Question

Prove the following identities. $$\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y$$

Short answer

Question: Prove the identity $\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y$. Solution: By simplifying both sides of the equation using the definitions of hyperbolic cosine and hyperbolic sine functions, we find that both sides become equal, proving the identity: $\frac{e^x e^y + e^{-x} e^{-y}}{2} = \frac{e^x e^y + e^{-x} e^{-y}}{2}$.

Step-by-step solution

Write down the given identity

Write down the identity to be proved: $$\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y$$

Express the hyperbolic functions in terms of exponentials

Use the definitions of the hyperbolic cosine and sine functions to rewrite both sides of the identity in terms of exponentials: $$\frac{e^{x+y}+e^{-(x+y)}}{2} = \left(\frac{e^x+e^{-x}}{2}\right)\left(\frac{e^y+e^{-y}}{2}\right)+\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^y-e^{-y}}{2}\right)$$

Simplify the left-hand side

Use basic algebra to simplify the left-hand side of the equation: $$\frac{e^{x+y}+e^{-(x+y)}}{2} = \frac{e^x \cdot e^y + e^{-(x+y)}}{2} = \frac{e^x e^y + e^{-x} e^{-y}}{2}$$

Simplify the right-hand side

Use basic algebra to simplify the right-hand side of the equation: $$\left(\frac{e^x+e^{-x}}{2}\right)\left(\frac{e^y+e^{-y}}{2}\right)+\left(\frac{e^x-e^{-x}}{2}\right)\left(\frac{e^y-e^{-y}}{2}\right) = \frac{e^x e^y + e^{-x} e^y + e^x e^{-y} + e^{-x} e^{-y}}{4} + \frac{e^x e^y - e^{-x} e^y + e^x e^{-y} - e^{-x} e^{-y}}{4}$$ Combine the terms in the numerator: $$= \frac{2 e^x e^y + 2 e^{-x} e^{-y}}{4} = \frac{e^x e^y + e^{-x} e^{-y}}{2}$$

Compare both sides

As you can see, both sides of the equation are the same: $$\frac{e^x e^y + e^{-x} e^{-y}}{2} = \frac{e^x e^y + e^{-x} e^{-y}}{2}$$ Therefore, the identity is proved: $$\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y$$

Key concepts

hyperbolic cosine

The hyperbolic cosine function, denoted as \(\cosh x\), is a fundamental concept in hyperbolic geometry. It is closely related to exponential functions. The hyperbolic cosine can be expressed using exponential functions as: \[ \cosh x = \frac{e^x + e^{-x}}{2} \] This equation shows that \(\cosh x\) is essentially the average of the exponential function \(e^x\) and its reciprocal \(e^{-x}\). This symmetric form is responsible for many of the properties that \(\cosh x\) possesses, such as being an even function, which means that \(\cosh(-x) = \cosh x\). Understanding this formula is crucial because it allows the conversion between hyperbolic and exponential forms, which is useful in proving identities or solving equations involving hyperbolic functions.

hyperbolic sine

The hyperbolic sine function, denoted as \(\sinh x\), is another key component of hyperbolic functions. It also can be described using exponential expressions. The formula for \(\sinh x\) is: \[ \sinh x = \frac{e^x - e^{-x}}{2} \] You will notice that this expression is similar to the hyperbolic cosine, except for the subtraction between \(e^x\) and \(e^{-x}\). This subtraction gives \(\sinh x\) distinctive properties—particularly, it is an odd function, which satisfies \(\sinh(-x) = -\sinh x\). Being able to switch to the exponential form helps in various calculations, including proving identities like \(\cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y\). By rewriting \(\sinh\) and \(\cosh\) using exponential functions, you can more easily simplify and verify complex expressions involving these functions.

exponential functions

Exponential functions are a cornerstone in mathematics and are widely used to describe a variety of phenomena, ranging from natural growth to complex wave patterns. The exponential function \(e^x\) is particularly significant because it has certain unique properties. It's the only function that is equal to its own derivative, which means it is self-replicating in the sense of growth. In terms of hyperbolic functions, both \(\cosh x\) and \(\sinh x\) utilize powers of \(e\). For instance:- \(\cosh x = \frac{e^x + e^{-x}}{2}\)- \(\sinh x = \frac{e^x - e^{-x}}{2}\)These expressions link hyperbolic functions directly with exponential growth and decay concepts.
Understanding these relationships can help unlock the beauty of the mathematics involved, simplifying the process of deriving and proving hyperbolic identities like the one given in the exercise. Knowing how exponential functions work allows us to convert easily between forms, facilitating easier simplification or calculation.