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Question Number 17279 by tawa tawa last updated on 03/Jul/17

Is cosh^2 (3x) = (1/2)[1 + cos(6x)] ??????

$$\mathrm{Is}\:\:\mathrm{cosh}^{\mathrm{2}} \left(\mathrm{3x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}\:+\:\mathrm{cos}\left(\mathrm{6x}\right)\right]\:\:?????? \\ $$

Commented by mrW1 last updated on 03/Jul/17

cosh^2 3x=(((e^(3x) +e^(−3x) )/2))^2 =((e^(6x) +e^(−6x) +2e^(3x) e^(−3x) )/4) =((e^(6x) +e^(−6x) +2)/4) =(1/2)(((e^(6x) +e^(−6x) )/2)+1) =(1/2)(cosh 6x+1)

$$\mathrm{cosh}^{\mathrm{2}} \:\mathrm{3x}=\left(\frac{\mathrm{e}^{\mathrm{3x}} +\mathrm{e}^{−\mathrm{3x}} }{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{e}^{\mathrm{6x}} +\mathrm{e}^{−\mathrm{6x}} +\mathrm{2e}^{\mathrm{3x}} \mathrm{e}^{−\mathrm{3x}} }{\mathrm{4}} \\ $$$$=\frac{\mathrm{e}^{\mathrm{6x}} +\mathrm{e}^{−\mathrm{6x}} +\mathrm{2}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{e}^{\mathrm{6x}} +\mathrm{e}^{−\mathrm{6x}} }{\mathrm{2}}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cosh}\:\mathrm{6x}+\mathrm{1}\right) \\ $$

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

$$ \\ $$

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

I really appreciate sir. God bless you.

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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