Question

Use the definitions of hyperbolic cosine and hyperbolic sine (their formulas) to prove the identity below. Show your work neatly in the space provided.
h2 − h2 = 1

Answer 1

Answer:

See below

Step-by-step explanation:

By definition:

$\sinh x = \frac{ {e}^{x} + {e}^{ - x} }{2} \\ \cosh x = \frac{ {e}^{x} - {e}^{ - x} }{2}$

We want to show that:

${ \sinh}^{2} x - { \cosh}^{2}x = 1$

We substitute into the left side to get:

$( \frac{ {e}^{x} + {e}^{ - x} }{2}) ^{2} - ( \frac{ {e}^{x} - {e}^{ - x} }{2}) ^{2}$

We apply exponential rule to get:

$= \frac{ ({e}^{x} + {e}^{ - x})^{2} }{4} - \frac{( {e}^{x} - {e}^{ - x} ) ^{2} }{4}$

We expand:

$= \frac{ {e}^{2x} + 2 {e}^{ - x} {e}^{x} + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 {e}^{ - x} {e}^{x} + {e}^{ - 2x} }{4}$

$= \frac{ {e}^{2x} + 2 + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 + {e}^{ - 2x} }{4}$

$= \frac{ {e}^{2x} + 2 + {e}^{ - 2x} - {e}^{2x} + 2 - {e}^{ - 2x}}{4}$

$= \frac{4}{4} \\ = 1$