show that $|\sin(1)-0.841|\leq10^{-3}$

Question

My textbook states that $|\sin(1)-0.841|\leq10^{-3}$ but I do not know how this could be true. It also gives a table showing values of ${1\over n!}\pm10^{-8}$ when $n$ is $1$ to $10$.

The reason why I think this statement is false:

$0.841$ is approximately $1-{1\over3!}$

Since the taylor series of $sin1$ is $1-{1\over3!}+{1\over5!}-{1\over7!}+\dots$, it is trivial that $|sin1-(1-{1\over3!})|={1\over5!}-{1\over7!}+\dots$

${1\over5!}$ is $0.00833333$ and ${1\over7!}$ is $0.00019841$

There is no way $|sin1-(1-{1\over3!})|$ can be less than or equal to $0.001$, which is precisely what the textbook's statement is.

Am I missing something here?

Answer 1

the statement is clearly true it states that $$0.84 \le \sin(1)\le 0.842$$

Remainder term of Taylor series $$R_n=\frac{f^{(n+1)}(z)(x-c)^n+1}{(n+1)!}$$

Expanding about $c=0$

put $x=1$ and calculate the $n$ required to get the remainder within $10^{-3}.$

Answer 2

We know that $\pi=3.14159...$ so that $\frac\pi3=1.04719...$, thus $1=\frac\pi3-a$ with $0<a<0.05$. Thus $a^3<0.000125$, the error of the quadratic Taylor polynomial is already small enough. Then \begin{align} \sin(1)&=\sin(\tfrac\pi3-a)=\sin\tfrac\pi3-\cos(\tfrac\pi3)\,a-\sin(\tfrac\pi3)\tfrac{a^2}2+\cos(\tfrac\pi3-\theta a)\tfrac{a^3}6 \\ &=0.86602540378(1-0.5\cdot 0.0022276)-0.5\cdot 0.04719+O(10^{-5}) \\ &=0.8660-0.0236-0.87\cdot 0.0011+O(10^{-5}) \\ &=0.8424-0.00096+O(10^{-5})=0.8414+O(10^{-5}) \end{align}

Answer 3

The Taylor series for $\sin x$ is an alternating series, so we have an estimate for the error when we use a finite expansion up to a certain order: the absolute value of the remainder is no more than the absolute value of its first term, and it has the same sign.

Thus, expanding at order $5$, we know that $$-\frac{x^7}{7!}< \sin x -\biggl(x-\frac{x^3}{3!}+\frac{x^5}{5!}\biggr)<0$$ and in particular $$-\frac{1}{7!}< \sin 1 -\biggl(1-\frac{1}{3!}+\frac{1}{5!}\biggr)<0.$$ Computing these values, we have that the approximate value $s_\text{approx}$ of $\sin1$ at order $5$ is $$1-\frac{1}{3!}+\frac{1}{5!}=\frac{101}{120}=0.841666\dots=0.841+\frac 2{3000}$$ and $0.841$ is its rounded down value $s_\text{dec}$ to $3$ decimal places, so that we have both $\; \sin 1, \,s_\text{dec} <s_\text{approx}$.

Now, there are two possibilities:

• either $\;\sin 1\le s_\text{dec} <s_\text{approx}$, and we have $$|\sin 1-s_\text{dec}| <|\sin 1-s_\text{approx}|<\frac 1{7!}<10^{-3},$$
• or $\;s_\text{dec} <\sin 1<s_\text{approx}$, and $$|\sin 1-s_\text{dec}| <|s_\text{approx}-s_\text{dec}|<\frac 2{3000}<10^{-3}.$$