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

We want to prove the identity:

\[
\sinh(x+y) = \sinh(x) \cosh(y) + \cosh(x) \sinh(y)
\]

### Proof:

#### Step 1: Use the definition of hyperbolic sine and cosine.
Recall the definitions of the hyperbolic sine and cosine functions:
\[
\sinh(z) = \frac{e^z - e^{-z}}{2}, \quad \cosh(z) = \frac{e^z + e^{-z}}{2}
\]

#### Step 2: Express \(\sinh(x + y)\) using the definition of \(\sinh\).
Substitute \(z = x + y\) into the definition of \(\sinh\):
\[
\sinh(x+y) = \frac{e^{x+y} - e^{-(x+y)}}{2}
\]

#### Step 3: Expand the exponential terms.
We can rewrite the exponentials as:
\[
\sinh(x+y) = \frac{e^x e^y - e^{-x} e^{-y}}{2}
\]

#### Step 4: Rewrite the right-hand side of the identity.
Now, let's substitute the definitions of \(\sinh(x)\), \(\sinh(y)\), \(\cosh(x)\), and \(\cosh(y)\) on the right-hand side of the original identity:
\[
\sinh(x)\cosh(y) + \cosh(x)\sinh(y) = \left(\frac{e^x - e^{-x}}{2}\right) \left(\frac{e^y + e^{-y}}{2}\right) + \left(\frac{e^x + e^{-x}}{2}\right) \left(\frac{e^y - e^{-y}}{2}\right)
\]

#### Step 5: Multiply out each term.
Expand both products:
\[
\sinh(x)\cosh(y) = \frac{e^x e^y + e^x e^{-y} - e^{-x} e^y - e^{-x} e^{-y}}{4}
\]
\[
\cosh(x)\sinh(y) = \frac{e^x e^y - e^x e^{-y} + e^{-x} e^y - e^{-x} e^{-y}}{4}
\]

#### Step 6: Combine like terms.
Now, add the two expressions:
\[
\sinh(x)\cosh(y) + \cosh(x)\sinh(y) = \frac{(e^x e^y - e^{-x} e^{-y}) + (e^{-x} e^y - e^x e^{-y})}{4}
\]
The terms \(e^{-x}e^y\) and \(-e^{-x}e^y\) cancel out, leaving:
\[
\sinh(x)\cosh(y) + \cosh(x)\sinh(y) = \frac{2(e^x e^y - e^{-x} e^{-y})}{4} = \frac{e^x e^y - e^{-x} e^{-y}}{2}
\]

#### Step 7: Conclusion.
This is exactly the same expression we derived for \(\sinh(x+y)\). Thus, we have proven the identity:
\[
\sinh(x + y) = \sinh(x)\cosh(y) + \cosh(x)\sinh(y)
\]

Would you like further details or have any questions?

Here are five related questions to deepen your understanding:
1. How would you prove \(\cosh(x+y) = \cosh(x)\cosh(y) + \sinh(x)\sinh(y)\)?
2. What are the hyperbolic identities analogous to the Pythagorean identities in trigonometry?
3. How are hyperbolic functions related to exponential functions in complex analysis?
4. How do you derive the derivative of \(\sinh(x)\) and \(\cosh(x)\)?
5. How does the identity change if you substitute complex values for \(x\) and \(y\)?

**Tip**: The hyperbolic functions \(\sinh(x)\) and \(\cosh(x)\) are closely related to the exponential function, and many properties of trigonometric functions have analogs in the hyperbolic world.

Discover how to prove the hyperbolic identity sinh(x+y) = sinh(x)cosh(y) + cosh(x)sinh(y). This solution involves key mathematical concepts such as hyperbolic and exponential functions, using step-by-step explanations suitable for Grades 11-12 and college-level math students.

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