Show that $cos(x)=0$ has a solution in $(\sqrt{2},1.6)$ using only the series definition. How do I show that $$\cos(x) = \sum\limits_{k=0}^{\infty} \left(\frac{(-1)^k x^{2k}}{(2k)!}\right) \qquad = 0$$ has a solution in the interval $(\sqrt{2},1.6)$ without using any 'external' results?
My thought is to show that it's positive at $x=\sqrt{2}$ and negative at $x = 1.6$ and use the IVT.  I think I can show it's positive at $\sqrt{2}$ (by grouping terms into positive pairs), but I'm struggling to show it's negative (in the vicinity of) 1.6.
 A: The series is alternating and decreasing for $k\ge -\frac34+\frac14\sqrt{1+4x^2}$.  Therefore, for $x\in[\sqrt{2},1.6]$, the series decreases for $k\ge 1$ and we have
$$1-\frac12 x^2\le \cos x\le 1-\frac12 x^2+\frac1{4!}x^4 \tag 1$$
From $(1)$ we have
$$0\le \cos \sqrt{2}\le \frac16$$
and
$$-\frac{13}{1,875}\le \cos (1.6)\le -\frac{7}{25} $$
Since the series representation is continuous, then there exists a number $x_0\in [\sqrt{2},1.6]$ such that $\cos x_0 = 0$.  And we are done!
A: It's a little nasty... But you can show by hand that the first three terms of the sum added together will be negative at $x = 1.6 = \frac{8}{5}$. Then you can write $$\cos(1.6) = A_3+\sum_{k=3}^\infty \frac{(-1)^k64^{k}}{(2k)!\cdot  25^k}$$ where $A_3$ is the first three terms of the sum, summed together into a negative quantity. Now if we let $a_k = \frac{(-1)^k64^{k}}{(2k)!\cdot  25^k}$, can you show that $\left|a_{2m-1} \right| > \left|a_{2m} \right|$ when $m>1$? Considering the quantity $\frac{\left|a_{2m} \right|}{\left|a_{2m-1} \right|}$ should make this easy. Then you can group the rest of the sum as a series of negative quantities.
A: I'm not sure what you mean by 'external' results - what counts?


*

*If you believe that $| \frac{d^n}{dx^n} \cos x| \leq 1$ then you can use a Taylor polynomial error bound to see that $|\cos(1.6) - (1- \frac{1.6^2}{2!})| \leq \frac{1}{4!}|1.6 - 0|^4 \approx 0.273$. Since $1 - \frac{1.6^2}{2!} = -0.28$ that means that $-0.553 \leq \cos(1.6) \leq -0.007$, so $\cos(1.6) < 0$.

*If you understand the proof of the alternating series test, you can see that since $\frac{1.6^{2k}}{(2k)!}$ is decreasing for $k > 0$, it is sufficient to show that $S_3 = 1 - \frac{1.6^2}{2} + \frac{1.6^4}{4!} = -0.006933 < 0$ to see that $\cos(1.6) < 0$.

A: I think the IVT is the right way to look at this problem. Your next step is correct: try to find a value you can plug in for $x$ to make the expression positive and another that makes the expression negative. 
The series expression for cosine makes it especially convenient to carry out this plan. If you're able to show plugging in $\sqrt{2}$ makes cosine positive, that's great. Here's another way to get a positive value.
Observe that the terms of the series are monotone decreasing in absolute values for $-1 \leq x \leq 1$. Moreover, the series alternates, and its first term is positive. This implies $\cos(x) > 0$ for $-1 \leq x \leq 1$. 
The trickier part is finding $x$ such that $\cos(x) < 0$. However, observe that, since factorials grow extremely rapidly (super-exponentially), the terms of the series are eventually monotone decreasing.
Given this fact, I think your next goal should be to find $x$ such that:
\begin{align}
1 + \frac{-x^2}{2} < 0, \\
1 + \frac{-x^2}{2} + \frac{x^4}{24} < 0, \, \mathrm{and} \\
\frac{x^{2k}}{(2k)!} \geq \frac{x^{2(k+1)}}{(2(k+1))!}
\end{align}
for $k \geq 3$. Such an $x$ will satisfy $\cos(x) < 0$ (why?). 
There are details to fill in, but hopefully this helps. 
