Which is larger: \(\sin(\cos x)\) or \(\cos(\sin x)\)?
Does it depend on \(x\)?
Let’s start by thinking about what we know about the behaviour of \(\sin x\) and \(\cos x.\)
Both \(\sin x\) and \(\cos x\) have period \(2\pi\) so \(\sin(\cos x)\) and \(\cos(\sin x)\) will be periodic. The period of each function can’t be more than \(2\pi\) so we only need to consider what happens for \(x\) in the interval\([0,2\pi].\)
\(\sin x\) and \(\cos x\) vary between \(-1\) and \(1.\) What does this tell us about the minimum and maximum values of \(\sin(\cos x)\) and \(\cos(\sin x)\)?
For \(x\) in \(\big[0,\tfrac{\pi}{2}\big]\) and \(\big[\tfrac{3\pi}{2},2\pi\big]\)\(\sin x\) increases and for \(x\) in \(\big[\tfrac{\pi}{2},\tfrac{3\pi}{2}\big]\) it decreases. Where does \(\cos x\) decrease or increase between \(0\) and \(2\pi\)?
We know that \(\cos x\) is an even function and \(\sin x\) is an odd function, i.e. \(\cos (-x) = \cos x\) and \(\sin(-x)=-\sin x.\)
Special points
We’ll consider some multiples of \(\tfrac{\pi}{2}\) to get a sense of how the two functions behave. You may find it helpful to start marking some points on a sketch graph as you read.
When \(x=0, \pi, 2\pi\) the value of \(\cos(\sin x)\) is \(1\), which must be the maximum value of the function. This means the period of \(\cos(\sin x)\) could be \(\pi.\)
\(\sin(\cos 2\pi)=\sin(\cos 0) = \sin 1.\) The value \(\sin 1\) must be positive, but as \(\sin x\) does not reach \(1\) until \(x=\tfrac{\pi}{2}\), \(\sin(\cos 0) <1.\) Similarly \(\sin (\cos \pi)=\sin (-1) = -\sin 1 >-1.\)
\(\sin \tfrac{\pi}{2}=1\), so \(\cos (\sin \tfrac{\pi}{2})=\cos 1\), which is positive, but less than \(1.\) Also, \(\cos (\sin \tfrac{3\pi}{2})=\cos 1\) because of the period of \(\cos (\sin x).\)
\(\cos x=0\) when \(x=\tfrac{\pi}{2}\) and \(x=\tfrac{3\pi}{2}\) so \(\sin(\cos x)=0\) at these points.
This means that \(\cos(\sin x) >\sin(\cos x)\) at \(0,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2}\) and \(2\pi.\) The values of \(\cos 1\) and \(\sin 1\) seem to be important. Without using a calculator or graphing software, how could you decide which of these values is bigger?
Behaviour between special points
As \(\cos x\) is negative between \(\tfrac{\pi}{2}\) and \(\tfrac{3\pi}{2}\) and \(\sin x\) is negative for \(x\) between \(-1\) and \(0\) the value of the function \(\sin (\cos x)\) must be negative for \(x\) between \(\tfrac{\pi}{2}\) and \(\tfrac{3\pi}{2}.\)
What does the same reasoning tell us about \(\cos(\sin x)\)?
As \(\cos x\) is an even function and \(\sin x\) is odd, \(\cos (\sin (-x))=\cos (-\sin x)=\cos (\sin x)\) and \(\sin (\cos (-x))=\sin (\cos x)\), so both functions are even. As well as telling us that the \(y\)-axis is a line of symmetry of the graphs of \(y=\sin (\cos x)\) and \(y=\cos (\sin x)\), this also shows that these graphs are symmetric about the line \(x=\pi.\) (For example, \(\sin (\cos (\pi-x))=\sin (\cos (x-\pi)).\)) This helps because we are interested in what happens between \(0\) and \(2\pi.\)
For \(x\) between \(0\) and \(\tfrac{\pi}{2}\), \(\sin x\) increases from \(0\) to \(1\) and \(\cos x\) decreases, so \(\cos(\sin x)\) decreases from \(1\) to \(\cos 1\). On \(\big[\tfrac{\pi}{2},\pi\big]\), \(\sin x\) decreases from \(1\) to \(0\) and \(\cos x\) decreases from \(0\) to \(-1\). This means that \(\cos(\sin x)\)increases from \(\cos 1\) to \(1\) and therefore has a minimum when \(x=\tfrac{\pi}{2}.\) What happens between \(\pi\) and \(2\pi\)?
These ideas, can be used to show that \(\cos(\sin x)\) has a period of \(\pi\) and its range is \([\cos 1,1]\) whereas \(\sin(\cos x)\) has period \(2\pi\) and its range is \([-\sin 1, \sin 1].\)
For all the points considered, \(\cos(\sin x) >\sin(\cos x)\) and we now know a lot about the graphs \(y=\cos (\sin x)\) and \(y=\sin (\cos x).\) It seems reasonable to conjecture that \(\cos(\sin x) >\sin(\cos x)\) for all \(x\), but to convince ourselves that this is true, we need to show that the graphs do not cross anywhere.
The graphs can’t cross between \(\tfrac{\pi}{2}\) and \(\tfrac{3\pi}{2}\) because \(\sin(\cos x)\) is negative here, but what about between \(0\) and \(\tfrac{\pi}{2}\) or between \(\tfrac{3\pi}{2}\) and \(2\pi\)?
We want to know whether \[\cos (\sin x) = \sin (\cos x)\] has any solutions for \(x\) in \(\left[0, \tfrac{\pi}{2}\right].\) Now, if \(\cos a = \sin b\) then \(a=\tfrac{\pi}{2}-b\), so \(a+b=\tfrac{\pi}{2}\), or in our case \(\sin x+ \cos x = \tfrac{\pi}{2}.\)
The diagram shows part of a circle of radius \(1\) and an angle \(x\) between \(0\) and \(\tfrac{\pi}{2}.\) The length of arc \(APB\) is \(\tfrac{\pi}{2},\) but \(\sin x\) is smaller than the length of arc \(AP\) and \(\cos x\) is smaller than the length of arc \(PB.\)
Therefore \[\sin x + \cos x < \tfrac{\pi}{2}\] so \(\cos (\sin x) = \sin (\cos x)\) does not have any solutions in \([0, \tfrac{\pi}{2}]\). The symmetry of the graphs about \(x=\pi\), means that there aren’t any solutions in \([\tfrac{3\pi}{2},2\pi]\) either. Therefore \(\cos (\sin x)\) is larger than \(\sin (\cos x)\) for all real \(x.\)
Another approach is to use the inequality \(\sin \theta < \theta\) for positive \(\theta.\) To see how this inequality arises, consider the diagram below. This shows the chord \(AB\) of a circle subtending an angle \(\theta\) at the centre of the circle, where \(0<\theta < \pi.\)
If we compare the areas of the triangle \(OAB\) and the minor sector \(OAB,\) we get \(\tfrac{1}{2}r^2 \sin \theta < \tfrac{1}{2}r^2 \theta\) and therefore \(\sin \theta < \theta\).
We can combine this inequality with the increasing and decreasing behaviour of \(\sin x\) and \(\cos x\) on \(\big[ 0, \tfrac{\pi}{2}\big]\) to show that \[\sin (\cos x) < \cos x \quad \text{and} \quad \cos (\sin x) > \cos x\]
Therefore \(\cos x\) is always sandwiched between \(\sin (\cos x)\) and \(\cos (\sin x)\) for \(x\) between \(0\) and \(\tfrac{\pi}{2}.\) As above, the symmetry of the graphs completes the proof that \(\cos (\sin x)\) is larger than \(\sin (\cos x)\) for all real \(x.\)
