Science:Math Exam Resources/Courses/MATH102/December 2014/Question A 08

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MATH102 December 2014

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Question A 08
In order to calculate $\arcsin(0.4)$ using Newton’s method, which of the following would be the appropriate formula? (A) $\arcsin(0.4)\approx 0+0.4$ (B) $\arcsin(0.4)\approx \arcsin(0)+\arccos(0)(0.4-0)$ (C) $x_{n+1}=x_{n}-{\dfrac {\sin(x_{n})-0.4}{\cos(x_{n})}}$ (D) $x_{n+1}=x_{n}-{\dfrac {\arcsin(x_{n})-0.4}{\arccos(x_{n})}}$ (E) None of the above.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Set $y=\sin(x)$ and use Newton’s method to find the root.

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Solution
Let $y=\sin(x)$ then the problem becomes finding $x$ such that $\sin(x)=0.4$ or $\sin(x)-0.4=0$. According to Newton’s method, to find roots of $f(x)=0$, the formula is $x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}.$ Here we have $f(x)=\sin(x)-0.4$, then $f'(x)=\cos(x)$ and the above formula becomes $x_{n+1}=x_{n}-{\frac {\sin(x_{n})-0.4}{\cos(x_{n})}},$ which is answer (C).

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