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Question Number 129634 by abdurehime last updated on 17/Jan/21

$${\mathit{s}\mathit{i}\mathit{n}}^{}\mathit{x}+{\mathit{c}\mathit{o}\mathit{s}}^{}\mathit{x}=1\mathrm{proof}\mathrm{by}\mathrm{step}\mathrm{by}\mathrm{step}\mathrm{or}\mathrm{by}\mathrm{showing}\mathrm{all}\mathrm{steps}$$$$\mathrm{b}/\mathrm{c}\mathrm{it}\mathrm{is}\mathrm{my}\mathrm{assignment}$$

Answered by MJS_new last updated on 17/Jan/21

$$\mathrm{the}\mathrm{easiest}\mathrm{way}\mathrm{to}\mathrm{prove}\mathrm{is}\mathrm{graphically}.$$$${\mathrm{here}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{different}\mathrm{proof}:$$$$\sin x=\frac{{\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x}}{2\mathrm{i}}\wedge \cos x=\frac{{\mathrm{e}}^{\mathrm{i}x}+{\mathrm{e}}^{-\mathrm{i}x}}{2}$$$$\mathrm{let}{\mathrm{e}}^{\mathrm{i}x}=p$$$${\sin }^{2}x+{\cos }^{2}x=\frac{{(p-\frac{1}{p})}^2}{{\left(2\mathrm{i}\right)}^2}+\frac{{(p+\frac{1}{p})}^2}{2^2}=$$$$=\frac{p^2-2+\frac{1}{p^2}}{-4}+\frac{p^2+2+\frac{1}{p^2}}{4}=\frac{4}{4}=1$$

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