Prove that cos x = 1- (x²)/(2!)+
Prove that $\cos x = 1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots+(-1)^{n-1}\dfrac{x^{2n-2}}{(2n-2)!}+(-1)^n\dfrac{x^{2n}}{(2n)!}\cos\theta x$, $0\lt \theta\lt 1$.
Explanation of Method
Stopping the expansion one pair of terms early is exactly what turns the last coefficient into a cosine.
Step 1: Derivatives of $\cos x$ at $x=0$
The derivatives of $f(x)=\cos x$ cycle with period 4: $\cos x,-\sin x,-\cos x,\sin x,\ldots$, so
\[ f^{(k)}(0) = \begin{cases} (-1)^{k/2} & k\text{ even} \\ 0 & k\text{ odd.} \end{cases} \tag{1} \]
Step 2: Taylor's formula stopped one pair early
The key is to apply Taylor's formula at $a=0$ only through degree $2n-2$ — not $2n$ — and attach the Lagrange remainder at order $2n$. Discarding the (already zero) odd-degree terms:
\[ \cos x = \sum_{j=0}^{n-1}\frac{(-1)^j}{(2j)!}x^{2j} + R_{2n-1}(x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots+(-1)^{n-1}\frac{x^{2n-2}}{(2n-2)!}+R_{2n-1}(x). \tag{2} \]
Step 3: The Lagrange remainder at order $2n$
By the Lagrange form, with $\xi=\theta x$ for some $\theta\in(0,1)$: \[R_{2n-1}(x) = \frac{f^{(2n)}(\theta x)}{(2n)!}x^{2n}. \tag{3}\]
Since $2n$ is even, (1) gives $f^{(2n)}(x)=(-1)^n\cos x$, so directly
\[ R_{2n-1}(x) = (-1)^n\frac{\cos(\theta x)}{(2n)!}x^{2n}. \tag{4} \]
Step 4: Verification
\[n=3,\ x=0.6\text{: solving }\cos(0.6)=1-\tfrac{0.36}{2}+\tfrac{0.6^4}{24}-\tfrac{0.6^6}{720}\cos(0.6\theta)\text{ for }\theta\text{ gives }\theta\approx0.189\in(0,1)\ \checkmark\]
Result
Final answer:
\[\boxed{\,\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots+(-1)^{n-1}\frac{x^{2n-2}}{(2n-2)!}+(-1)^n\frac{x^{2n}}{(2n)!}\cos\theta x, \qquad 0<\theta<1\,}\]
Substituting (4) into (2) gives exactly the stated formula:
The two facts driving the proof: all odd-order derivatives of $\cos x$ vanish at $0$, and $f^{(2n)}(x)=(-1)^n\cos x$ is itself a cosine, so the Lagrange remainder at order $2n$ automatically takes the form of the $(-1)^n x^{2n}/(2n)!$ term with a $\cos(\theta x)$ in place of the missing $\cos0=1$.
Remark
Remark. An easy way to lose the thread here is to start Taylor's formula through degree $2n$ (rather than $2n-2$) and then hunt for a way to fold the resulting order-$(2n+1)$ or order-$(2n+2)$ remainder back into a cosine at order $2n$ — that path forces an awkward detour through the Cauchy remainder. Starting one pair of terms earlier, at degree $2n-2$, makes the order-$2n$ Lagrange remainder produce the cosine directly, with no extra manoeuvre.