ACSEE (Form 6) Hyperbolic Functions — practice questions

1. Write $\cosh x=\tfrac12(e^{x}+e^{-x})$ and $\sinh x=\tfrac12(e^{x}-e^{-x})$. 2. Substituting into $m\sinh x+n\cosh x=h$ and clearing denominators gives $(m+n)e^{2x}-2he^{x}+(n-m)=0$. 3. This is a quadratic in $e^{x}$. Equal roots require the discriminant to vanish: $(-2h)^2-4(m+n)(n-m)=0$. 4. So $4h^2=4(n^2-m^2)$, giving $h^2=n^2-m^2$ and $h=\pm\sqrt{n^2-m^2}$. 5. The option with $m^2-n^2$ reverses the sign of the product $(m+n)(n-m)$; the two with a plus sign treat the constant term as $(n+m)$. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 2: Hyperbolic Functions_

1. Apply the factor formulae: $\cosh x-\cosh y=2\sinh\left(\frac{x+y}{2}\right)\sinh\left(\frac{x-y}{2}\right)$. 2. Likewise $\sinh x-\sinh y=2\cosh\left(\frac{x+y}{2}\right)\sinh\left(\frac{x-y}{2}\right)$. 3. Squaring and subtracting leaves $4\sinh^2\left(\frac{x-y}{2}\right)\left[\sinh^2\left(\frac{x+y}{2}\right)-\cosh^2\left(\frac{x+y}{2}\right)\right]$. 4. The bracket is $-1$ by the identity $\cosh^2\theta-\sinh^2\theta=1$. 5. Hence the value is $-4\sinh^2\left(\frac{x-y}{2}\right)$. Dropping the minus sign is the usual slip. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 2: Hyperbolic Functions_