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

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

Commented by mrW1 last updated on 03/Jul/17

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

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

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

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

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