Use the definitions of sinh(x) and cosh(x) in terms of exponential functions to prove that cosh(2x) = 2cosh²(x) - 1.

We are asked to prove the identity:

cosh(2x) = 2cosh²(x) - 1

Step 1: Recall the definitions of sinh(x) and cosh(x)
The hyperbolic sine and cosine functions are defined in terms of exponential functions as follows:

sinh(x) = (e^x - e^(-x)) / 2
cosh(x) = (e^x + e^(-x)) / 2
Step 2: Express cosh(2x) using the definition of cosh
We need to find cosh(2x). Using the definition of cosh(x), we get:

cosh(2x) = (e^(2x) + e^(-2x)) / 2

Step 3: Express cosh²(x)
Now, we calculate cosh²(x) from its definition:

cosh(x) = (e^x + e^(-x)) / 2 Therefore, cosh²(x) = [(e^x + e^(-x)) / 2]² cosh²(x) = (e^(2x) + 2 + e^(-2x)) / 4

Step 4: Multiply cosh²(x) by 2
To find 2cosh²(x), we multiply cosh²(x) by 2:

2cosh²(x) = 2 * (e^(2x) + 2 + e^(-2x)) / 4 2cosh²(x) = (e^(2x) + 2 + e^(-2x)) / 2

Step 5: Compare cosh(2x) with 2cosh²(x) - 1
Now, we compare cosh(2x) and 2cosh²(x) - 1:

cosh(2x) = (e^(2x) + e^(-2x)) / 2 2cosh²(x) - 1 = (e^(2x) + 2 + e^(-2x)) / 2 - 1

Simplifying the right-hand side:

2cosh²(x) - 1 = (e^(2x) + e^(-2x)) / 2 + 1 - 1 2cosh²(x) - 1 = (e^(2x) + e^(-2x)) / 2

Step 6: Conclusion
We see that both sides of the equation are equal:

cosh(2x) = 2cosh²(x) - 1

Thus, the identity is proved.