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Prove the identities $\begin{aligned} \sinh (x+y) &=\sinh x \cosh y+\cosh x \sinh y \\ \cosh (x+y) &=\cosh x \cosh y+\sinh x \sinh y \end{aligned}$Then use them to show thata. \sinh 2 x=2 \sinh x \cosh xb. \cosh 2 x=\cosh ^{2} x+\sinh ^{2} x

(a) $x=2 \sinh x \cosh x$(b) $x=\cosh ^{2} x+\sinh ^{2} x$

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Transcendental Functions

Section 7

Hyperbolic Functions

Functions

Derivatives

Differentiation

Differential Equations

Campbell University

Harvey Mudd College

Baylor University

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in this example, we're going to be proving several hyperbolic identities, starting with one we see displayed here to prove this identity for a sign h x plus y. Let's begin with the right hand side. Our goal is to express the right hand side in a way that would simplify and turn into the left hand side. So start by using the definition of hyperbolic sign. Hyperbolic sign of X will be either the ex with a minus e to the negative X divide by two. Next, when we do hyperbolic co sign of why we get nearly the same quantity. But it will read you the X and this here is our one change plus Oh, and the variable here is why not X then plus E to the negative y divide by two. I'll copy the plus sign in between and two more fractions. We have hyperbolic co signing again, but this time at X. So have either the ex ah minus E to the negative X, then divide by two. Now, one more function. This time it's hyperbolic sign of why, and this reads either y minus e to the negative y divide by two and This is our full expansion, using the definitions of the hyperbolic. So next we can combine this as one overall fraction with the least common denominator for as well as go to all the numerator and begin multiplying each one of these groups that we see displayed here. Let's start with either the ex distributed into the next group, and we'll have it Duthie X Plus y then plus E to the X minus y. Now let's do minus E to the power Negative X will have minus to the power of negative X plus y and then a minus you the power of negative X minus y, and that completes the first group. Let's now start distributing over here with Eve Power X will get a plus, even the power of X plus y that ah minus to the power of X minus y next, distribute in negative each of the power of negative X. We'll get negative either the power of negative X plus y and a positive either the power off Negative X minus Y. Oh, and one small correction here. This sign should be a positive since it came from hyperbolic co sign and what that means is thes. Last two signs here and here must now be written in as a positive and minus. Okay, let's proceed then to the next step. What we need to do now is simplified by combining our like terms that we see in the numerator and see what this will all turn into. First, we have either the X plus y and it looks like we have two copies of that term. It's all right to either the X plus y now for E to the X minus y notice that we have a cancellation. This term will cancel with this term e to the negative each of the power of negative X plus y. We'll cancel with this term here now that we have the correct it sign. But this term at this term will combine to a quantity of negative, too each the power of negative. Let's write this as X plus y in anticipation of the next steps, and then the whole quantity is now divided by four. Now we can reduce by dividing everything by two, and this becomes each of the X plus y minus each of the power of negative X plus y divide by two, but notice that this is the same as the hyperbolic sign of X plus y, because hyper block sine X plus y creates the fraction we see here where X plus y must go into those two positions. So this verifies our first identity. Now we're going to skip to a new page where we have a new identity, and this time we're proving this identity here for hyper bulk co sign the strategy for proving this identity is entirely the similar to what we did on the first identity. Let's begin with the most complicated side. The right hand side will express it in terms of the Basie exponential and then simplify first hyper balk Coastline of X is either the X plus each of the negative X divide by two. And for hyper bark co sign a why we have the same thing. Except we put a very belie in exponents position. Now it's moved to hyperbolic sign for hyperbolic sign of X. We have. If the X we place a minus instead of a plus and still use either the power of negative X on the second term, divide by two, and for hyperbolic sign. It's the same thing, but with the variable. Why? So we'll have either the Y minus e to the power negative y divide by two. Next, as before, we can combine the fractions into one overall fraction, with least commented on a nominator of four. And on the next steps, we need to now distribute by multiplying out each one of these groups. First, let's focus on either the ex that will have either the X Times either the why, which is either the X Plus y then plus the X minus y and distribute next to the negative X will have plus E to the negative X plus y and then plus you to the negative X minus y. That takes us now to the second pair or will foil these groups out first with either the ex, we have plus either the X plus y and then a minus e to the X minus y distributing in this second quantity and noting the negative sign here gives us a minus e to the power negative X plus y and then finally plus here the power negative x minus y. So our distributions complete and our goal now is to begin simplifying by combining like terms are at least common denominator. Still four and I have a sheet of the X plus y here and here. So there's to either the X plus wise next e to the X minus Y cancels with here and here, since their coefficients are opposites. Likewise e to the power negative X plus y here cancels with this quantity. And so what? We have left our looks like e to the power of negative X minus y, which is here in here. It's all right to each of the power of factor out the negative sign, as we did before leaving X Plus y. And now we can cancel by dividing the entire fraction top of bottom By two. We'll have either the X plus y plus either the power of negative X plus y divide by two. And because we're able to force an X plus y here and here, this then implies that this is equal to hyperbolic co sign at X plus y. So the proof for the second identity is now complete. We're going to do two more identities, so let's turn the page next to the third identity, which is a double angle formula for hyperbolic sign. In order to prove this identity, we're going to take a different approach from the others and begin with the left hand side. The left hand side, hyperbolic sign of two X can be written as sign H of X plus X. If we use this formula, then noticed that this first identity can now be used where we're going to let the second very bold be X. So be here, here and here with the necks. This then will turn into hyperbolic sign of X times hyper balk co sign at X and then this quantity is really just the same expression in a different order. So we have two copies altogether, and this is the right hand side. This identity was proved very quickly because we had a powerful identity. That's one here which supported us. Now let's do something similar and prove 1/4 identity, where the identity we need to prove the 4th 1 will be the 2nd 1 for hyperbolic co sign X plus y. So using this identity here, we're going to prove the doubling glide Danny for hyperbolic co sign. So we'll start with the left hand side, which is Kash of two X. But we can write this as X plus X Then in this identity provided here, we're going to let this being X and this Benex as well as this here that would give us first hyperbolic co sign of X times hyperbolic coastline of X which is cosh squared of X. Likewise for this quantity here with sign. Since the inputs are both ex, we get plus hyperbolic sign squared at X and this is the right hand side. And so our last identity here has been verified.

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