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Question Number 33688 by mondodotto@gmail.com last updated on 22/Apr/18

$$\mathbf{given}\mathbf{that}$$$$\mathbf{f}\left(\mathbf{x}\right)=\frac{1}{2}({10}^{\mathit{x}}+{10}^{-\mathit{x}})\mathbf{prove}\mathbf{that}$$$$2\mathbf{f}\left(\mathbf{x}\right)\mathbf{f}\left(\mathbf{y}\right)=\mathbf{f}(\mathbf{x}+\mathbf{y})+\mathbf{f}(\mathbf{x}-\mathbf{y})$$

Answered by Rasheed.Sindhi last updated on 22/Apr/18

$$\mathbf{f}\left(\mathbf{x}\right)=\frac{1}{2}({10}^{\mathit{x}}+{10}^{-\mathit{x}})$$$$\mathbf{f}\left(\mathrm{y}\right)=\frac{1}{2}({10}^{\mathrm{y}}+{10}^{-\mathrm{y}})$$$$2\mathbf{f}\left(\mathbf{x}\right)\mathbf{f}\left(\mathbf{y}\right)=2\times \frac{1}{2}({10}^{\mathit{x}}+{10}^{-\mathit{x}})\times \frac{1}{2}({10}^{\mathrm{y}}+{10}^{-\mathrm{y}})$$$$=\frac{1}{2}({10}^{\mathrm{x}+\mathrm{y}}+{10}^{-\mathrm{x}-\mathrm{y}}+{10}^{\mathrm{x}-\mathrm{y}}+{10}^{\mathrm{y}-\mathrm{x}})$$$$=\frac{1}{2}({10}^{\mathrm{x}+\mathrm{y}}+{10}^{-(\mathrm{x}+\mathrm{y})}+{10}^{\mathrm{x}-\mathrm{y}}+{10}^{-(\mathrm{x}-\mathrm{y})})$$$$=\frac{1}{2}({10}^{\mathrm{x}+\mathrm{y}}+{10}^{-(\mathrm{x}+\mathrm{y})})+\frac{1}{2}({10}^{\mathrm{x}-\mathrm{y}}+{10}^{-(\mathrm{x}-\mathrm{y})})$$$$=\mathbf{f}(\mathbf{x}+\mathbf{y})+\mathbf{f}(\mathbf{x}-\mathbf{y})$$

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