How to show that $\cos(x_0)\leq \cos(x)+(x-x_0)\left[\sin(x_0)+\frac{(x-x_0)}{2}\right]$?

Question

How to show that $$\cos(x_0)\leq \cos(x)+(x-x_0)\left[\sin(x_0)+\frac{(x-x_0)}{2}\right]$$ where $x$ is a variable and $x_0$ constant?

Answer 1

HINT

Use Taylor's expansion for $\cos x$ at $x=x_0$.

Answer 2

Hint. Consider the function $$f(x)=\cos(x)+(x-x_0)\left[\sin(x_0)+\frac{(x-x_0)}{2}\right]-\cos(x_0).$$ Then $$f'(x)=-\sin(x)+\sin(x_0)+(x-x_0)\quad \text{and}\quad f''(x)=-\cos(x)+1\geq 0$$ and we may conclude that $f$ is convex in $\mathbb{R}$. Now note that $f(x_0)=0$, $f'(x_0)=0$. Show that $f(x)\geq 0$ that is $f$ has a minimum point at $x_0$.