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Hyperbolic Identities

Calculus 2 · Chapter 3: Hyperbolic FunctionsSection 3.2 · lesson 13 of 5411 min · free

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In this lesson

The trigonometric catalogue, with predictable sign changes.

  • Derive the fundamental identity and its two companions
  • Apply the addition and double-argument formulas
  • Track every sign difference to a single source
  • Fall back on the exponential definitions when memory fails

Transcript

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MAT-265 - Calculus 2 · Section 3.2

Hyperbolic Identities

Stand beneath a power line strung between two poles. Or look at the main cable of a suspension bridge, before the roadway is attached. You see a distinctive sag. Not a parabola, as many people assume — a catenary. The curve traced by y equals cosh x. Engineers who compute the tension, the length, and the load of such cables rely on the hyperbolic sine and cosine. And every one of those calculations leans on a small set of algebraic identities that these functions obey. Remarkably, those identities mirror the ones for ordinary sine and cosine almost exactly. With a few sign changes — and every one of them traces back to a single source. Mastering these turns long exponential computations into one-line simplifications. Exactly as the trig identities do for circular functions.

The Setting

Everything Reduces to Exponentials

Before we start, one point that makes this whole section less intimidating. The hyperbolic functions are defined directly from the exponential function. Sinh x is e to the x minus e to the minus x, over two. Cosh x is the sum version, over two. Because both are built from e to the x, every hyperbolic identity can ultimately be verified by expanding into exponentials and simplifying. That's not a fallback. That's the engine behind all of it. So if you ever forget an identity in this section, you're not stuck. You can rebuild it.

The Fundamental Identity

Where the Minus Sign Comes From

Start with the single most important identity. The analogue of the Pythagorean identity. Cosh squared x minus sinh squared x equals one. Let's derive it, because the derivation tells you where that minus sign comes from. Expand each square separately. For cosh squared, you square the sum, and get e to the two x, plus two e to the x times e to the minus x, plus e to the minus two x. All over four. The middle term simplifies beautifully, because e to the x times e to the minus x is e to the zero, which is one. So cosh squared is e to the two x, plus two, plus e to the minus two x, over four. Similarly, sinh squared is e to the two x, minus two, plus e to the minus two x, over four. Now subtract the second from the first. The e to the two x terms cancel. The e to the minus two x terms cancel. And you're left with two minus minus-two, over four. Which is four over four. One. Now here's the part worth sitting with. Notice precisely where that minus sign originates. Subtracting the two squares cancels the exponentials, and doubles the constant. Producing plus one. But had the squares been added instead, the constants would cancel and the exponentials would survive. No clean identity at all. That is the structural reason the hyperbolic identity carries a minus, where the circular one carries a plus. Two companion forms follow immediately. Divide the fundamental identity by cosh squared, and you get one minus tanh squared equals sech squared. Divide by sinh squared instead, and you get coth squared minus one equals csch squared.

Worked Example 1

A Sign Decision Circular Trig Never Gives You

Let's use the identity. Given that sinh x is three quarters, find cosh x. Rearrange the fundamental identity: cosh squared x equals one plus sinh squared x. So cosh squared is one plus nine sixteenths. Which is twenty-five sixteenths. Now take the square root. And here's where something interesting happens. Normally, taking a square root leaves you with a sign ambiguity. Plus or minus. But cosh x is always positive. Its definition is a sum of two positive exponentials — it cannot be negative. So only the positive root is valid. Cosh x is five quarters. And that sign decision is a key difference from circular trigonometry. When you solve for cosine from the Pythagorean identity, you have to think about which quadrant you're in, because cosine can be negative. With cosh, there's nothing to decide. It's always positive. One less thing to get wrong.

Addition Formulas

Nearly Identical - One Sign Apart

Next, the addition formulas. These express sinh and cosh of a sum, in terms of the hyperbolic functions of x and y separately. And they're nearly identical to the circular addition formulas. Again differing in exactly one sign. Sinh of x plus y equals sinh x cosh y, plus cosh x sinh y. That's identical to the circular version. No sign change at all. Cosh of x plus y equals cosh x cosh y, plus sinh x sinh y. And there's the difference. Circular cosine has a minus there. Hyperbolic cosine has a plus. Why? The derivation shows it clearly. Expand sinh x cosh y plus cosh x sinh y into exponentials. When you add the two results, the e to the x minus y terms and the e to the minus x plus y terms cancel in pairs. While the e to the x plus y and e to the minus x minus y terms reinforce. What's left is exactly the definition of sinh of x plus y. The cosine formula is proved the same way — except there the cross terms reinforce instead of canceling. Which is precisely why its right-hand side carries a plus rather than a minus. Let's apply one. Express cosh of x plus log two, in terms of sinh x and cosh x. Use the cosine addition formula with y equals log two. Then evaluate the constants from the definitions, using e to the log two equals two, and e to the minus log two equals one half. Cosh of log two is five fourths. Sinh of log two is three fourths. Substitute, and you get five fourths cosh x, plus three fourths sinh x.

Double-Argument Formulas

Set y Equal to x

Setting y equal to x in the addition formulas immediately produces the double-argument formulas. Sometimes called double-angle formulas — though of course hyperbolic arguments aren't angles at all. These are the workhorses for simplifying powers, and for integration. For sinh, set y equals x and you get sinh two x equals two sinh x cosh x. Identical in form to the circular version. For cosh, you get cosh two x equals cosh squared x plus sinh squared x. And again — circular cosine has a minus there. Hyperbolic has a plus. Two alternative forms come from substituting the fundamental identity. Since sinh squared equals cosh squared minus one, replacing sinh squared gives cosh two x equals two cosh squared x, minus one. And replacing cosh squared instead gives one plus two sinh squared x. These rearrangements parallel the power-reduction identities of circular trigonometry, and they're equally useful for integrals of cosh squared and sinh squared. Let's use them. Simplify sinh two x, over one plus cosh two x. In the numerator, use sinh two x equals two sinh x cosh x. In the denominator, use the form cosh two x equals two cosh squared x minus one. So one plus cosh two x becomes two cosh squared x. The twos cancel. One factor of cosh cancels. And you're left with sinh x over cosh x. Which is tanh x. The whole expression collapses to a single function — the hyperbolic analogue of the circular half-angle simplification.

Reference

Every Sign Difference in One Table

Let's put every sign difference in one place. The Pythagorean identity: circular has plus, hyperbolic has minus. The sine addition formula: no sign change at all. Identical. The cosine addition formula: circular has minus, hyperbolic has plus. The cosine double-argument formula: circular has minus, hyperbolic has plus. So the pattern is: sine formulas are unchanged, cosine and squared formulas flip. Now — here's something genuinely worth thinking about. Every one of those sign differences can be traced to a single source. There's a deep connection, revealed through complex numbers, that cosh of i x equals cosine x. And sinh of i x equals i times sine x. Using only those two relationships — can you see why squaring sinh of i x equals i sine x introduces a factor of i squared? And i squared is minus one. So ask yourself how that single factor accounts for the entire plus-becomes-minus pattern across the whole table. You don't need to prove the complex relationships. Just trace where the i squared would land.

Key Takeaways

You Only Have to Remember Two Definitions

One last practical note, and it's the most reassuring thing in this section. Because every hyperbolic function reduces to exponentials, there is always a fallback when memory fails. Expand everything into e to the x and e to the minus x, and simplify. It's slower than applying a known identity. But it's foolproof, and it requires nothing beyond algebra. A useful habit: verify any unfamiliar hyperbolic identity this way once, then rely on the identity itself thereafter. The exponential definitions are the bedrock. The identities are convenient shortcuts built on top of them. Which means a student who remembers only the two definitions of sinh and cosh can reconstruct every result in this section from scratch. That makes these identities far less burdensome to memorize than they first appear. So — five takeaways. The fundamental identity is cosh squared minus sinh squared equals one, and the minus arises from canceling exponentials when the squares are subtracted. Cosh x is always positive, so take only the positive square root when solving for it. The addition formulas match the circular ones, except cosh of x plus y carries a plus where circular cosine carries a minus. The double-argument formulas follow from setting y equal to x, with two alternative forms via the fundamental identity. And every identity can be re-derived from the exponential definitions. Next up is section three point three, Hyperbolic Derivatives — where the same near-parallel with circular trigonometry appears yet again.

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FAQ

Questions students ask us

About Hyperbolic Identities

Why does \cosh^2 x - \sinh^2 x = 1 have a minus sign?
The minus sign comes from the exponential definitions. Squaring gives \cosh^2 x = (e^{2x} + 2 + e^{-2x})/4 and \sinh^2 x = (e^{2x} - 2 + e^{-2x})/4. Subtracting cancels the exponentials and leaves (2-(-2))/4 = 1. Adding the squares would cancel the constants instead and leave the exponentials. Dividing by \cosh^2 x gives 1 - \tanh^2 x = \operatorname{sech}^2 x, and dividing by \sinh^2 x gives \coth^2 x - 1 = \operatorname{csch}^2 x.
How do you find cosh x if you know sinh x?
Use \cosh^2 x = 1 + \sinh^2 x and take the positive square root. For example, if \sinh x = 3/4, then \cosh^2 x = 1 + 9/16 = 25/16, so \cosh x = 5/4. Reject the negative root because \cosh x = (e^x + e^{-x})/2 is half the sum of two positive exponentials, so it is always positive. Unlike solving for \cos x, there is no quadrant to check.
What are the hyperbolic addition and double-angle formulas?
The addition formulas are \sinh(x+y) = \sinh x\cosh y + \cosh x\sinh y and \cosh(x+y) = \cosh x\cosh y + \sinh x\sinh y. Setting y = x gives \sinh 2x = 2\sinh x\cosh x and \cosh 2x = \cosh^2 x + \sinh^2 x. The second can be rewritten as 2\cosh^2 x - 1 or 1 + 2\sinh^2 x. For instance, \sinh 2x/(1+\cosh 2x) = 2\sinh x\cosh x/(2\cosh^2 x) = \tanh x.
Why do some hyperbolic identities have opposite signs from trig identities?
A sign flips wherever a product of two sines becomes a product of two sinh's. The complex relations \cosh(ix) = \cos x and \sinh(ix) = i\sin x explain this, since (i\sin x)(i\sin y) = -\sin x\sin y. That is why \cos^2 x + \sin^2 x = 1 becomes \cosh^2 x - \sinh^2 x = 1. It is also why \cos 2x = 1 - 2\sin^2 x becomes \cosh 2x = 1 + 2\sinh^2 x, while \sinh(x+y) keeps the sine pattern.

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