Proving Hyperbolic Identities Coth²x-1=cosech²x And Sinh(x+y)=sinhx Coshy+coshx Sinhy

by Chloe Fitzgerald 86 views

Hey math enthusiasts! Today, we're going to dive deep into the fascinating world of hyperbolic functions and tackle two important identities. We'll break down the proofs step by step, making sure you grasp not just the how but also the why behind these equations. So, grab your thinking caps, and let's get started!

a) Proving the Identity: coth²x - 1 = cosech²x

Alright, let's kick things off with our first identity: coth²x - 1 = cosech²x. This identity might look a bit intimidating at first, but trust me, it's easier than it seems once we break it down. The key here is to remember the fundamental definitions of hyperbolic functions.

To prove this identity, we'll start by revisiting the definitions of coth x and cosech x. Remember, these are hyperbolic functions, which are counterparts to trigonometric functions but defined using exponential functions. Specifically:

  • coth x = cosh x / sinh x
  • cosech x = 1 / sinh x

Now that we have these definitions fresh in our minds, let's take a look at the left-hand side (LHS) of our identity, which is coth²x - 1. Our goal is to manipulate this expression until it matches the right-hand side (RHS), which is cosech²x.

So, let's substitute the definition of coth x into the LHS:

coth²x - 1 = (cosh x / sinh x)² - 1

Next, we square the fraction:

= cosh²x / sinh²x - 1

Now, to combine the terms, we need a common denominator. We can rewrite 1 as sinh²x / sinh²x:

= cosh²x / sinh²x - sinh²x / sinh²x

Now we can combine the fractions:

= (cosh²x - sinh²x) / sinh²x

Here comes a crucial step! Remember the fundamental hyperbolic identity: cosh²x - sinh²x = 1. This identity is the cornerstone of many hyperbolic function proofs, and it's essential to have it memorized. So, let's substitute 1 for cosh²x - sinh²x in our expression:

= 1 / sinh²x

Now, look closely! We're almost there. Recall the definition of cosech x: cosech x = 1 / sinh x. Therefore, 1 / sinh²x is simply cosech²x:

= cosech²x

And there you have it! We've successfully transformed the left-hand side (coth²x - 1) into the right-hand side (cosech²x). This completes the proof of the identity:

coth²x - 1 = cosech²x

Key Takeaways and Insights

  • This proof beautifully demonstrates the power of definitions in mathematics. By starting with the fundamental definitions of coth x and cosech x, we were able to systematically manipulate the expression and arrive at our desired result.
  • The identity cosh²x - sinh²x = 1 is a cornerstone of hyperbolic function identities, much like the Pythagorean identity sin²x + cos²x = 1 in trigonometry. It's crucial to recognize and apply this identity whenever possible.
  • This identity highlights the relationship between hyperbolic cotangent and hyperbolic cosecant, showing how they are interconnected through this elegant equation.

b) Proving the Identity: sinh(x + y) = sinh x cosh y + cosh x sinh y

Okay, let's move on to our second identity: sinh(x + y) = sinh x cosh y + cosh x sinh y. This one involves the hyperbolic sine of a sum, and it's analogous to the trigonometric identity for sin(x + y). To tackle this, we'll again rely on the exponential definitions of hyperbolic functions.

Before we dive into the proof, let's quickly recap the definitions of sinh x and cosh x:

  • sinh x = (e^x - e^(-x)) / 2
  • cosh x = (e^x + e^(-x)) / 2

Now, let's focus on the left-hand side (LHS) of our identity: sinh(x + y). To get started, we'll substitute the definition of sinh, but this time with (x + y) as the argument:

sinh(x + y) = (e^(x + y) - e^(-(x + y))) / 2

Using the properties of exponents, we can rewrite e^(x + y) as e^x * e^y and e^(-(x + y)) as e^(-x) * e^(-y):

= (e^x * e^y - e^(-x) * e^(-y)) / 2

Now, let's shift our attention to the right-hand side (RHS) of the identity: sinh x cosh y + cosh x sinh y. Our goal is to show that this expression is equivalent to what we derived for sinh(x + y). So, let's substitute the definitions of sinh x, cosh y, cosh x, and sinh y:

sinh x cosh y + cosh x sinh y = [(e^x - e^(-x)) / 2] * [(e^y + e^(-y)) / 2] + [(e^x + e^(-x)) / 2] * [(e^y - e^(-y)) / 2]

This looks a bit messy, but don't worry! We'll simplify it step by step. First, let's multiply the fractions:

= [(e^x - e(-x))(ey + e^(-y))] / 4 + [(e^x + e(-x))(ey - e^(-y))] / 4

Now, let's expand the products in the numerators:

= [e^x * e^y + e^x * e^(-y) - e^(-x) * e^y - e^(-x) * e^(-y)] / 4 + [e^x * e^y - e^x * e^(-y) + e^(-x) * e^y - e^(-x) * e^(-y)] / 4

Now we have two fractions with the same denominator, so we can combine them:

= [e^x * e^y + e^x * e^(-y) - e^(-x) * e^y - e^(-x) * e^(-y) + e^x * e^y - e^x * e^(-y) + e^(-x) * e^y - e^(-x) * e^(-y)] / 4

Notice that some terms cancel out! Specifically, e^x * e^(-y) and -e^x * e^(-y) cancel, and -e^(-x) * e^y and e^(-x) * e^y cancel. This leaves us with:

= [2 * e^x * e^y - 2 * e^(-x) * e^(-y)] / 4

We can simplify this by dividing both the numerator and the denominator by 2:

= [e^x * e^y - e^(-x) * e^(-y)] / 2

Using the properties of exponents again, we can rewrite this as:

= [e^(x + y) - e^(-(x + y))] / 2

Wait a minute... doesn't this look familiar? This is exactly the expression we derived for sinh(x + y)! So, we've shown that:

sinh x cosh y + cosh x sinh y = [e^(x + y) - e^(-(x + y))] / 2 = sinh(x + y)

This completes the proof of the identity:

sinh(x + y) = sinh x cosh y + cosh x sinh y

Key Insights and Takeaways

  • This proof beautifully illustrates how the exponential definitions of hyperbolic functions can be used to derive more complex identities.
  • The process of expanding the products and simplifying the expression highlights the importance of careful algebraic manipulation.
  • This identity is a fundamental result in hyperbolic function theory and has applications in various areas of mathematics and physics.

Final Thoughts

So, there you have it, guys! We've successfully proven two important hyperbolic identities. Remember, the key to mastering these proofs is to understand the definitions of the hyperbolic functions and to be comfortable with algebraic manipulation. Practice makes perfect, so keep working on these, and you'll be a hyperbolic function pro in no time! These identities aren't just abstract equations; they're powerful tools that can be used to solve a wide range of problems in mathematics, physics, and engineering. So, keep exploring, keep learning, and most importantly, keep having fun with math!