# Approximate cos(58) to four decimal place accuracy using taylor's theorem

Approximate cos(58) to four decimal place accuracy using taylor's theorem.

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Hint: I imagine the $58$ is in degrees. This is almost $60$, so in radians it is $\pi/3 -2\pi/180$. Expand $\cos x$ about $a=\pi/3$. We know explicitly $\cos$ and its derivatives at $\pi/3$. –  André Nicolas Apr 12 '12 at 19:48
No thanks. You can do your own homework. –  user5137 Apr 12 '12 at 19:54
(^surely there's a way to have this easily "copied to clipboard" for quick reply to such questions? I had to look up AM's recent activity...) –  The Chaz 2.0 Apr 12 '12 at 20:02
I have a text file where I keep all of my standard math.SE responses. –  Zev Chonoles Apr 12 '12 at 22:17

Since we have to use Taylor's Theorem to solve this problem, let's first recall what it (or at least one formulation) is:

Taylor's Theorem:
Let $n\in\Bbb N$, let $I=[a,b]$, and let $f:I\rightarrow \Bbb R$ be such that $f$ and its derivatives $f'$, $f''$, $\ldots\,$, $f^{(n)}$ are continuous on $I$ and that $f^{(n+1)}$ exists on $(a,b)$. If $x_0\in I$, then for any $x$ in $I$ there exists a point $c$ between $x$ and $x_0$ such that $$f(x)=\color{darkgreen}{f(x_0)+f'(x_0)(x-x_0)+{f''(x_0)\over 2!}(x-x_0)^2+ {f'''(x_0)\over 3!}(x-x_0)^3}$$ $$\color{darkgreen} { +\cdots +{f^{(n)}(x_0)\over n!}(x-x_0)^n} +\color{maroon}{{f^{(n+1)}(c)\over (n+1)!}(x-x_0)^{n+1}}$$

The darkgreen term in Taylor's Theorem $$P_n(x)=\textstyle \color{darkgreen}{f(x_0)+f'(x_0)(x-x_0)+{f''(x_0)\over 2!}(x-x_0)^2+ {f'''(x_0)\over 3!}(x-x_0)^3} \color{darkgreen} { +\cdots +{f^{(n)}(x_0)\over n!}(x-x_0)^n}$$ is the Taylor polynomial of the function $f$ of degree $n$ centered at $x_0$.

The maroon term in Taylor's Theorem, $$\tag{1} E_n(x)=\color{maroon}{{f^{(n+1)}(c)\over (n+1)!} (x-x_0)^{n+1}}$$ is called the the error term. Note that the "$c$" here depends on both $n$ and $x$.

If we estimate the function $f(x)$ with the value of $P_n(x)$, then the error in approximation is given by $E_n(x)$. Now, of course, we can't in practice hope to know what $E_n(x)$ is exactly; but if we can find a number $M$ so that $$\tag{2} \max_{t\text{ btw }x_0\text{ and }x } \Biggl|\, {f^{(n+1)}(t)\over (n+1)! (x-x_0)^{n+1}} \, \Biggr | \le M,$$ then we would know from Taylor's Theorem that $$\tag{3} \bigl|\,f(x)-P_n(x)\,\bigr|\le M.$$ The general procedure for approximating a function's value $f(x)$ with a Taylor polynomial $P_n(x)$ to within a certain degree of accuracy, say $\epsilon$, is to first determine a value of $n$, the degree of the Taylor Polynomial, needed so that the right hand side of $(2)$ is less than or equal to $\epsilon$. Then $(3)$ would tell us that the Taylor polynomial of degree $n$ will approximate the value of $f(x)$ to within $\epsilon$.

Now on to your problem. This is calculus, we will use radians for angular measurements.

We wish to estimate $\cos (29\pi/90)$ to within $1/10^4$ using Taylor's Theorem. Since $29\pi/90\approx\pi/3$, we will do so by using a Taylor polynomial of $f(x)=\cos x$ centered at $x_0=\pi/3$ and using $x=29\pi/90$ in Taylor's Theorem.

But what should should the degree $n$ of the polynomial be? We can use the remarks above to determine this:

We want the error in approximation, $E_n(29\pi/90)$ to satisfy $$\tag{4} \bigl|\,E_n(29\pi/90)\,\bigr|\le {1\over 10^4}.$$ Using $(1)$ to write the term $E_n(29\pi/90)$ more explicitly, we want

$$\Biggl|\,{ f^{(n+1)} (c)\over (n+1)!} \bigl({\textstyle{29\pi\over 90}-{\pi\over3}}\bigr)^{n+1}\,\Biggr|\le {1\over 10^4};$$ which simplifies to

$$\tag{5} \Biggl|\,{ f^{(n+1)} (c)\over (n+1)!}\bigl({\textstyle{- \pi\over 90}}\bigr )^{n+1}\,\Biggr|\le {1\over 10^4}.$$ Since we do not know what $c$ is, we find an upper bound for the expression on the left hand side of $(5)$; that is, we find an $M$ as in $(2)$. We can be slack here and use the fact that all derivatives of $f(x)=\cos x$ produce functions with absolute value at most 1. So, we have $$\tag{6} \bigl|E_n(29\pi/90)\bigr|= \Biggl|\,{ f^{(n+1)} (c)\over (n+1)!}\bigl({\textstyle{- \pi\over 90}}\bigr )^{n+1}\,\Biggr|\le \Biggl|\, {1\over (n+1)!}({ -\pi\over 90} )^{n+1}\,\Biggr|.$$ Now we find an $n$ so that the right hand side of $(6)$ is less than or equal to $1/10^4$. Then it will follow that inequality $(4)$ holds for this value of $n$, and consequently that $P_n(29\pi/90)$ approximates $\cos(29\pi/90)$ to within $1/10^4$. It turns out that (nicely) $n=2$ does the job, as direct verification will attest (you can plug values of $n$ into the right hand side of $(6)$ until you see the inequality is satisfied).

So, $n=2$, and the desired approximation is $P_2(29\pi/90)$.

We, of course, need to find $P_2(29\pi/90)$. Towards this end we first find the general form of $P_2(x)$ which is: $$P_2(x)=f(\pi/3)+f'(\pi/3)(x-\pi/3)+{f''(\pi/3)\over2!}(x-\pi/3)^2.$$

We have: \eqalign{ f(\pi/3) &= \cos(\pi/3)=1/2\cr f'(\pi/3)&=-\sin(\pi/3)=\sqrt3/2\cr f''(\pi/3)&= -\cos(\pi/3)=-1/2\cr }

So $$P_2(x) = {1\over 2}-{\sqrt3\over2}(x-{\pi/3})-{1\over 2\cdot2!}(x-{\pi/3})^2.$$

Finally, the desired approximation is: $$P_2(29\pi/90) = {1\over 2}-{\sqrt3\over2}(-\pi/90)-{1\over 2\cdot2!}(-\pi/90)^2.$$

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I can't decide whether to love or to hate this answer! –  Chris Taylor Apr 12 '12 at 23:06