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Question Number 97157 by mhmd last updated on 06/Jun/20

Commented by mhmd last updated on 06/Jun/20

$$sirpleasehelpmeiwantEx\left(3\right)andEx\left(4\right)$$

Answered by Sourav mridha last updated on 06/Jun/20

$$\mathit{E}\mathrm{x}\left(2\right).\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{x}=\frac{{\mathit{e}}^{\mathit{x}}+{\mathit{e}}^{-\mathit{x}}}{2}$$$$\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{x}=\frac{{\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{-\mathit{x}}}{2}$$$${(\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{x}+\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{x})}^{\mathit{n}}={\mathit{e}}^{\mathit{n}\mathit{x}}$$$$=[\frac{{\mathit{e}}^{\mathit{n}\mathit{x}}+{\mathit{e}}^{-\mathit{n}\mathit{x}}}{2}+\frac{{\mathit{e}}^{\mathit{n}\mathit{x}}-{\mathit{e}}^{-\mathit{n}\mathit{x}}}{2}]$$$$=\mathit{c}\mathit{o}\mathit{s}\mathit{h}\left(\mathit{n}\mathit{x}\right)+\mathit{s}\mathit{i}\mathit{n}\mathit{h}\left(\mathit{n}\mathit{x}\right)$$

Commented by mhmd last updated on 06/Jun/20

$$thankyousirpleasehelpmeinEx\left(3\right)$$

Answered by Sourav mridha last updated on 06/Jun/20

$$\mathit{E}\mathit{x}\left(4\right)\mathit{s}\mathit{i}\mathit{n}\mathit{h}\left(3\mathit{x}\right)=\frac{{\mathit{e}}^{3\mathit{x}}-{\mathit{e}}^{-3\mathit{x}}}{2}$$$$=\frac{1}{2}[{({\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{-\mathit{x}})}^3+3{\mathit{e}}^{\mathit{x}}.{\mathit{e}}^{-\mathit{x}}({\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{-\mathit{x}})]$$$$=3\left[\frac{{\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{-\mathit{x}}}{2}\right]+\frac{2^3}{2}{\left[\frac{{\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{-\mathit{x}}}{2}\right]}^3$$$$=3\mathit{s}\mathit{i}\mathit{n}\mathit{h}\left(\mathit{x}\right)+4{\mathit{s}\mathit{i}\mathit{n}\mathit{h}}^3\left(\mathit{x}\right).$$

Answered by mathmax by abdo last updated on 06/Jun/20

$$3)\mathrm{we}\mathrm{have}{\mathrm{ch}}^2\mathrm{x}-{\mathrm{sh}}^2\mathrm{x}=1\Rightarrow {\mathrm{ch}}^2\mathrm{x}=1+{\mathrm{sh}}^2\mathrm{x}=1+{\tan }^2\theta$$$$\mathrm{but}\mathrm{chx}>0\Rightarrow \mathrm{ch}\left(\mathrm{x}\right)=\sqrt{1+{\tan }^2\theta }=\frac{1}{\cos \theta }$$$$\mathrm{thx}=\frac{\mathrm{shx}}{\mathrm{chx}}=\frac{\tan \theta }{\frac{1}{\cos \theta }}=\cos \theta \times \frac{\sin \theta }{\cos \theta }=\sin \theta$$

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