📐 Level 3 - Topic 7: Advanced Trigonometry - 1. Hyperbolic Functions (Sinh, Cosh, Tanh, etc.) 🚀✏️

1) Introduction to Hyperbolic Trigonometric Functions - Beyond Circular Trigonometry 🚀

Welcome to **Level 3 - Topic 7: Advanced Trigonometry**. We're starting with Hyperbolic Trigonometric Functions. You're already familiar with circular trigonometric functions (sine, cosine, tangent, etc.) based on the unit circle. Hyperbolic functions, on the other hand, are based on the hyperbola, specifically the unit hyperbola \( x^2 - y^2 = 1 \). These functions, while related to exponential functions, share many properties and identities analogous to their circular counterparts. They are essential in various areas of physics, engineering, and advanced mathematics.

Definitions of Hyperbolic Functions:

Hyperbolic functions are defined in terms of the exponential function \( e^x \).

  • Hyperbolic Sine (sinh):

    \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)

  • Hyperbolic Cosine (cosh):

    \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

  • Hyperbolic Tangent (tanh):

    \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

  • Hyperbolic Cotangent (coth):

    \( \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{e^x + e^{-x}}{e^x - e^{-x}} = \frac{1}{\tanh(x)}, \quad x \neq 0 \)

  • Hyperbolic Secant (sech):

    \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} = \frac{2}{e^x + e^{-x}} \)

  • Hyperbolic Cosecant (csch):

    \( \operatorname{csch}(x) = \frac{1}{\sinh(x)} = \frac{2}{e^x - e^{-x}}, \quad x \neq 0 \)

Analogies and Differences with Circular Trigonometric Functions:

  • Analogies: Many identities for hyperbolic functions are similar to trigonometric identities, but with sign changes in some cases.
  • Differences: Hyperbolic functions are not periodic. They are defined based on exponential functions, not circles. Geometrically, circular functions relate to angles and circles, while hyperbolic functions relate to areas and hyperbolas.

Key Relationship:

\( \cosh^2(x) - \sinh^2(x) = 1 \)

This identity is analogous to \( \cos^2(\theta) + \sin^2(\theta) = 1 \) in circular trigonometry, but with a minus sign, reflecting the hyperbola \( x^2 - y^2 = 1 \).

Let's explore the properties, identities, and algebraic applications of these fascinating functions! 🚀📐✏️

2) Identities of Hyperbolic Functions - Exploring Analogous Formulas 🧮

Hyperbolic functions satisfy a set of identities that are strikingly similar to, yet sometimes subtly different from, those of circular trigonometric functions. Understanding these identities is crucial for simplifying expressions and solving problems involving hyperbolic functions algebraically.

Fundamental Identities of Hyperbolic Functions

  • Pythagorean Identity:

    \( \cosh^2(x) - \sinh^2(x) = 1 \)

    Dividing by \( \cosh^2(x) \) and \( \sinh^2(x) \) yields:

    \( 1 - \tanh^2(x) = \operatorname{sech}^2(x) \)

    \( \coth^2(x) - 1 = \operatorname{csch}^2(x) \)

  • Even and Odd Functions:

    \( \cosh(-x) = \cosh(x) \) (Even function)

    \( \sinh(-x) = -\sinh(x) \) (Odd function)

    \( \tanh(-x) = -\tanh(x) \) (Odd function)

  • Sum and Difference Formulas:

    \( \sinh(x \pm y) = \sinh(x)\cosh(y) \pm \cosh(x)\sinh(y) \)

    \( \cosh(x \pm y) = \cosh(x)\cosh(y) \pm \sinh(x)\sinh(y) \)

    \( \tanh(x \pm y) = \frac{\tanh(x) \pm \tanh(y)}{1 \pm \tanh(x)\tanh(y)} \)

  • Double Angle Formulas:

    \( \sinh(2x) = 2\sinh(x)\cosh(x) \)

    \( \cosh(2x) = \cosh^2(x) + \sinh^2(x) = 2\cosh^2(x) - 1 = 1 + 2\sinh^2(x) \)

    \( \tanh(2x) = \frac{2\tanh(x)}{1 + \tanh^2(x)} \)

  • Half Angle Formulas:

    \( \sinh^2\left(\frac{x}{2}\right) = \frac{\cosh(x) - 1}{2} \Rightarrow \sinh\left(\frac{x}{2}\right) = \pm\sqrt{\frac{\cosh(x) - 1}{2}} \)

    \( \cosh^2\left(\frac{x}{2}\right) = \frac{\cosh(x) + 1}{2} \Rightarrow \cosh\left(\frac{x}{2}\right) = \sqrt{\frac{\cosh(x) + 1}{2}} \) (cosh is always positive)

    \( \tanh\left(\frac{x}{2}\right) = \frac{\sinh(x)}{\cosh(x) + 1} = \frac{\cosh(x) - 1}{\sinh(x)} \)

Example 1: Verifying the Identity \( \cosh^2(x) - \sinh^2(x) = 1 \)

Prove algebraically that \( \cosh^2(x) - \sinh^2(x) = 1 \) using the exponential definitions.

  1. 1. Substitute exponential definitions: \[ \cosh^2(x) - \sinh^2(x) = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2 \]
  2. 2. Expand the squares: \[ = \frac{(e^x + e^{-x})^2}{4} - \frac{(e^x - e^{-x})^2}{4} = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} \]
  3. 3. Simplify: \[ = \frac{e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}}{4} = \frac{4}{4} = 1 \]

Verification: We have algebraically shown that \( \cosh^2(x) - \sinh^2(x) = 1 \).

Example 2: Using Sum Formula for \( \cosh(x+y) \)

Calculate \( \cosh(\ln 2 + \ln 3) \) algebraically using the sum formula.

  1. 1. Apply sum formula: \( \cosh(x + y) = \cosh(x)\cosh(y) + \sinh(x)\sinh(y) \) with \( x = \ln 2 \), \( y = \ln 3 \). \[ \cosh(\ln 2 + \ln 3) = \cosh(\ln 2)\cosh(\ln 3) + \sinh(\ln 2)\sinh(\ln 3) \]
  2. 2. Evaluate hyperbolic functions using exponential definitions: \[ \cosh(\ln a) = \frac{e^{\ln a} + e^{-\ln a}}{2} = \frac{a + \frac{1}{a}}{2}, \quad \sinh(\ln a) = \frac{e^{\ln a} - e^{-\ln a}}{2} = \frac{a - \frac{1}{a}}{2} \] For \( a = 2 \): \( \cosh(\ln 2) = \frac{2 + \frac{1}{2}}{2} = \frac{5}{4} \), \( \sinh(\ln 2) = \frac{2 - \frac{1}{2}}{2} = \frac{3}{4} \). For \( a = 3 \): \( \cosh(\ln 3) = \frac{3 + \frac{1}{3}}{2} = \frac{10}{6} = \frac{5}{3} \), \( \sinh(\ln 3) = \frac{3 - \frac{1}{3}}{2} = \frac{8}{6} = \frac{4}{3} \).
  3. 3. Substitute and calculate: \[ \cosh(\ln 2 + \ln 3) = \left(\frac{5}{4}\right)\left(\frac{5}{3}\right) + \left(\frac{3}{4}\right)\left(\frac{4}{3}\right) = \frac{25}{12} + \frac{12}{12} = \frac{37}{12} \]

Solution: \( \cosh(\ln 2 + \ln 3) = \frac{37}{12} \).

3) Inverse Hyperbolic Functions - Reversing the Hyperbolic Operations 🔄

Since hyperbolic functions are defined based on exponential functions, their inverses can be expressed in terms of logarithms. Inverse hyperbolic functions are useful for solving equations algebraically.

Inverse Hyperbolic Functions and their Logarithmic Forms

  • Inverse Hyperbolic Sine (arcsinh or \( \sinh^{-1} \)):

    \( \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \), for all \( x \in \mathbb{R} \)

  • Inverse Hyperbolic Cosine (arccosh or \( \cosh^{-1} \)):

    \( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \), for \( x \geq 1 \), \( y \geq 0 \)

    Note: \( \cosh^{-1}(x) \) is defined only for \( x \geq 1 \) and is non-negative.

  • Inverse Hyperbolic Tangent (arctanh or \( \tanh^{-1} \)):

    \( \tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right) \), for \( |x| < 1 \)

    Defined for \( -1 < x < 1 \).

  • Inverse Hyperbolic Cotangent (arccoth or \( \coth^{-1} \)):

    \( \coth^{-1}(x) = \frac{1}{2} \ln\left(\frac{x + 1}{x - 1}\right) \), for \( |x| > 1 \)

    Defined for \( |x| > 1 \).

  • Inverse Hyperbolic Secant (arcsech or \( \operatorname{sech}^{-1} \)):

    \( \operatorname{sech}^{-1}(x) = \ln\left(\frac{1 + \sqrt{1 - x^2}}{x}\right) \), for \( 0 < x \leq 1 \), \( y \geq 0 \)

    Defined for \( 0 < x \leq 1 \), and \( \operatorname{sech}^{-1}(x) \geq 0 \).

  • Inverse Hyperbolic Cosecant (arccsch or \( \operatorname{csch}^{-1} \)):

    \( \operatorname{csch}^{-1}(x) = \ln\left(\frac{1}{x} + \frac{\sqrt{1 + x^2}}{|x|}\right) = \ln\left(\frac{1}{x} + \sqrt{\frac{1}{x^2} + 1}\right), \quad x \neq 0 \)

    For \( x > 0 \): \( \operatorname{csch}^{-1}(x) = \ln\left(\frac{1 + \sqrt{1 + x^2}}{x}\right) \). For \( x < 0 \): \( \operatorname{csch}^{-1}(x) = \ln\left(\frac{1 - \sqrt{1 + x^2}}{x}\right) \).

Example 3: Evaluating \( \sinh^{-1}(2) \)

Evaluate \( \sinh^{-1}(2) \) algebraically using its logarithmic form.

  1. 1. Use logarithmic formula for \( \sinh^{-1}(x) \): \( \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \).
  2. 2. Substitute \( x = 2 \): \[ \sinh^{-1}(2) = \ln(2 + \sqrt{2^2 + 1}) = \ln(2 + \sqrt{5}) \]
  3. 3. Approximate value (optional): \( \ln(2 + \sqrt{5}) \approx \ln(2 + 2.236) = \ln(4.236) \approx 1.444 \).

Solution: \( \sinh^{-1}(2) = \ln(2 + \sqrt{5}) \approx 1.444 \).

Example 4: Solving Equation using Inverse Hyperbolic Cosine

Solve the equation \( \cosh(x) = 3 \) for \( x \) algebraically.

  1. 1. Apply inverse hyperbolic cosine: If \( \cosh(x) = 3 \), then \( x = \cosh^{-1}(3) \) (or \( x = -\cosh^{-1}(3) \) because \( \cosh(x) \) is even). We take the positive solution \( x = \cosh^{-1}(3) \).
  2. 2. Use logarithmic form for \( \cosh^{-1}(x) \): \( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \).
  3. 3. Substitute \( x = 3 \): \[ x = \cosh^{-1}(3) = \ln(3 + \sqrt{3^2 - 1}) = \ln(3 + \sqrt{8}) = \ln(3 + 2\sqrt{2}) \]
  4. 4. Approximate value (optional): \( \ln(3 + 2\sqrt{2}) \approx \ln(3 + 2 \times 1.414) = \ln(3 + 2.828) = \ln(5.828) \approx 1.763 \).

Solution: \( x = \pm \cosh^{-1}(3) = \pm \ln(3 + 2\sqrt{2}) \approx \pm 1.763 \).

4) Applications of Hyperbolic Functions - Catenary and Beyond 🔗

Hyperbolic functions have significant algebraic applications in various fields, particularly in physics and engineering. One classic example is the shape of a hanging chain or cable.

Applications of Hyperbolic Functions

  • Catenary Curve: The shape of a hanging flexible chain or cable suspended between two points, under its own weight, is described by the hyperbolic cosine function. The equation of a catenary is:

    \( y = a \cosh\left(\frac{x}{a}\right) + b \)

    where \( a \) and \( b \) are constants determined by the physical setup. Catenaries appear in architecture (arches), physics (equilibrium of chains), and engineering (overhead power lines). Algebraically, the hyperbolic cosine function dictates the curve's form.
  • Physics - Relativity: Hyperbolic functions appear in Lorentz transformations in special relativity, describing space and time transformations between moving frames of reference. These transformations are expressed algebraically using hyperbolic functions.
  • Engineering - Transmission Lines: In the algebraic analysis of long transmission lines, hyperbolic functions are used to describe voltage and current distribution along the line.
  • Mathematics - Geometry: Hyperbolic geometry, a non-Euclidean geometry, extensively uses hyperbolic functions algebraically.
  • Complex Numbers and Transformations: Hyperbolic functions appear in transformations in the complex plane and have algebraic relationships with complex trigonometric functions.

Example 5: Catenary Equation for a Hanging Cable

A cable is hung between two poles 100m apart, and the lowest point of the cable is 20m below the level of suspension. Model the shape of the cable using a catenary equation of the form \( y = a \cosh\left(\frac{x}{a}\right) + c \), with the lowest point at \( x = 0 \), \( y = 20 \), and poles at \( x = \pm 50 \).

  1. 1. Set up conditions: Lowest point at \( (0, 20) \) and points at \( (\pm 50, y_1) \) (same height due to symmetry). For lowest point, \( x=0 \), \( y = 20 \Rightarrow 20 = a \cosh(0) + c = a + c \Rightarrow c = 20 - a \). So equation is \( y = a \cosh\left(\frac{x}{a}\right) + 20 - a \).
  2. 2. Use pole position: Let's assume height at poles is, say, 70m at \( x = \pm 50 \). Then at \( x = 50 \), \( y = 70 \). \[ 70 = a \cosh\left(\frac{50}{a}\right) + 20 - a \Rightarrow 50 + a = a \cosh\left(\frac{50}{a}\right) \Rightarrow \cosh\left(\frac{50}{a}\right) = \frac{50}{a} + 1 \] This is an equation for \( a \) that can be solved numerically or graphically. For demonstration, we will not solve for \(a\) here but show how the equation is formed.
  3. 3. Catenary equation form: Based on lowest point at \( (0, 20) \), the catenary equation is algebraically represented as \( y = a \cosh\left(\frac{x}{a}\right) + (20 - a) \). To find a specific value for \( a \), additional information is used to solve the algebraic equation derived in step 2.

Solution: The catenary equation is algebraically of the form \( y = a \cosh\left(\frac{x}{a}\right) + (20 - a) \), where \( a \) can be determined by solving \( \cosh\left(\frac{50}{a}\right) = \frac{50}{a} + 1 \).

Example 6: Simplifying an Expression using Hyperbolic Functions

Simplify the expression \( \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \) algebraically.

  1. 1. Recognize hyperbolic tangent definition: \( \tanh(u) = \frac{e^u - e^{-u}}{e^u + e^{-u}} \).
  2. 2. Substitute \( u = 2x \): Let \( u = 2x \). Then the expression becomes \( \frac{e^{u} - e^{-u}}{e^{u} + e^{-u}} \).
  3. 3. Express in terms of \( \tanh \): Using the definition, this is algebraically equivalent to \( \tanh(u) \). Substitute back \( u = 2x \). \[ \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} = \tanh(2x) \]

Solution: The algebraic simplification of the expression is \( \tanh(2x) \).

5) Advanced Applications of Trigonometric Identities - Beyond Basic Simplification 🚀

Beyond basic simplification, trigonometric identities are essential algebraic tools in solving more complex problems in various fields. Advanced applications often involve strategic algebraic manipulation of identities to transform expressions and solve equations.

Advanced Applications of Trigonometric Identities

  • Simplifying Complex Trigonometric Expressions: Using sum-to-product, product-to-sum, double and half-angle formulas to algebraically reduce complexity of expressions, making them easier to analyze or solve algebraically.
  • Solving Trigonometric Equations: Advanced equations may require identities to algebraically convert equations into factorable or standard forms. For example, converting sums to products can help in finding general algebraic solutions.
  • Geometric and Vector Problems: Trigonometric identities are used in geometric proofs and vector algebra, for example, in simplifying expressions involving dot and cross products in 2D and 3D space.
  • Algebraic Proofs and Derivations: Trigonometric identities form the basis for many algebraic proofs in trigonometry and related fields, allowing for the derivation of new formulas and relationships.
  • Transformations in Coordinate Systems: In coordinate geometry and linear algebra, trigonometric identities are crucial for transformations like rotation and reflection of coordinate axes, which are algebraically described by trigonometric functions.

Example 7: Simplifying a Trigonometric Expression Algebraically

Simplify the trigonometric expression \( \sin(3x)\cos(2x) + \cos(3x)\sin(2x) \) algebraically.

  1. 1. Apply sum identity: Recognize the sum formula \( \sin(A)\cos(B) + \cos(A)\sin(B) = \sin(A + B) \). Here \( A = 3x \), \( B = 2x \). \[ \sin(3x)\cos(2x) + \cos(3x)\sin(2x) = \sin(3x + 2x) \]
  2. 2. Simplify: \[ = \sin(5x) \]

Solution: The algebraic simplification is \( \sin(5x) \).

Example 8: Solving Trigonometric Equation Algebraically using Sum-to-Product

Solve the equation \( \sin(3x) + \sin(x) = 0 \) algebraically for \( 0 \leq x \leq \pi \).

  1. 1. Apply sum-to-product identity: \( \sin(A) + \sin(B) = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right) \). Here \( A = 3x \), \( B = x \). \[ \sin(3x) + \sin(x) = 2\sin\left(\frac{3x + x}{2}\right)\cos\left(\frac{3x - x}{2}\right) = 2\sin(2x)\cos(x) \]
  2. 2. Set product to zero: \( 2\sin(2x)\cos(x) = 0 \Rightarrow \sin(2x) = 0 \) or \( \cos(x) = 0 \).
  3. 3. Solve for \( x \) algebraically: For \( \sin(2x) = 0 \), \( 2x = n\pi \Rightarrow x = \frac{n\pi}{2} \), for integer \( n \). In \( [0, \pi] \), \( x = 0, \frac{\pi}{2}, \pi \). For \( \cos(x) = 0 \), \( x = \frac{\pi}{2} + k\pi \). In \( [0, \pi] \), \( x = \frac{\pi}{2} \).
  4. 4. Combine solutions: Algebraic solutions in \( [0, \pi] \) are \( x = 0, \frac{\pi}{2}, \pi \).

Solution: Algebraic solutions are \( x = 0, \frac{\pi}{2}, \pi \) in the interval \( [0, \pi] \).

6) Polar Coordinates and Trigonometry - Algebraic Description of Radial Systems 🔄

Polar coordinates provide an alternative algebraic system to describe points in a plane using a radial distance and an angle. Trigonometry is fundamentally linked to polar coordinates, providing the algebraic conversion between polar and Cartesian coordinate systems.

Polar Coordinates and Trigonometry

  • Polar Coordinates \( (r, \theta) \): A point in the plane is represented by a radial distance \( r \) from the origin and an angle \( \theta \) measured counterclockwise from the positive x-axis.
  • Conversion from Polar to Cartesian \( (x, y) \): The algebraic conversion formulas from polar coordinates \( (r, \theta) \) to Cartesian coordinates \( (x, y) \) are given by:

    \( x = r \cos(\theta) \)

    \( y = r \sin(\theta) \)

  • Conversion from Cartesian to Polar \( (r, \theta) \): The algebraic conversion formulas from Cartesian coordinates \( (x, y) \) to polar coordinates \( (r, \theta) \) are given by:

    \( r = \sqrt{x^2 + y^2} \)

    \( \tan(\theta) = \frac{y}{x} \), with quadrant of \( \theta \) determined by signs of \( x \) and \( y \). Typically, \( \theta = \arctan\left(\frac{y}{x}\right) \) adjusted for the correct quadrant.

  • Algebraic Representations in Polar Coordinates: Many curves and shapes can be described more simply in polar coordinates algebraically. For example, a circle centered at the origin with radius \( a \) is simply \( r = a \). A spiral can be represented as \( r = a\theta \) or \( r = ae^{b\theta} \).
  • Applications in Algebra and Geometry: Polar coordinates simplify the algebraic analysis of problems involving radial symmetry, rotational transformations, and curves defined by radial distances and angles. They are essential in complex number algebra, vector algebra in polar form, and geometric descriptions of spiral paths and circular motions.

Example 9: Converting Polar to Cartesian Coordinates Algebraically

Convert the polar coordinates \( (r, \theta) = (4, \frac{\pi}{3}) \) to Cartesian coordinates \( (x, y) \) algebraically.

  1. 1. Apply conversion formulas: Use \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \).
  2. 2. Substitute \( r = 4 \) and \( \theta = \frac{\pi}{3} \): \[ x = 4 \cos\left(\frac{\pi}{3}\right) = 4 \times \frac{1}{2} = 2 \] \[ y = 4 \sin\left(\frac{\pi}{3}\right) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \]

Solution: The Cartesian coordinates are \( (x, y) = (2, 2\sqrt{3}) \).

Example 10: Converting Cartesian to Polar Coordinates Algebraically

Convert the Cartesian coordinates \( (x, y) = (-1, 1) \) to polar coordinates \( (r, \theta) \) algebraically.

  1. 1. Calculate \( r \): Use \( r = \sqrt{x^2 + y^2} \). \[ r = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]
  2. 2. Calculate \( \theta \): Use \( \tan(\theta) = \frac{y}{x} = \frac{1}{-1} = -1 \). Since \( x < 0 \) and \( y > 0 \), point is in the second quadrant. The reference angle is \( \arctan(1) = \frac{\pi}{4} \). So, \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).

Solution: The polar coordinates are \( (r, \theta) = (\sqrt{2}, \frac{3\pi}{4}) \).

6) Practice Questions 🎯

6.1 Fundamental Questions – Hyperbolic and Trigonometric Algebra

1. Algebraically verify the identity \( \coth^2(x) - 1 = \operatorname{csch}^2(x) \) using exponential definitions.

2. Use sum formula to expand and simplify \( \sinh(x + \ln 2) \) algebraically.

3. Evaluate \( \cosh(\ln 3) \) algebraically without a calculator.

4. Solve for \( x \) algebraically: \( \tanh(x) = \frac{1}{2} \).

5. Simplify \( 2\cosh^2(x) - \cosh(2x) \) algebraically.

6. Convert polar coordinates \( (r, \theta) = (2, \pi) \) to Cartesian coordinates algebraically.

7. Convert Cartesian coordinates \( (x, y) = (0, -3) \) to polar coordinates algebraically.

8. Simplify \( \sin(4x)\cos(3x) - \cos(4x)\sin(3x) \) using sum/difference identities.

9. Solve algebraically for \( x \in [0, \pi] \): \( \cos(2x) + \cos(x) = 0 \).

10. Verify algebraically: \( \tanh(2x) = \frac{2\tanh(x)}{1 + \tanh^2(x)} \) using definitions of \( \tanh \).

6.2 Challenging Questions – Advanced Algebraic Manipulations 💪🚀

1. Derive the logarithmic form of \( \tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right) \) algebraically.

2. Prove algebraically: \( \sinh(3x) = 3\sinh(x) + 4\sinh^3(x) \).

3. Solve the equation \( \sinh(2x) = \cosh(x) \) algebraically for \( x \).

4. Convert the polar equation \( r = 2\cos(\theta) \) to Cartesian coordinates and identify the shape.

5. Algebraically simplify: \( \frac{\sin(3x) + \sin(5x)}{\cos(3x) + \cos(5x)} \).

7) Summary - Hyperbolic Functions and Advanced Trigonometric Algebra 🎉

  • Hyperbolic Functions Defined Algebraically: Explored sinh, cosh, tanh, coth, sech, csch using exponential definitions, focusing on algebraic properties.
  • Key Identities: Mastered and algebraically verified fundamental hyperbolic identities like Pythagorean, sum/difference, double/half-angle formulas.
  • Inverse Hyperbolic Functions: Understood inverse functions (arcsinh, arccosh, etc.) and their logarithmic forms for algebraic solutions.
  • Algebraic Applications: Examined applications including catenary curves and expression simplification.
  • Advanced Trig Identities: Applied trigonometric identities for simplifying expressions and solving equations algebraically.
  • Polar Coordinates: Reviewed algebraic conversions between polar and Cartesian coordinates.
  • Practice Questions: Solved fundamental and challenging questions to strengthen algebraic skills in hyperbolic and trigonometric functions.

Excellent work! You've now navigated the algebraic landscape of hyperbolic functions and advanced trigonometric identities. You've honed your skills in algebraic manipulation, equation solving, and coordinate conversions. This topic sets a strong algebraic foundation for further studies in advanced mathematics and physics. Continue to the next topic to expand your mathematical toolkit! 🚀📐✏️🌟

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