# Find Taylor series for $\sin x$ at $\frac{\pi}{6}$ and estimate $\sin 38^\circ$

I want to see if I understand Taylor series correctly. Mind checking this work?

(a) So $$f(x) = \sin{x}$$ and $$a = \frac{\pi}{6}$$, $$n = 4$$ (I want to find $$T_4(x)$$ and I want to find the accuracy of the approximation when $$x$$ lies in the interval $$[0,\frac{\pi}{3}]$$.

So I write:

$$f'(x) = \cos{x},\quad f''(x) = -\sin{x},\quad f''(x) = -\cos{x},$$ $$f^{(4)}(x) = \sin{x},\quad f^{(5)}(x) = \cos{x}$$

and thus $$f'(\frac{\pi}{6}) = \frac{\sqrt{3}}{2},\quad f''(\frac{\pi}{6}) = \frac{-1}{2},\quad f'''(\frac{\pi}{6}) = \frac{-\sqrt{3}}{2}$$ $$f^{(4)}(\frac{\pi}{6}) = \frac{1}{2},\quad f^{(5)}(\frac{\pi}{6}) = \frac{\sqrt{3}}{2},$$ which gives $$T_4(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}(x-\frac{\pi}{6}) + \frac{\frac{-1}{2}(x-\frac{\pi}{6})^2}{2!} + \frac{\frac{-\sqrt{3}}{2}(x-\frac{\pi}{6})^3}{3!} + \frac{\frac{1}{2}(x-\frac{\pi}{6})^4}{4!}.$$

Question 1: Is the Taylor polynomial of the 4th degree above correct?

(b) Now I try to find the error in approximating $$\sin\frac{\pi}{6}$$ by the first 4 terms using Taylor's Inequality in my textbook: First of all, $$f^{(5)}(x) = \cos{x}$$. For $$x$$ with $$0 \leq x \leq \frac{\pi}{3}$$, $$\cos x$$ is largest at $$0$$ so $$f^{(5)}(x) \leq 1$$ and $$M = 1$$.

So $$R_5(x) = \frac{1}{5!}(x-a)^5$$

I'm stuck here.

Question 2: Can I say that $$|x-a| < d$$ where d = $$\frac{\pi}{6}$$?

(c) To estimate $$\sin 38^{\circ}$$, I write:

$$\sin 38^{\circ} = \sin(\frac{38}{180}\pi) = \sin(\frac{19}{90}\pi)$$

Question 3: Do I just plugin that value to the $$T_4$$ expression?

Question 4: More generally, what does it mean to be a Taylor series about $$\frac{\pi}{6}$$?

• a) Looks good. c) Remember that $\frac{\pi}{6}=\frac{15\pi}{90}$, then yes. b) $|R_4(x)|\leq\frac{1}{5!}|x-\frac\pi 6|^5$ The RHS is maximal when $x=\frac \pi 3 \to R_4(x)\leq \frac{\pi^5}{5!\cdot 6^5}\approx 3.280\cdot 10^{-4}$ – Rhys Hughes Apr 19 at 0:49
• For b, I got the same answer, but I have that the 5th deriative of f = cosx is maximal when x= 0 right? @RhysHughes – Jwan622 Apr 19 at 0:54
• Yeah you're right I should've used $0$ and it will get the same answer because $|\frac{\pi}{3}-\frac \pi 6|=|0-\frac\pi 6|$ – Rhys Hughes Apr 19 at 0:57
• @RhysHughes more generally, waht does it mean to be a taylor series "about" $\frac{\pi}{6}$? – Jwan622 Apr 19 at 1:18
• This graph should make that clear. The taylor approximation $T_n(x)$ about $P$ of $f(x)$ has the property that $T_n(P)=f(P)$ and $T_n(x)\approx f(x)$ for $x\neq P$,with better approximation as $n$ increases. – Rhys Hughes Apr 19 at 13:06

Normally, we have a Taylor Series for $$\sin(x)$$ centered around $$x=0$$: $$\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ or written more obviously, this is $$\sum_{n=0}^\infty (-1)^n\frac{(x-0)^{2n+1}}{(2n+1)!}$$ We know this is centered at zero because our computations of the derivatives and the $$(x-0)$$ term. But what does this mean? Well, it means that we approximate the curve $$\sin(x)$$ using higher and higher order derivatives at zero, so like the tangent line, the "tangent cubic", the "tangent quintic", and so on, as you can see here: As you can see, at $$x=0$$, all the curves approximate $$\sin(x)$$ really well. The farther you get from $$x=0$$, the worse the approximation gets. Well, what if I want a value pretty far away? I would have to use many terms if I restricted myself to be centered at zero, but I can always center my approximation around a new point point, as you can see in this picture here: This new Taylor series is the one you can compute using derivatives evaluated at $$x=5\pi/6$$ and the terms $$(x-5\pi/6)^n$$. All that we are doing is calculating the tangent line, the "tangent cubic", the "tangent quintic", and so on at the new point, allowing for more accurate calculations at and around this new point of $$x=5\pi/6$$ without having a gajillion terms in our sum.

Let $$x=y+\frac \pi 6$$ making $$\sin(x)=\sin \left(y+\frac{\pi }{6}\right)=\frac 12 \big( \sqrt{3} \sin (y)+\cos (y)\big)$$ for which the derivatives at $$y=0$$ are very simple.
As a result $$\frac 12 \big( \sqrt{3} \sin (y)+\cos (y)\big)= \sum_{n=0}^\infty \frac{\sqrt{3} \sin \left(\frac{\pi n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)}{2 (n!)}\, y^n$$
Replace $$y$$ by $$x-\frac{\pi }{6}$$.