Problem: by finding the Taylor series about an appropriate point “a”, calculate an approximation for sin(3).

I understand that i need to use remainder, but i don't know what exact answer I should get


For any suitably differentiable function $f$, the Taylor Series of $f$ expanded at $a$ is defined to be

$$f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k=f(a)+f'(a)(x-a)+\frac 12 f''(a)(x-a)^2+\cdots$$

In this case, since you want to calculate the value of $\sin (3)$, it is more appropriate to expand at $a=\pi$ because $\pi$ is close to $3$, so we get

$$\sin (3)=\sum_{k=0}^\infty \frac{1}{k!}\biggl(\frac{d^k}{dx^k}\sin(x)\biggr)_{x=\pi}(3-\pi)^k$$

But observe that all the even terms in this sum are $0$ because $\frac{d^k}{dx^k}\sin(x)= \pm \sin (x)$ for even $k$, which evaluated at $\pi$ gives $0$.

On the other hand, for $k$ being odd, $\frac{d^k}{dx^k}\sin(x)= \pm \cos (x)$ and $\cos(\pi)=-1$. Hence


You can truncate this sum at anywhere to obtain an approximation. Truncating at a later place would obviously yield a more accurate approximation. eg. the first few approximations are:

$$\pi-3 \; \; ,\; \; -\frac{\pi^3}{6}+\frac{3\pi^2}{2}-\frac{7\pi}{2}+\frac 32 \; \; ,\; \; \cdots$$

If having your answer in terms of $\pi$ is not accepted, then expand about $0$ instead:


Giving the approximations

$$3 \; \; ,\; \; -\frac 32 \; \; ,\; \; \frac{21}{40} \; \; ,\; \; \frac{51}{560} \; \; ,\; \; \cdots $$


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