2(tan⁡−1(2x))f(x) = \tan^2(\tan^{-1}(2x))

    • Since the function and inverse match, the triangle is not needed.

    • f(x)=(2x)2=4x2f(x) = (2x)^2 = 4x^2.

  • Example 2: Different type nesting (Sine of Arccosine)

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

    • Let m=cos⁡−1(x)m = \cos^{-1}(x), which implies cos⁡(m)=x\cos(m) = x.

    • Construct a right triangle where angle is mm, adjacent is xx, and hypotenuse is 11.

    • By Pythagorean theorem, opposite side is 1−x2\sqrt{1 - x^2}.

    • f(x)=sin⁡(m)=oppositehypotenuse=1−x21=1−x2f(x) = \sin(m) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}.

  • Example 3: Secant of Arcsine

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

    • Let m=sin⁡−1(x)m = \sin^{-1}(x), which implies sin⁡(m)=x\sin(m) = x.

    • Construct a triangle where angle is mm, opposite is xx, and hypotenuse is 11.

    • Adjacent side is 1−x2\sqrt{1 - x^2}.

    • f(x)=sec⁡(m)=1cos⁡(m)=11−x2f(x) = \sec(m) = \frac{1}{\cos(m)} = \frac{1}{\sqrt{1 - x^2}}.

  • Example 4: Cosine of Arccosecant

    • f(x)=cos⁡(cosec−1(x))f(x) = \cos(\text{cosec}^{-1}(x))

    • Let m=cosec−1(x)m = \text{cosec}^{-1}(x), which implies cosec(m)=x\text{cosec}(m) = x.

    • Since sin⁡(m)=1x\sin(m) = \frac{1}{x}, opposite is 11 and hypotenuse is xx.

    • Adjacent side is x2−1\sqrt{x^2 - 1}.

    • f(x)=cos⁡(m)=adjacenthypotenuse=x2−1xf(x) = \cos(m) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{x^2 - 1}}{x}.