From the definition of sinhx and coshx, in terms of exponentials show that,
sinhx+sinhy=2sinh((x+y)/2).cosh((x-y)/2)
Can anyone help explain how to solve this? Cheers.
2 \times \frac{e^{\frac{x+y}{2}}-e^{-\frac{x+y}{2}}}{2} \times \frac{e^{\frac{x-y}{2}}+e^{-\frac{x-y}{2}}}{2}
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\frac{e^x-e^{-x}}{2} + \frac{e^y-e^{-y}}{2} = 2 \times \frac{e^{\frac{x+y}{2}}-e^{-\frac{x+y}{2}}}{2} \times \frac{e^{\frac{x-y}{2}}+e^{-\frac{x-y}{2}}}{2}
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How did the 2 (highlighted in red) cancel out?