Question

Express \(\sinh ^{4} x\) in terms of hyperbolic cosines of multiples of \(x\), and hence solve $$ 2 \cosh 4 x-8 \cosh 2 x+5=0 $$

Short answer

Using \cosh^{-1}, the values are \frac{\cosh^{-1}(1.5)}{2} and \frac{\cosh^{-1}(0.5)}{2}.

Step-by-step solution

- Use Hyperbolic Identity for Squared Terms

Recall the identity for hyperbolic sine squared: \[ \sinh^{2}x = \frac{\cosh(2x) -1}{2} \]To express \sinh^{4}x, we first express \sinh^{2}x and then square that expression.

- Express \sinh^{4}x in Terms of \cosh(2x)

Using the identity from Step 1, we write: \[ \sinh^{4}x = \left(\frac{\cosh(2x) - 1}{2} \right)^2 \]Simplify this to: \[ \sinh^{4}x = \frac{(\cosh(2x) - 1)^2}{4} \]

- Expand and Simplify

Expand the fraction: \[ \sinh^{4}x = \frac{\cosh^2(2x) - 2 \cosh(2x) + 1}{4} \]

- Use the Given Equation

Given: \[ 2 \cosh(4x) - 8 \cosh(2x) + 5 = 0 \]Express \cosh(4x) in terms of \cosh^2(2x) using the identity \[ \cosh(4x) = 2 \cosh^2(2x) - 1 \]Substitute into the equation: \[ 2 \left( 2 \cosh^2(2x) - 1 \right) - 8 \cosh(2x) + 5 = 0 \]

- Simplify the New Equation

Distribute and simplify: \[ 4 \cosh^2(2x) - 2 - 8 \cosh(2x) + 5 = 0 \]Combine like terms: \[ 4 \cosh^2(2x) - 8 \cosh(2x) + 3 = 0 \]

- Solve the Quadratic Equation

Let \cosh(2x) = y. Then the equation becomes: \[ 4y^2 - 8y + 3 = 0 \]Solve this quadratic using the quadratic formula: \[ y = \frac{8 \pm \sqrt{64 - 48}}{8} \] This simplifies to: \[ y = \frac{8 \pm \sqrt{16}}{8} \] \[ y = \frac{8 \pm 4}{8} \] \[ y = 1.5 \ or \ y = 0.5 \]

- Back Substitute for \cosh(2x)

We now have: \[ \cosh(2x) = 1.5 \] or \[ \cosh(2x) = 0.5 \]Solve for \cosh(x) if initial substitution is required, but \cosh(x) values are obtained directly.

- Final Values

Undo \cosh using \cosh^{-1} for each value: For \cosh(2x) = 1.5: \[ 2x = \cosh^{-1}(1.5) \] \[ x = \frac{\cosh^{-1}(1.5)}{2} \]For \cosh(2x) = 0.5: \[ 2x = \cosh^{-1}(0.5) \] \[ x = \frac{\cosh^{-1}(0.5)}{2} \]

Key concepts

Hyperbolic Identities

Hyperbolic identities are essential tools in understanding and solving problems involving hyperbolic functions. Similar to trigonometric identities, hyperbolic identities relate the hyperbolic sine and cosine functions to each other. One fundamental identity is the hyperbolic sine squared identity:
\[ \sinh^{2}x = \frac{\cosh(2x) -1}{2} \]
This identity allows us to express \sinh^{2}x in terms of \cosh(2x). By squaring both sides of this identity, we can derive an expression for \sinh^{4}x:
\[ \sinh^{4}x = \left(\frac{\cosh(2x) - 1}{2} \right)^2 \]
Simplifying further, we get:
\[ \sinh^{4}x = \frac{(\cosh(2x) - 1)^2}{4} \]
Expanding the square:
\[ \sinh^{4}x = \frac{\cosh^2(2x) - 2 \cosh(2x) + 1}{4} \]
This final form helps to bridge the gap between hyperbolic sines and cosines, useful for solving equations like the one in the exercise.

Quadratic Equations

A quadratic equation is any equation that can be written in the form
\[ ax^2 + bx + c = 0 \]
where \a, \b, and \c are constants. In the given problem, after substituting \cosh(2x) with \y, we obtain:

\[ 4y^2 - 8y + 3 = 0 \]
We solve this using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plug in the values \a = 4, \b = -8, and \c = 3:
\[ y = \frac{8 \pm \sqrt{64 - 48}}{8} \]
This simplifies to:
\[ y = \frac{8 \pm \sqrt{16}}{8} \] \[ y = \frac{8 \pm 4}{8} \] \[ y = 1.5 \text{{ or }} y = 0.5 \]
Thus, we find two possible values for \cosh(2x). Solving quadratic equations like this is crucial in various areas of mathematics.

Inverse Hyperbolic Functions

Inverse hyperbolic functions allow us to solve for the variable inside a hyperbolic function. The inverse hyperbolic cosine function, denoted as \cosh^{-1}(x), helps in finding the original input from \cosh(x).
Given the solutions \cosh(2x) = 1.5 and \cosh(2x) = 0.5, we can find \x by applying the inverse function:
For \cosh(2x) = 1.5:
\[ 2x = \cosh^{-1}(1.5) \]
\[ x = \frac{\cosh^{-1}(1.5)}{2} \]
For \cosh(2x) = 0.5:
\[ 2x = \cosh^{-1}(0.5) \]
\[ x = \frac{\cosh^{-1}(0.5)}{2} \]
These steps revert the hyperbolic function to its original input value. Understanding and using inverse hyperbolic functions is essential when working backwards from the result to the variable.