Question

Use a Taylor polynomial to approximate the function with an error of less than 0.001.sin 95$^{\circ}$

Solution

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Answered 2 years ago
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First, we will find the series for the function $g(x)= \sin x$ and then we will substitute $x$ with $95^{\circ}$ to get the series for our function $\sin 95^{\circ}$.

Let's convert $\textit{degrees}$ to $\textit{radians}$.

$95^{\circ}=1.66 rad$

From the table of $\textit{Power Series for Elementary Functions}$ we see that

$g(x)=\sin x=\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$

So, the series is

\begin{align*} \boldsymbol{\sin 95^{\circ}} &= \sin 1.66 \\ &= \boldsymbol{\sum_{n=0}^{\infty} \dfrac{(-1)^n 1.66^{2n+1}}{(2n+1)!}} \end{align*}

According to $\textit{Theorem 8.15}$ the given inequality is

$\left|S-S_N \right| = \left|R_N \right| \leq a_{N+1 }=\dfrac{(-1)^{N+1} 1.66^{2(N+1)+1}}{(2(N+1)+1)!}$

Hence, since the error is less than $0.001$ we have the new inequality

$\dfrac{(-1)^{N+1} 1.66^{2(N+1)+1}}{(2(N+1)+1)!}<0.001$

It is correct for $N=3$.

Therefore, the required number for correct approximation is $\boldsymbol{3}$.

Therefore,

\begin{align*} \boldsymbol{\sin 95^{\circ}} &= \sin 1.66 \\ &= \sum_{n=0}^{3} \dfrac{(-1)^n 1.66^{2n+1}}{(2n+1)!} \\ &= \dfrac{(-1)^0 1.66^{2\cdot 0+1}}{(2\cdot 0+1)!}+ \dfrac{(-1)^1 1.66^{2\cdot 1+1}}{(2\cdot 1+1)!}+ \dfrac{(-1)^2 1.66^{2\cdot 2+1}}{(2\cdot 2+1)!}+ \dfrac{(-1)^3 1.66^{2\cdot 3+1}}{(2\cdot 3+1)!} \\ &\approx \boldsymbol{0.995767} \end{align*}

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