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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/4Determine the derivatives of the given functions \( \mathrm{{\sin{\phi}}} \)and \( \mathrm{{\cos{\phi}}} \)as shown below:\( \mathrm{\frac{{{d}{\left\lbrace{\sin{\phi}}\right\rbrace}}}{{{d}\phi}}={\cos{\phi}}} \)\( \mathrm{\frac{{{d}{\left\lbrace{\cos{\phi}}\right\rbrace}}}{{{d}\phi}}=-{\sin{\phi}}} \)Explanation:Please refer to solution in this step.Step2/4Now evaluate the obtained derivatives at the point \( \mathrm{\phi={0}} \).\( \mathrm{{\sin{{\left({0}\right)}}}={0}} \)\( \mathrm{{\cos{{\left({0}\right)}}}={1}} \)Therefore, the linear approximation for \( \mathrm{{\sin{\phi}}} \)is determined as follows:\( \begin{align*} \mathrm{{\sin{\phi}}} &\approx \mathrm{{\sin{{\left({0}\right)}}}+\frac{{{d}{\left\lbrace{\sin{\phi}}\right\rbrace}}}{{{d}\phi}}{\left(\phi-{0}\right)}}\\[3pt] &= \mathrm{{\sin{{\left({0}\right)}}}+{\cos{{\left({0}\right)}}}{\left(\phi-{0}\right)}}\\[3pt] &= \mathrm{{0}+{1}{\left(\phi-{0}\right)}}\\[3pt] &= \mathrm{\phi} \end{align*} \)Therefore, \( \mathrm{{\sin{\phi}}\approx\phi.} \)Explanation:Please refer to solution in this step.Step3/4Now, determin ... See the full answer