Question ​​​​​​​ Problem 2 [10 Points] One of Einsteins most amazing predictions was that light traveling from distant stars would bend around the sun on the way to earth. His calculations involved solving for \( \phi \) in the equation \[ \sin (\phi)+b\left(1+\cos ^{2}(\phi)+\cos (\phi)\right)=0 \] (A) Using derivatives and the linear approximation, estimate the values of \( \sin (\phi) \) and \( \cos (\phi) \) when \( \phi \approx 0 \). (B) Approximate the above equation by substituting the approximations for sin and cos. (C) Solve for \( \phi \) approximately. Explain your answers in detail.

RU0CYG The Asker · Advanced Mathematics

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Transcribed Image Text: Problem 2 [10 Points] One of Einsteins most amazing predictions was that light traveling from distant stars would bend around the sun on the way to earth. His calculations involved solving for \( \phi \) in the equation \[ \sin (\phi)+b\left(1+\cos ^{2}(\phi)+\cos (\phi)\right)=0 \] (A) Using derivatives and the linear approximation, estimate the values of \( \sin (\phi) \) and \( \cos (\phi) \) when \( \phi \approx 0 \). (B) Approximate the above equation by substituting the approximations for sin and cos. (C) Solve for \( \phi \) approximately. Explain your answers in detail.
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Transcribed Image Text: Problem 2 [10 Points] One of Einsteins most amazing predictions was that light traveling from distant stars would bend around the sun on the way to earth. His calculations involved solving for \( \phi \) in the equation \[ \sin (\phi)+b\left(1+\cos ^{2}(\phi)+\cos (\phi)\right)=0 \] (A) Using derivatives and the linear approximation, estimate the values of \( \sin (\phi) \) and \( \cos (\phi) \) when \( \phi \approx 0 \). (B) Approximate the above equation by substituting the approximations for sin and cos. (C) Solve for \( \phi \) approximately. Explain your answers in detail.
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CUGY81

【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