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In the exam 2 of this page the first question asks to approximate $\sin(\pi + \frac 1 {100})$ to two decimal places. The solution they give seems incorrect: they simply say $\sin(\pi+\frac 1 {100}) \approx \sin(\pi)+\cos(\pi)\frac 1 {100} = 0-1\frac 1{100} = -0.01$. Nowhere did they prove that this approximation is good enough to guarantee two decimal places of correctness. Am i missing something?

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  • $\begingroup$ It doesn't sound like the question asks for a proof, so I don't see why the lack of a proof makes the answer incorrect. $\endgroup$ Apr 4, 2023 at 20:07
  • $\begingroup$ @MishaLavrov Of course it makes it incorrect. Without a proof how would you know whether the approximation is correct or not? The question explicitly asks to estimate it to two digits. This process does not guarantee that the approximation obtained is correct to two digits. It would be like someone asking you to calculate $1+1$ and you answering $1+1 = 3 = 2$. You get the correct answer but the reasoning is wrong. $\endgroup$
    – Carla_
    Apr 4, 2023 at 20:11
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    $\begingroup$ I would describe that as "the answer is correct, and the reasoning is not provided". If I were writing the exam, I might ask for justification (and then say that an answer with no justification is incorrect). It sounds like you would as well. But the provided answer is correct to the question as I see it. $\endgroup$ Apr 4, 2023 at 20:22

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Apparently they used the angle addition identity $$\sin(x+y) = \sin x \cos y + \cos x \sin y.$$ This gives $$\sin(\pi + 0.01) = \sin \pi \cos (0.01) + \cos \pi \sin (0.01) = - \sin (0.01).$$

This much is algebraically valid, no approximation is used here. Then to obtain $\sin (0.01)$, the approximation $$\sin x \approx x + O(x^3)$$ is used. More precisely, $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$ being the series expansion about $x = 0$, it is clear that for $|x| < 10^{-2}$, the error term will be less than $(10^{-2})^3 = 10^{-6}$. This is the missing justification. Hence $$\sin (0.01) \approx 0.01 + \epsilon$$ where $|\epsilon| < 0.000001.$

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  • $\begingroup$ I think they've just used the quadratic taylor expansion, however in their notes they don't prove any bounds for the error of the taylor expansion which makes it suspicious that they ask this question in the exam but your reasoning seems correct. In fact they don't even talk about taylor series at that point. $\endgroup$
    – Carla_
    Apr 4, 2023 at 20:17
  • $\begingroup$ How do you connect the taylor expansion and the error term? I would say it helps that the series is alternating and decreasing in modulus so that one can build lower and upper bounds easily?... could you clarify the missing justification? $\endgroup$
    – Thomas
    Apr 4, 2023 at 20:39
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    $\begingroup$ The identity $\sin(\pi +x)=-\sin x$ is elementary and does not require the formula for $\sin(x+y).$ $\endgroup$ Apr 4, 2023 at 22:54

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