Question

Show that \(\sin (x+i y)=\sin x \cosh y+i \cos x \sinh y\) using trigonometric identities and the exponential forms of these functions.

Short answer

Through converting the trigonometric function to exponential forms and applying hyperbolic definitions, it has been shown that \(\sin (x+i y)=\sin x \cosh y+i \cos x \sinh y\).

Step-by-step solution

Recall the Exponential Form of Sine Function

To begin, we must remember how to write the sine function in exponential form using Euler's formula. Applying Euler's formula gives the exponential form of sine as: \(\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}\).

Substitution of \(x + iy\) into the Exponential Form of sine

Let's substitute \(x + iy\) into the exponential form of sine. Hence, \(\sin(x + iy) = \frac{e^{i(x +iy)} - e^{-i(x +iy)}}{2i}\).

Simplify the Exponent

We simplify the exponent by multiplying \(i\) through the brackets, which results in \(e^{ix-y} - e^{-ix+y}\). Therefore \(\sin(x + iy) = \frac{e^{ix-y} - e^{-ix+y}}{2i}\).

Use the Properties of Hyperbolic Functions and Simplify Further

Recall that the hyperbolic cosine, \(\cosh y\), and hyperbolic sine, \(\sinh y\), are given by \(\cosh y = \frac{e^y + e^{-y}}{2}\) and \(\sinh y = \frac{e^y - e^{-y}}{2}\) respectively. Hence, \(e^y\) and \(e^{-y}\) can be expressed in terms of \(\cosh y\) and \(\sinh y\). The expression \(e^{ix-y} - e^{-ix+y}\) can therefore be written as \(2i\sin x \cosh y - 2 \cos x \sinh y\) given that \(\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}\) and \(\cos(x) = \frac{e^{ix} + e^{-ix}}{2}\). Hence, \(\sin(x + iy) = i\sin x \cosh y - \cos x \sinh y\).

Simplify the Expression

Rearrange the simplified expression to obtain \(\sin(x + iy) = \sin x \cosh y + i \cos x \sinh y\).

Key concepts

Trigonometric identities

Trigonometric identities are mathematical equations that express relationships between trigonometric functions, which are functions based on angles. These identities are essential for simplifying expressions and solving equations in trigonometry. For example, the Pythagorean identity expresses the basic relationship between sine and cosine:

• \( \sin^2 x + \cos^2 x = 1 \).

Using such identities allows us to rewrite expressions in more manageable forms. In complex analysis, these identities help us extend the definitions of trigonometric functions to complex numbers. For example, the identity \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \) can be adapted for complex arguments. This adaptability is crucial when dealing with expressions like \( \sin(x + iy) \), where \( x \) and \( y \) are real numbers, and \( i \) is the imaginary unit. Trigonometric identities thus form the backbone of solving problems involving complex trigonometric functions.

Exponential functions

Exponential functions are mathematical functions of the form \( f(x) = a^{x} \), where \( a \) is a positive constant. In complex analysis, they play an integral role in expressing trigonometric functions through Euler's formula. Euler's formula states:

• \( e^{ix} = \cos x + i\sin x \).

By utilizing this relationship, we can express sine and cosine in terms of exponential functions. The sine function, for instance, can be written as:

• \( \sin x = \frac{e^{ix} - e^{-ix}}{2i} \).

This representation is particularly useful when extending trigonometric functions to complex domains, such as in the expression \( \sin(x + iy) \). Here, substituting \( x + iy \) into the exponential form enables simplification and connection with trigonometric and hyperbolic functions. Exponential functions, therefore, provide a powerful toolkit for bridging real and complex mathematical worlds.

Hyperbolic functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola, similar to how trigonometric functions relate to a circle. They are defined using exponential functions, allowing them to be used effectively in complex analysis. The hyperbolic sine and cosine functions are defined as:

• \( \cosh y = \frac{e^y + e^{-y}}{2} \)
• \( \sinh y = \frac{e^y - e^{-y}}{2} \).

These functions share many properties with sine and cosine, such as identities that are useful in simplifying expressions. In the context of our problem, they allow us to simplify the expression \( e^{ix-y} - e^{-ix+y} \), incorporating \( \sinh \) and \( \cosh \) to express \( \sin(x + iy) \) in terms of familiar functions. Thus, hyperbolic functions are crucial for solving and understanding equations where complex numbers are involved, bridging the gap between trigonometric and exponential functions in new and useful ways.