Using $ T_{7} $ of a Taylor series to estimate the value of $ \sin(3^{\circ}) $.

Question

This is Part (b) of a problem. Part (a) consisted of finding $ {T_{7}}(x) $ for $ \sin(x) $ at $ a = 0 $, and I got

$ {T_{7}}(x) = x - x^{3} + x^{5} - x^{7} $.

Now, I need to use $ T_{7} $ to estimate the value of $ \sin(3^{\circ}) $. Do I essentially plug $ 3^{\circ} $ into my answer from Part (a)? I’m confused because my teacher said that $ 3^{\circ} $ must be converted into radians, so if I plug $ 3 $ in, then I’ll have a constant instead of something in radians.

Any help on how to execute this (seemingly easy) problem would be great!

Answer 1

When you take derivatives of trig functions line $\sin x$ the argument $(x)$ is assumed to be expressed in radians. A radian is an angle corresponding to an arc length in a circle that is equal to the radius of that circle; thus since the circumference of the circle is $2\pi r$, $2\pi$ radians is $360^\circ$.

So $3^\circ$ is $3 \times \frac{2\pi}{360} = \frac{\pi}{60}$ radians.

As the comment pointed out, you have the wrong Taylor series (you forgot the $n!$ in the denominator of each term.

Answer 2

Here is the correct Taylor polynomial: $$ {T_{7}}(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!}. $$ Furthermore, for $ n \in \{ 7,8 \} $, we have $$ \forall x \in \mathbb{R}: \qquad \sin(x) = {T_{7}}(x) + x^{n} \cdot \epsilon(x), $$ where $ \epsilon: \mathbb{R} \to \mathbb{R} $ is some function satisfying $ \displaystyle \lim_{x \to 0} \epsilon(x) = 0 $. Therefore, $ \sin(3^{\circ}) \approx T_{7} \! \left( \dfrac{3 \pi}{180} \right) $.