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Question Number 17280 by tawa tawa last updated on 03/Jul/17

prove that: cosh(2x) = 2cosh^2 (x) − 1

$$\mathrm{prove}\:\mathrm{that}:\:\:\mathrm{cosh}\left(\mathrm{2x}\right)\:=\:\mathrm{2cosh}^{\mathrm{2}} \left(\mathrm{x}\right)\:−\:\mathrm{1} \\ $$

Commented by mrW1 last updated on 03/Jul/17

cosh (2x)=((e^(2x) +e^(−2x) )/2) =((e^(2x) +e^(−2x) +2−2)/2) =(((e^x )^2 +2 e^x e^(−x) +(e^(−x) )^2 )/2)−1 =(((e^x +e^(−x) )^2 )/2)−1 =2(((e^x +e^(−x) )/2))^2 −1 =2 cosh^2 x−1

$$\mathrm{cosh}\:\left(\mathrm{2x}\right)=\frac{\mathrm{e}^{\mathrm{2x}} +\mathrm{e}^{−\mathrm{2x}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{e}^{\mathrm{2x}} +\mathrm{e}^{−\mathrm{2x}} +\mathrm{2}−\mathrm{2}}{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{e}^{\mathrm{x}} \right)^{\mathrm{2}} +\mathrm{2}\:\mathrm{e}^{\mathrm{x}} \:\mathrm{e}^{−\mathrm{x}} +\left(\mathrm{e}^{−\mathrm{x}} \right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1} \\ $$$$=\frac{\left(\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} \right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1} \\ $$$$=\mathrm{2}\left(\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$=\mathrm{2}\:\mathrm{cosh}^{\mathrm{2}} \:\mathrm{x}−\mathrm{1} \\ $$

Commented by tawa tawa last updated on 03/Jul/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

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