sinh2x identity

\sinh2x\equiv2\sinh x\cdot\cosh x

Proof

flashcards

QuestionAnswer
What is the double angle identity for \sinh 2x?\sinh 2x \equiv 2 \sinh x \cdot \cosh x
How is \sinh x defined in terms of exponentials?\sinh x = \frac{e^x - e^{-x}}{2}
How is \cosh x defined in terms of exponentials?\cosh x = \frac{e^x + e^{-x}}{2}
Starting from 2 \sinh x \cdot \cosh x, what is the first step of the proof using exponential definitions?2\cdot\frac{e^x - e^{-x}}{2}\cdot\frac{e^x + e^{-x}}{2}
After simplifying 2\cdot\frac{e^x - e^{-x}}{2}\cdot\frac{e^x + e^{-x}}{2}, what expression is obtained?\frac{(e^x - e^{-x})(e^x + e^{-x})}{2}
What is the result of expanding (e^x - e^{-x})(e^x + e^{-x})?e^{2x} - e^{-2x}
What final expression does \frac{e^{2x} - e^{-2x}}{2} equal?\sinh 2x