Use the Taylor polynomials for the sine and cosine functions to find a rational function with a degree 5 numerator and no fractional coefficients that approximates the tangent function near 0.
A) $R(x)=\frac{x^{5}+24 x^{4}-120 x^{3}}{x^{4}-6 x^{2}+120}$
B) $R(x)=\frac{x^{5}-20 x^{3}+120 x}{5 x^{4}-60 x^{2}+120}$
C) $R(x)=\frac{30 x^{4}-6 x^{2}+120}{-120 x^{3}+x^{5}}$
D) $R(x)=\frac{120 x^{5}-24 x^{3}+x}{30 x^{4}-60 x^{2}+120}$

Question

Quizplus · Accepted Answer

The Taylor series expansions for sine and cosine functions around 0 are $sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ and $cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$. The tangent function, $tan(x) = \frac{sin(x)}{cos(x)}$, can be approximated by dividing the polynomials of sine and cosine. For a degree 5 numerator, we use up to $x^5$ for sine and $x^4$ for cosine. Simplifying the division gives $R(x)=\frac{x - \frac{x^3}{6} + \frac{x^5}{120}}{1 - \frac{x^2}{2} + \frac{x^4}{24}}$, which simplifies to $R(x)=\frac{x^{5}-20 x^{3}+120 x}{5 x^{4}-60 x^{2}+120}$ after multiplying both numerator and denominator by 120 to eliminate fractional coefficients.

Use the Taylor Polynomials for the Sine and Cosine Functions R(x)=x5+24x4−120x3x4−6x2+120R(x)=\frac{x^{5}+24 x^{4}-120 x^{3}}{x^{4}-6 x^{2}+120}R(x)=x4−6x2+120x5+24x4−120x3​

Use the Taylor Polynomials for the Sine and Cosine Functions $R(x)=\frac{x^{5}+24 x^{4}-120 x^{3}}{x^{4}-6 x^{2}+120}$