Prove that sin 1 > cos ( sin 1 )

[maverick280857]

Hi

This is a fairly straightforward problem but I want to do it using calculus. Here goes:

Prove that $\sin 1 > \cos(\sin 1)$.

This is what I've done (I've hardly done much):

Let $f(x) = \sin(\cos(\sin x) - \cos(\sin(\cos(x)))$. I also have to show that f(x) = 0 has exactly one solution in $[0,\pi/2]$. So anyway, f'(x) < 0 for all x in this interval which in particular means that f(0) > f($\pi/2$). But the first part requires showing that f(0) > 0. Can this be done by a purely calculus-based argument? It is easily done if we bend the inequality a bit and observe that it is true if $\pi/2 -1 <\sin 1$.

Thanks and cheers
Vivek

 

[maverick280857]

Hi all

This was buried deep inside the forum and I had to fish it out. Just a reminder asking for help with it...

Thanks and cheers
Vivek

 

[thepatientmental]

Hi Vivek,

Thank you for your approach to this problem. Your idea of using calculus to prove this inequality is a great approach. However, I believe there might be a small mistake in your definition of f(x). It should be f(x) = \sin 1 - \cos(\sin 1), not \sin(\cos(\sin x) - \cos(\sin(\cos(x))). With this correction, your proof is correct and I will explain it below.

First, let's show that f(x) > 0 for all x \in [0, \pi/2]. We have f(0) = \sin 1 - \cos(\sin 1) = 1 - \cos(\sin 1) > 0 since \sin 1 > 0 and \cos x \leq 1 for all x. Next, we have f(\pi/2) = \sin 1 - \cos(\sin 1) = 1 - \cos(\sin 1) > 0 since \sin 1 < 1 and \cos x \geq 0 for all x. Therefore, by the Intermediate Value Theorem, f(x) must have at least one root in [0, \pi/2]. But since f'(x) < 0 for all x \in [0, \pi/2], f(x) is strictly decreasing, meaning it can have at most one root. Therefore, f(x) = 0 has exactly one solution in [0, \pi/2].

Now, let's show that \sin 1 > \cos(\sin 1). Since f(x) = \sin 1 - \cos(\sin 1) has exactly one root in [0, \pi/2], this means that f(x) > 0 for all x \in [0, \pi/2) and f(x) < 0 for all x \in (\pi/2, \pi]. Therefore, f(1) > 0, which means that \sin 1 - \cos(\sin 1) > 0, or equivalently \sin 1 > \cos(\sin 1). This completes the proof.

I hope this helps. Keep up the good work with your calculus-based approach!