Hyperbolic functions — practice questions

We start with the left-hand side (LHS) and substitute the definitions.

LHS = $\cosh^2(x) - \sinh^2(x)$

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

Expand the numerators: $= \frac{(e^x)^2 + 2(e^x)(e^{-x}) + (e^{-x})^2}{4} - \frac{(e^x)^2 - 2(e^x)(e^{-x}) + (e^{-x})^2}{4}$

$= \frac{e^{2x} + 2e^0 + e^{-2x}}{4} - \frac{e^{2x} - 2e^0 + e^{-2x}}{4}$

Since $e^0 = 1$: $= \frac{e^{2x} + 2 + e^{-2x} - (e^{2x} - 2 + e^{-2x})}{4}$

Combine the fractions: $= \frac{e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}}{4}$

Cancel terms in the numerator: $= \frac{4}{4} = 1$

LHS = 1 = RHS. The identity is proven.

We will use the exponential definitions for $\cosh(x)$ and $\sinh(x)$.

$3\left(\frac{e^x + e^{-x}}{2}\right) + 2\left(\frac{e^x - e^{-x}}{2}\right) = 4$

Multiply the entire equation by 2 to clear the denominators: $3(e^x + e^{-x}) + 2(e^x - e^{-x}) = 8$

Expand the brackets: $3e^x + 3e^{-x} + 2e^x - 2e^{-x} = 8$

Collect like terms: $5e^x + e^{-x} = 8$

To form a quadratic, multiply by $e^x$ and recall that $e^{-x} \times e^x = e^0 = 1$: $5(e^x)^2 + 1 = 8e^x$

Rearrange into a standard quadratic form: $5(e^x)^2 - 8e^x + 1 = 0$

Let $u = e^x$. The equation becomes: $5u^2 - 8u + 1 = 0$

This does not factorise easily, so we use the quadratic formula: $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. $u = \frac{8 \pm \sqrt{(-8)^2 - 4(5)(1)}}{2(5)}$ $u = \frac{8 \pm \sqrt{64 - 20}}{10}$ $u = \frac{8 \pm \sqrt{44}}{10} = \frac{8 \pm 2\sqrt{11}}{10} = \frac{4 \pm \sqrt{11}}{5}$

Now, substitute back $u = e^x$. We have two potential solutions: $e^x = \frac{4 + \sqrt{11}}{5}$ or $e^x = \frac{4 - \sqrt{11}}{5}$

Since $\sqrt{11}$ is between 3 and 4 (as $3^2=9, 4^2=16$), both numerators are positive. Therefore, both values of $u$ are positive and valid solutions for $e^x$.

Taking the natural logarithm of both sides for each solution: $x = \ln\left(\frac{4 + \sqrt{11}}{5}\right)$ or $x = \ln\left(\frac{4 - \sqrt{11}}{5}\right)$

These are the two exact solutions.