$\cos(a)+\cos(b)-\cos(a+b)\geq 1$

Question

I am trying to prove that $$\cos(a)+\cos(b)-\cos(a+b)\geq 1$$ For $a,b \geq 0$ and $0\leq a+b\leq 180^°$

I have checked in Wolfram Alpha that the inequality is true, but I am not able to prove it. The trigonometric identitiy I have tried to apply (basically, $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b))$$ does not seem useful as it is, so if there is some other one that you think it could help... any hint would be welcomed.

Thanks!

Answer 1

$$\cos a + \cos b - \cos(a+b) - 1 = 2\cos\frac{a+b}{2}\cos\frac{a-b}{2} - 2\cos^2\frac{a+b}{2} = 2\cos\frac{a+b}{2}\left(\cos\frac{a-b}{2} - \cos\frac{a+b}{2}\right)\geq 0.$$ The thing in the parenthesis is non-negative because $\cos$ decreases on $[0,\pi/2].$

Answer 2

As an alternative, assume $a$ fixed and $b=x\in\left[0, \pi-a\right]$ then we have by convexity

$$f(x)=\cos(a)+\cos(x)-\cos(a+x)-1\ge 0$$

indeed $f(0)=f(\pi -a)=0$ and for $x\in\left(0, \pi-a\right)$ we have

$$f''(x) =\cos (a+x)-\cos x\le 0$$

since $\cos$ is decreasing in $[0,\pi]$.

Answer 3

We have: $\cos a + \cos b -\cos(a+b) - 1 = 2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right) - 2\cos^2\left(\frac{a+b}{2}\right)+1 -1= 2x\cos\left(\frac{a-b}{2}\right)-2x^2=f(x), x = \cos\left(\frac{a+b}{2}\right)$. We now show: $f(x) \ge 0$, or $2x^2 - 2x\cos\left(\frac{a-b}{2}\right) \le 0$. Observe that $0 \le x \le \cos\left(\frac{a-b}{2}\right)$, as a quadratic function in $x$, $f(x) \ge 0$.

Answer 4

If you expand $\cos(a+b)=\cos a\cos b-\sin a\sin b$, we can rewrite

$$\begin{array}{crcl} & \cos a+\cos b-\cos(a+b) & \le & 1 \\[2pt] \iff & \cos a+\cos b-[\cos a\cos b-\sin a\sin b] & \le & 1 \\[2pt] \iff & \sin a\sin b & \le & 1-\cos a-\cos b+\cos a\cos b \\[2pt] \iff & \sin a\sin b & \le & (1-\cos a)(1-\cos b) \\[3pt] \iff & 1 & \le & \displaystyle\frac{1-\cos a}{\sin a}\,\frac{1-\cos b}{\sin b} \\[3pt] \iff & 1 & \le & \,\,\tan(a/2)\,\tan(b/2) \end{array} $$

Since $0^\circ\le a/2+b/2\le90^\circ$ and $\tan$ is increasing on $[0^\circ,90^\circ]$, we can reason

$$ \tan(a/2)\tan(b/2) \,~\le~\, \tan(a/2)\tan(90^\circ-a/2) \,~=~\, 1, $$

so the claim is proved. (I guess handle the edge cases separately tho.)