### Calculus of Trigonometry & Logarithms

Problem requiring decisions

## Things you might have noticed

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

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.

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?

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.
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 $360^{\circ}$.