Math Problem Statement

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

Solution

We want to prove the identity:

sinh(x+y)=sinh(x)cosh(y)+cosh(x)sinh(y)\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)=ezez2,cosh(z)=ez+ez2\sinh(z) = \frac{e^z - e^{-z}}{2}, \quad \cosh(z) = \frac{e^z + e^{-z}}{2}

Step 2: Express sinh(x+y)\sinh(x + y) using the definition of sinh\sinh.

Substitute z=x+yz = x + y into the definition of sinh\sinh: sinh(x+y)=ex+ye(x+y)2\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)=exeyexey2\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(x), sinh(y)\sinh(y), cosh(x)\cosh(x), and cosh(y)\cosh(y) on the right-hand side of the original identity: sinh(x)cosh(y)+cosh(x)sinh(y)=(exex2)(ey+ey2)+(ex+ex2)(eyey2)\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)=exey+exeyexeyexey4\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)=exeyexey+exeyexey4\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)=(exeyexey)+(exeyexey)4\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 exeye^{-x}e^y and exey-e^{-x}e^y cancel out, leaving: sinh(x)cosh(y)+cosh(x)sinh(y)=2(exeyexey)4=exeyexey2\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)\sinh(x+y). Thus, we have proven the identity: sinh(x+y)=sinh(x)cosh(y)+cosh(x)sinh(y)\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)\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)\sinh(x) and cosh(x)\cosh(x)?
  5. How does the identity change if you substitute complex values for xx and yy?

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

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Math Problem Analysis

Mathematical Concepts

Hyperbolic Functions
Exponential Functions
Trigonometry

Formulas

sinh(z) = (e^z - e^{-z}) / 2
cosh(z) = (e^z + e^{-z}) / 2
sinh(x + y) = sinh(x)cosh(y) + cosh(x)sinh(y)

Theorems

Addition Formulas for Hyperbolic Functions
Exponential Function Properties

Suitable Grade Level

Grades 11-12, College-level Mathematics