Lower bound for $2+ \cos(t)$

Question

I was doing this question on convergence of improper integrals where in our book they have used the fact that $2+ \cos(t) \ge1$. Can somebody prove this?

The minimum value for $\cos t$ is $-1$, so $2+\cos t$ has minimum of $1$ — lEm, Apr 15 '16 at 08:39

score 2 · Answer 1 · answered Apr 15 '16 at 08:45

2

Since $\cos (t) \ge -1$, we have $2 + \cos (t) \ge 2 - 1 = 1$.

answered Apr 15 '16 at 08:45

MathMajor

6,478

score 2 · Answer 2 · answered Apr 15 '16 at 08:45

2

More formally:

$$\begin{array}{r c l} -1 \le & \cos t &\le 1,\quad \forall t\\ 2 - 1 \le & 2+\cos t &\le 2+ 1\\ 1 \le& 2+\cos t &\le 3 \end{array} $$

answered Apr 15 '16 at 08:45

thanasissdr

6,348

Lower bound for $2+ \cos(t)$

2 Answers2