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Simplified using factor formulae, $(\cosh x-\cosh y)^2-(\sinh x-\sinh y)^2$ equals:
A$4\cosh^2\left(\frac{x+y}{2}\right)$
B$4\sinh^2\left(\frac{x-y}{2}\right)$
C$-4\cosh^2\left(\frac{x-y}{2}\right)$
D$-4\sinh^2\left(\frac{x-y}{2}\right)$
Answer & Solution
Correct answer: D. $-4\sinh^2\left(\frac{x-y}{2}\right)$
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_