Question

prove sinh2x=2sinhx.coshx

Answer 1

Ooo! hyperbolic! I love these!

Answer 2

Let's use our definitions:$$\sinh x=\frac12(e^x-e^{-x})\\\cosh x=\frac12(e^x+e^{-x})$$

Answer 3

... so $$\sinh(2x)=\frac12(e^{2x}-e^{-2x})=\frac12\left((e^{x})^2-(e^{-x})^2\right)=\frac12(e^x-e^{-x})(e^x+e^{-x})$$Notice we just used difference of squares to factor that

Answer 4

Recall $$\cosh x=\frac12(e^x+e^{-x})$$so we know \(e^x+e^{-x}=2\cosh x\). Similarly, we know \(\dfrac12(e^x-e^{-x})=\sinh x\). Replacing this all we have:$$\frac12(e^x-e^{-x})(e^x+e^{-x})=(\sinh x)(2\cosh x)=2\sinh x\cosh x$$

Answer 5

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