Find the coordinates of the turning points on the curve \(y = \sin x + \cos x\) for \(0 < x < 2\pi.\)

Method 1; using calculus

The turning points of the curve occur where the gradient is zero.

It is possible for the gradient of the curve to be zero and for this not to be a turning point, if we have a point of inflection. A sketch of the graph indicates that we do not have any points of inflection.

Differentiating, we have \(\dfrac{dy}{dx} = \cos x - \sin x\). This is zero if and only if \(\sin x = \cos x\). This equation is the same as \(\tan x = 1\).

We divided by \(\cos x\). This cannot be zero because then \(\sin x\) would be non-zero, so this would not be a solution to the equation.

Thus the solutions we seek in the range are \(x = \dfrac{\pi}{4}\) and \(\dfrac{5\pi}{4}\).

By substituting we find the coordinates of the turning points are a maximum at \(\left(\dfrac{\pi}{4}, \sqrt{2}\right)\) and a minimum at \(\left(\dfrac{5\pi}{4}, -\sqrt{2}\right)\).

Method 2; using trigonometry

There is an alternative method that does not use calculus but relies on the trigonometric identity \[\sin(x+\alpha) = \sin x \cos \alpha + \cos x \sin \alpha.\]

This holds for all real values of \(x\) and \(\alpha\).

So we can say that for some value of \(R\) and \(\alpha\) \[R\sin(x+\alpha) = R\sin x \cos \alpha + R\cos x \sin \alpha = \sin x + \cos x.\] Comparing coefficients on the right hand side, we want \(R\cos \alpha =1\), \(R\sin \alpha = 1\).

Squaring and adding these equations gives us \(R^2 = 2 \implies R = \sqrt{2}\), and dividing the two equations gives us \(\tan \alpha = 1 \implies \alpha = \dfrac{\pi}{4}\).

So \[\sin x + \cos x = \sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right),\] and the turning points of the curve are where \(\sin\left(x+\dfrac{\pi}{4}\right)\) is \(1\) or \(-1\).

This happens when \(x = \dfrac{\pi}{4}\) (a maximum) or when \(x = \dfrac{5\pi}{4}\) (a minimum), and the full coordinates of the turning points are as before.

In this case, using calculus gives an easiest solution, but there are some questions that can more sensibly be tackled using the second method.