# MIT old Calculus Exam Question about estimating $\sin(\pi+\frac 1 {100})$ to two decimal places

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?

• 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. Apr 4, 2023 at 20:07
• @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. Apr 4, 2023 at 20:11
• 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. Apr 4, 2023 at 20:22

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 $$|\epsilon| < 0.000001.$$
• The identity $\sin(\pi +x)=-\sin x$ is elementary and does not require the formula for $\sin(x+y).$ Apr 4, 2023 at 22:54