Prove the inequalities $1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$ [closed]

$1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$

I think I should check the two sides. but how can I show that $1-\frac{x^2}{2}- \cos(x)\le 0$? And on the other side $0\le-\cos x+1-\frac{x^2}{2}+\frac{x^4}{24}$

Solution 1:

This is a straight application of the Leibniz rule for the convergence of an alternating series.

The cosine series is alternating and for all alternating series you have that the partial sums are alternating upper and lower bounds for the value of the series.

If $a_n$ is decreasing to zero and $s_n=\sum_{k=0}^n(-1)^ka_k$ then $$ s_{2m+1}\le s_\infty\le s_{2n} $$ for all $m,n\in\Bbb N$.

Here $a_n=\frac{x^{2n+2}}{(2n+2)!}$, so $s_\infty=1-\cos x$ and this sequence is decreasing for $x^2<12$.

Solution 2:

By Taylor-Lagrange, there is $\theta_x\in (0,1)$ s.t. $$\cos(x)=1-\frac{x^2}{2}\cos(\theta_x x).$$ Since $-\cos(\theta_x x)\geq -1$, we get $$\cos(x)\geq 1-\frac{x^2}{2}$$

In the same way, there is $\eta_x\in (0,1)$ s.t.

$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}\cos(\eta_x x).$$

Since $\cos(\eta_x x)\leq 1$, we get $$\cos(x)\leq 1-\frac{x^2}{2}+\frac{x^4}{24}.$$