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How one can prove by pen and paper that $0.98089<\sin\left (\frac{11}{8}\right )<0.9809$? I was thinking some series proof but I'm not sure how to prove that the error is small enough. I also thought to use the formulas $\sin 3x=3 \sin x - 4 \sin^3 x$ and $\sin x\approx x$ for small $x$ but I don't know how to prove that the error is small enough.

studying
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2 Answers2

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I would not know about completely manually, but using $x=\frac\pi2-\frac{11}8$ and $\sin(\frac{11}8)=\cos x$ you could try if $$ 1-\frac12 x^2\le\cos x\le 1-\frac12x^2+\frac1{24}x^4 $$ is already precise enough.

In [42]: 1-x**2/2
Out[42]: 0.980831899206813

In [43]: 1-x**2/2+x**4/24
Out[43]: 0.9808931352214827

In [46]: cos(x)
Out[46]: 0.9808930570231557

In [47]: 1-x**2/2+x**4/24-x**6/720
Out[47]: 0.9808930569696093

And no, the second partial sum of the alternating cosine series on the left is not precise enough, one needs the 4 term sum up to degree 6.

Lutz Lehmann
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The series $$\sin x=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}x^{2k+1}=x-\frac16x^3+\frac1{120}x^5-\frac1{5040}x^7+\frac1{362880}x^9\mp\ldots$$ is alternating for $x>0$ and the summands are decreasing in value as soon as $x^2\le{2k(2k+1)}$. This means that two consecutive partial sums can be used as bounds on $\sin x$ provided that $x^2\le{2k(2k+1)}$ holds for the first summand left out from the second sum. For $x=\frac{11}8$, this is the case already for $k=1$, i.e., this poses no restriction. One can (by hand, but awfully laborously) verify that $$ x - \frac 16x^3 + \frac1{120}x^5 - \frac1{5040}x^7 + \frac1{362880}x^9 =\frac{6824909553341}{6957847019520}<\frac{9809}{10000}$$ and that after subtracting $\frac1{39916800}x^{11}=\frac{25937424601}{31171154647449600}$ we are still $>\frac{98089}{100000}$. That's it.

The main problem is that the desired relative error is quite small ($\approx 3\cdot 10^{-6}$).

Hagen von Eitzen
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