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It might be helpful to note that 1 < \frac {\pi} 2.
From your sketches, you should be able to see that if you can prove the result for x \in [0, \pi] then it will be true for all x. (You might like to compare your sketches with these graphs).
You should have already shown that \cos(\sin x) > \sin(\cos x) when x=0, \frac{\pi} 2 and \pi.
Can you use your sketches to help you explain why \cos(\sin x) > \sin(\cos x) when x \in \left(\frac{\pi} 2, \pi \right)?
The argument for x\in \left( 0, \frac{\pi} 2\right) is a bit harder. One way to approach this is to try and prove the stronger result \cos(\sin x) > \cos x > \sin(\cos x) for x\in \left( 0, \frac{\pi} 2\right) instead.
You might like to sketch y=\cos x alongside the other two graphs using Desmos to convince yourself this is true before trying to prove it!
Method 1: Calculus and increasing/decreasing functions
Lemma: x > \sin x for x\in \left( 0, \frac{\pi} 2\right).
We know that \cos x is a decreasing function for x\in \left( 0, \frac{\pi} 2\right). Explain why this, along with the lemma result, means that we have \cos(\sin x) > \cos x.
Explain why the lemma means that we have \cos x > \sin(\cos x).
A lemma is an intermediate theorem in the proof of another theorem.
Method 2: Geometry
Imagine a quarter-circle where the radius is equal to 1.
The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?
Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.