Easy -1.2 This is a straightforward proof requiring only substitution of the exponential definitions of cosh x and sinh x, followed by basic algebraic manipulation. It's a standard first exercise when introducing hyperbolic identities, requiring no problem-solving insight—just direct application of definitions.
Recalls exponential definitions and substitutes \(\frac{e^x+e^{-x}}{2}\) and \(\frac{e^x-e^{-x}}{2}\). Condone cosh \(x\) replaced with \(\frac{e^x-e^{-x}}{2}\) and sinh \(x\) replaced with \(\frac{e^x+e^{-x}}{2}\).
\(= \frac{4}{4} = 1\)
R1 (AO2.1)
Completes a fully correct proof to reach the required result.
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cosh^2x - \sinh^2x = \left(\frac{e^x+e^{-x}}{2}\right)^2 - \left(\frac{e^x-e^{-x}}{2}\right)^2 = \frac{e^{2x}+2+e^{-2x}}{4} - \frac{e^{2x}-2+e^{-2x}}{4}$ | M1 (AO1.2) | Recalls exponential definitions and substitutes $\frac{e^x+e^{-x}}{2}$ and $\frac{e^x-e^{-x}}{2}$. Condone cosh $x$ replaced with $\frac{e^x-e^{-x}}{2}$ and sinh $x$ replaced with $\frac{e^x+e^{-x}}{2}$. |
| $= \frac{4}{4} = 1$ | R1 (AO2.1) | Completes a fully correct proof to reach the required result. |
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