Things you might have noticed

Estimate the gradient of \(f(x) = \sin x\) at the points A, B and C.

A sine graph with three points labelled. A at a maximum, B on an x intercept and C between a minimum and the x axis

If we draw in a tangent line at A, we will see that the gradient is, or is close to, zero.

At B we have to do a little more work. If we draw a tangent line, then we can form a right-angled triangle to help us estimate a gradient.

a tangent is drawn on the sine curve at B with the tangent becoming the hypotenuse in a right angled triangle which will help us to estimate the gradient

Why has a triangle of this size been chosen? Does it make our estimation easier?

We know the maxima and minima of a sine graph are \(1\) and \(-1\), so we could estimate the vertical height of the triangle as about 3. The base of the triangle goes from the \(x\) value of the maximum to the \(x\) value of the minimum. What is this distance?

This will depend on the scale we have imagined on the \(x\) axis. Have we thought in radians or degrees? In radians the distance between the maximum and minimum point is \(\pi\), and in degrees it is \(180^{\circ}\). This gives us two very different possible values for the gradient.

Can you produce a sketch of the gradient function of \(f(x) = \sin x\) when

  1. \(x\) is in radians?

  2. \(x\) is in degrees?
  1. Depending on the accuracy of the method used, in radians we should get gradients of approximately \(0\), \(-1\) and \(0.7\) at points A, B and C. We can use the values and the symmetry of the function \(\sin x\), to work out the gradient at some other points, before sketching a graph. What function does the sketch look like?

  2. In degrees, the gradient at A is still zero, but is much smaller at B and C: around \(0.017\) at B, and around \(0.012\) at C. How is this gradient function similar to the function in radians? What is the link between them?

Thinking about radians and degrees

Although it is hard to see when the \(y\) axes both have the same scale, the gradient functions both have the same features. Why must this be the case?

In radians, the gradient function looks like a \(\cos\) graph, and in fact it is. If \(f(x) = \sin x\) then \(f'(x) = \cos x\) when \(x\) is in radians. To think more about why this is true, see Rotating derivatives.

In degrees, when we enlarge the scale of the \(y\)-axis then we see the familiar shape appearing.

zoomed in on the gradient function in degrees so the cosine shape of the graph can be seen

The gradient function of \(f(x) = \sin x\) is \(f'(x) = a\cos x\), where \(a\) is a value (quite a lot) smaller than \(1\). To think about what the value of \(a\) might be, consider how you would transform a sine graph with period of \(2\pi\) to a graph with period of \(360^{\circ}\).

It is important to reflect on the fact that the derivative of sine depends on the what unit our angles are in. We have seen that in degrees we get an additional scale factor appearing in the derivative which does not appear in radians. This is why when we do calculus with trigonometric functions we use radians and not degrees.