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Show that $\sin (x) \leq x$ for $x \geq 0$ and $\sin (x) \geq x$ for $x \leq 0$. Also show that $\cos (x) \geq 1-\frac{1}{2} x^{2}$ for all $x \in \mathbb{R}$.
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Proof.

Consider $f(x)=\sin (x)-x$. Then $f^{\prime}(x)=\cos (x)-1 \leq 0$. This means that $f(x)$ is weakly decreasing on $\mathbb{R}$. Since $f(0)=0$, we have $f(x) \leq 0$ for all $x \geq 0$ and $f(x) \geq 0$ for all $x \leq 0$. Now consider $g(x)=\cos (x)-1+\frac{1}{2} x^{2}$. we have $g^{\prime}(x)=-\sin (x)+x=-f(x)$. Therefore $g^{\prime}(x) \geq 0$ for $x \geq 0$ and $g^{\prime}(x) \leq 0$ for $x \leq 0$. It follows that $g$ is weakly increasing for $x \geq 0$ and $g$ is weakly decreasing for $x \leq 0$. Since $g(0)=0$ we have $g(x) \geq 0$ for all $x \in \mathbb{R}$

by Diamond (71,587 points)

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