The trigonometric identity $\cos^2 x + \sin^2 x = 1$ comes from the unit circle: $x^2 + y^2 = 1$.
Hyperbolic functions satisfy an analogous identity from the unit hyperbola: $x^2 - y^2 = 1$.
This single sign change, from $+$ to $-$, ripples through all the identities. Understanding this pattern lets you reconstruct hyperbolic identities from their trig counterparts without memorizing a separate list.
Answer: $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$. If you couldn't recall these, review Hyperbolic Function Definitions.
Answer: $\cos^2 x + \sin^2 x = 1$. The hyperbolic version has a key sign difference!
| Property | Value |
|---|---|
| Concept | Hyperbolic Functions |
| Chapter | 6.7 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
$$\boxed{\cosh^2 x - \sinh^2 x = 1}$$
This is the hyperbolic Pythagorean identity. Compare with trig: $\cos^2 x + \sin^2 x = 1$.
Why it works: Substitute the definitions:
$$\cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2$$
$$= \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{4}{4} = 1$$
Dividing the fundamental identity by $\cosh^2 x$ or $\sinh^2 x$:
| Identity | Derivation |
|---|---|
| $1 - \tanh^2 x = \text{sech}^2 x$ | Divide by $\cosh^2 x$ |
| $\coth^2 x - 1 = \text{csch}^2 x$ | Divide by $\sinh^2 x$ |
| Function | Property | Because |
|---|---|---|
| $\sinh(-x) = -\sinh x$ | Odd | $(e^{-x} - e^x)/2 = -(e^x - e^{-x})/2$ |
| $\cosh(-x) = \cosh x$ | Even | $(e^{-x} + e^x)/2 = (e^x + e^{-x})/2$ |
| $\tanh(-x) = -\tanh x$ | Odd | Odd divided by even |
$$\sinh(x + y) = \sinh x \cosh y + \cosh x \sinh y$$ $$\cosh(x + y) = \cosh x \cosh y + \sinh x \sinh y$$
Notice: The $\sinh$ formula looks exactly like $\sin(x+y)$, but the $\cosh$ formula has $+$ where $\cos(x+y)$ has $-$.
Setting $y = x$ in the addition formulas:
$$\sinh 2x = 2\sinh x \cosh x$$ $$\cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1 = 1 + 2\sinh^2 x$$
$$\cosh x + \sinh x = e^x$$ $$\cosh x - \sinh x = e^{-x}$$
These are extremely useful for simplification. They follow directly from adding or subtracting the definitions.
| Trig Identity | Hyperbolic Identity |
|---|---|
| $\cos^2 x + \sin^2 x = 1$ | $\cosh^2 x - \sinh^2 x = 1$ |
| $1 + \tan^2 x = \sec^2 x$ | $1 - \tanh^2 x = \text{sech}^2 x$ |
| $\sin(x+y) = \sin x \cos y + \cos x \sin y$ | $\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$ |
| $\cos(x+y) = \cos x \cos y - \sin x \sin y$ | $\cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y$ |
Pattern: When both terms in a trig identity have the same sign, the hyperbolic version may flip one sign.
Simplify $\cosh^2(3x) - \sinh^2(3x)$.
Verify that $\tanh(-x) = -\tanh x$ (i.e., $\tanh$ is an odd function) by:
(a) Computing $\tanh(-\ln 3)$ and $-\tanh(\ln 3)$ and showing they are equal.
(b) Explaining why $\tanh$ must be odd based on the even/odd properties of $\sinh$ and $\cosh$.
If $\tanh x = \frac{4}{5}$ and $x > 0$, find $\sinh x$ and $\cosh x$.
(a) Use the identity $\cosh 2x = 2\cosh^2 x - 1$ to find the exact value of $\cosh^2(\ln 2)$.
(b) Verify your answer by computing $\cosh(\ln 2)$ directly from the definition and squaring it.
The identity $\cosh x + \sinh x = e^x$ implies that $(\cosh x + \sinh x)^n = e^{nx}$ for any $n$.
(a) Use this to show that $(\cosh x + \sinh x)^3$ can be written as $\cosh(3x) + \sinh(3x)$.
(b) Expand $(\cosh x + \sinh x)^3$ using the binomial theorem and use hyperbolic identities to simplify. Verify you get the same result.
(c) Find the value of $(\cosh(\ln 2) + \sinh(\ln 2))^5$.
Question 1: The identity $\cosh^2 x - \sinh^2 x = 1$ differs from $\cos^2 x + \sin^2 x = 1$ by a sign. This sign difference reflects:
(A) Hyperbolic functions are defined using subtraction instead of addition (B) Points $(\cosh t, \sinh t)$ lie on a hyperbola rather than a circle (C) Hyperbolic functions are unbounded while trig functions are bounded (D) There is no geometric significance to the sign difference
(B) The circle $x^2 + y^2 = 1$ gives $\cos^2 + \sin^2 = 1$. The hyperbola $x^2 - y^2 = 1$ gives $\cosh^2 - \sinh^2 = 1$. The sign change directly reflects the different conic section.
Question 2: If you know that $\cosh x = 2$, how many possible values are there for $\sinh x$?
(A) 0 (B) 1 (C) 2 (D) Infinitely many
(C) From $\cosh^2 x - \sinh^2 x = 1$: $\sinh^2 x = 4 - 1 = 3$, so $\sinh x = \pm\sqrt{3}$. Since $\sinh$ is odd and $\cosh$ is even, $x$ and $-x$ give the same $\cosh$ but opposite $\sinh$ values.
Question 3 (Explain Why): The identity $\cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y$ has a plus sign between the two terms, while the trig version $\cos(x+y) = \cos x \cos y - \sin x \sin y$ has a minus sign. Why does this sign difference occur?
The sign difference traces back to the fundamental identities. For trig functions, we have $\cos^2 + \sin^2 = 1$ (addition), while for hyperbolic functions, we have $\cosh^2 - \sinh^2 = 1$ (subtraction).
When you expand $\cosh(x+y)$ using exponential definitions, the cross terms involving products of $e^x$ and $e^{-y}$ (and vice versa) combine constructively rather than destructively. The "subtraction" in the fundamental hyperbolic identity means that products of two $\sinh$ terms (which would appear with minus signs in the trig case) appear with plus signs in the hyperbolic case.
Key insight: Whenever a trig identity involves a product of two sine terms, the hyperbolic analog flips that sign.
The Sign-Flip Pattern:
To convert a trig identity to its hyperbolic analog:
This works because $\sinh^2$ terms come with the "wrong" sign compared to $\sin^2$, due to the hyperbola vs. circle relationship.
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|---|---|---|
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Last updated: 2026-01-22