# Using Mean Value Theorem to solve an inequality involving $\cos^{-1}(x)$

Question:

Using the Mean Value Theorem, show that for all $$-\frac{1}{2}\lt a,b \lt \frac{1}{2}$$ with $$a\lt b$$

$$\lvert \cos^{-1}(a)-\cos^{-1}(b)\rvert \lt \frac{2\sqrt{3}}{3}\lvert a-b\rvert$$

My attempt:

Let $$f(x)= \cos^{-1}(x)$$

$$\cos^{-1}(x)\ \text{is continuous and differentiable on the interval so by the MVT,}$$

$$\lvert \cos^{-1}(b)-\cos^{-1}(a) \rvert= \lvert b-a\rvert.\lvert f'(c)\rvert$$

$$\lvert f'(c) \rvert= \frac{1}{\sqrt{1-c^2}} \lt \frac{2\sqrt{3}}{3}$$

Therefore we have,

$$\lvert \cos^{-1}(a)-\cos^{-1}(b) \rvert= \lvert a-b\rvert.\lvert f'(c) \rvert \le 1.\lvert a-b\rvert \lt \lvert b-a\rvert \frac{2\sqrt{3}}{3}$$

Because $$\lvert a-b\rvert= \lvert b -a\rvert$$

Would this be correct?

No, this wouldn't be correct. The equality you wrote from the MVT is not right; if you take absolute values, then you must also have $|f'(c)|$ in the RHS; and just because $f'(c) < 1$ it doesn't mean its absolute value is also $<1$.

• I have now edited it Commented May 24, 2016 at 14:48
• You still don't have $|f'(c)|$ in the RHS right at the top, so even if the rest is correct, this step breaks the whole proof. Also, what's tan(a) and tan(b) doing there? What's tan got to do with anything?
– user325968
Commented May 24, 2016 at 14:49
• I have just changed that, possibly refresh the page? Commented May 24, 2016 at 14:50
• Is it now correct? Commented May 24, 2016 at 14:53
• Not yet. $c^2$ can be as high as $1/4$, so $\dfrac 1 {\sqrt{1-c^2}}$ could be as high as $\dfrac {2\sqrt 3} 3$, where did $\sqrt{15}$ come from?
– user325968
Commented May 24, 2016 at 15:11

$|f'(c)| = \dfrac{1}{\sqrt{1 - c^2}} < \dfrac{1}{\sqrt{1 - (1/2)^2}}$ since $c^2 < (1/2)^2$.

• But c can equal 0 can it not? Commented May 24, 2016 at 14:47
• Yes, yet $c$ could be $1/4$, in which case $|f'(1/4)| = \dfrac{1}{\sqrt{1 - (1/4)^2}} = \dfrac{4}{\sqrt{15}} > 1$. Commented May 24, 2016 at 14:55
• I see! Thanks for that Commented May 24, 2016 at 14:57
• So how would I conclude the final answer? Could I simply have $\frac{4}{\sqrt{15}} \lt \frac{2\sqrt{3}}{3}$ Commented May 24, 2016 at 14:59
• Not quite like that. Instead, we know that $-1/2 < c < 1/2$, so $c^2 < 1/4$. Therefore $1 - c^2 > 3/4$ or rather $\sqrt{1 - c^2} > \frac{\sqrt{3}}{2}$. Then we see that $\dfrac{1}{\sqrt{1 - c^2}} < \frac{2}{\sqrt{3}}$ as desired. Commented May 24, 2016 at 15:03