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)=sinxg(x)= \sin x and then we will substitute xx with 9595^{\circ} to get the series for our function sin95\sin 95^{\circ}.

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

95=1.66rad95^{\circ}=1.66 rad

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

g(x)=sinx=n=0(1)nx2n+1(2n+1)!g(x)=\sin x=\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}

So, the series is

sin95=sin1.66=n=0(1)n1.662n+1(2n+1)!\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 Theorem 8.15\textit{Theorem 8.15} the given inequality is

SSN=RNaN+1=(1)N+11.662(N+1)+1(2(N+1)+1)!\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 0.0010.001 we have the new inequality

(1)N+11.662(N+1)+1(2(N+1)+1)!<0.001\dfrac{(-1)^{N+1} 1.66^{2(N+1)+1}}{(2(N+1)+1)!}<0.001

It is correct for N=3N=3.

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

Therefore,

sin95=sin1.66=n=03(1)n1.662n+1(2n+1)!=(1)01.6620+1(20+1)!+(1)11.6621+1(21+1)!+(1)21.6622+1(22+1)!+(1)31.6623+1(23+1)!0.995767\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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