Math: Hyperbolic Functions and Their Inverses
Using definitions, identities, graphs, and inverse functions
Math: Hyperbolic Functions and Their Inverses
Using definitions, identities, graphs, and inverse functions
Math - Grade 9-12
- 1
Use the definitions sinh x = (e^x - e^(-x))/2 and cosh x = (e^x + e^(-x))/2 to find exact expressions for sinh(ln 3) and cosh(ln 3).
Rewrite e^(-ln 3) as 1/e^(ln 3).
Since e^(ln 3) = 3 and e^(-ln 3) = 1/3, sinh(ln 3) = (3 - 1/3)/2 = 4/3. Also, cosh(ln 3) = (3 + 1/3)/2 = 5/3. - 2
Prove the identity cosh^2 x - sinh^2 x = 1 using the exponential definitions of sinh x and cosh x.
Expand both squares carefully and combine like terms.
Using the definitions, cosh^2 x - sinh^2 x = [(e^x + e^(-x))/2]^2 - [(e^x - e^(-x))/2]^2. Expanding gives [e^(2x) + 2 + e^(-2x)]/4 - [e^(2x) - 2 + e^(-2x)]/4 = 4/4 = 1. - 3
Evaluate tanh 0, sinh 0, and cosh 0. Explain what each value means on the graphs of the functions.
sinh 0 = 0, cosh 0 = 1, and tanh 0 = 0. These values mean that the graph of sinh x passes through (0, 0), the graph of cosh x has the point (0, 1), and the graph of tanh x passes through (0, 0). - 4
Use a calculator to approximate sinh 2, cosh 2, and tanh 2 to three decimal places.
Use the hyperbolic function keys if your calculator has them, or use the exponential definitions.
sinh 2 is approximately 3.627, cosh 2 is approximately 3.762, and tanh 2 is approximately 0.964. - 5
Show that tanh x = sinh x/cosh x can be written as tanh x = (e^(2x) - 1)/(e^(2x) + 1).
After substituting, multiply the top and bottom of the fraction by e^x.
Starting with tanh x = sinh x/cosh x, substitute the definitions to get tanh x = (e^x - e^(-x))/(e^x + e^(-x)). Multiplying the numerator and denominator by e^x gives tanh x = (e^(2x) - 1)/(e^(2x) + 1). - 6
Describe the domain, range, and symmetry of y = cosh x.
The domain of y = cosh x is all real numbers. The range is y >= 1 because cosh x has a minimum value of 1 at x = 0. The function is even because cosh(-x) = cosh x, so its graph is symmetric across the y-axis. - 7
Describe the domain, range, horizontal asymptotes, and symmetry of y = tanh x.
Think about what happens to (e^(2x) - 1)/(e^(2x) + 1) as x becomes very large or very negative.
The domain of y = tanh x is all real numbers. The range is -1 < y < 1. The graph has horizontal asymptotes y = 1 and y = -1. The function is odd because tanh(-x) = -tanh x, so its graph has origin symmetry. - 8
Solve sinh x = 4 exactly. Write your answer using an inverse hyperbolic function and then as a natural logarithm.
Use the identity arcsinh u = ln(u + sqrt(u^2 + 1)).
The solution can be written as x = arcsinh 4. Using the formula arcsinh u = ln(u + sqrt(u^2 + 1)), x = ln(4 + sqrt(17)). - 9
Solve cosh x = 3. Give the exact solutions.
Since cosh x is even, there are two solutions. Using arcosh u = ln(u + sqrt(u^2 - 1)), the positive solution is ln(3 + sqrt(8)), so the exact solutions are x = ln(3 + sqrt(8)) and x = -ln(3 + sqrt(8)). - 10
Find the exact value of arcosh(5/3).
First simplify sqrt(25/9 - 9/9).
Using arcosh u = ln(u + sqrt(u^2 - 1)), arcosh(5/3) = ln(5/3 + sqrt(25/9 - 1)) = ln(5/3 + 4/3) = ln 3. - 11
A hanging cable can be modeled by y = 2 cosh(x/2). Find the lowest point of the cable and the height of the cable at x = 4. Round the height to three decimal places.
The lowest point occurs when x = 0 because cosh(x/2) has its minimum when x/2 = 0. The lowest point is (0, 2). At x = 4, y = 2 cosh(2), which is approximately 2(3.762) = 7.524. - 12
Find the derivative of f(x) = arctanh(3x). State the interval of x-values where the function is defined.
Use the chain rule and remember that arctanh u is defined only for -1 < u < 1.
The derivative of arctanh u is u'/(1 - u^2). With u = 3x, f'(x) = 3/(1 - 9x^2). The function is defined when -1 < 3x < 1, so the domain is -1/3 < x < 1/3.