Revision:OCR Core 4 - Differentiating trigonometric functions

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1 1. Differentiating trigonometric functions

1. Differentiating trigonometric functions

The advantage of using radians

Until this point the use of degrees and radians is largely interchangeable, it is possible to use either, this is not the same for calculus involving trigonometric functions.

The derivative of several different functions can be based upon the derivative of a linked function using a rule (such as the chain rule, the product rule, or the addition rule) if there is a known identity regarding the linkage of one function to another.

A candidate at this point might realise that the derivative of tangent can be found from the derivatives of the sine and cosine functions, and the quotient rule, as:

$\tan{ \theta } \equiv \frac{ \sin{ \theta }}{ \cos{ \theta } }$ .

(Note that the derivative of cosine could be derived from that of the sine function, however it is simple to deal with both of these functions simultaneously due to their similarities).

In radians, the graphs of the sine and cosine functions both have gradient 1 when the graphs cross the x-axis, this is only true for radians, and is helpful in the foundations of calculus of trigonometric functions.

Consider $\frac{ \sin{ \theta }}{ \theta }$ for small $\theta$ .

As the value of $\theta$ tends to 0, this expression tends to 1.

Hence:

$\lim _{ \theta \to 0 } \frac{ \sin{ \theta } }{ \theta } = 1$ .

Some inequalities and limits

Consider a sector of a circle whose angle at the centre of the circle is less than $\frac{ \pi }{2}$ . Let the centre of the circle be "O", and the two other corners of this sector be "A", and "B". If "A", and "B" are directly joined then the area of the triangle OAB is less than the area of the sector of the circle. This is because the straight line will always be the shortest distance between two points, and therefore the triangle will have one boundary which is within the sector, and two other boundaries that are shared. If OA is extended such that a tangent to the circle at "B" will meet the line (let the intersection be at "C"), then as the line OB is a radius, the triangle OBC is right-angled (due to the radius meeting a tangent to the circle being at right-angles to it). Hence, the area of this triangle is greater than that of the area of the sector. The above information can be summed up (for a circle of radius "r"):

$\frac{r^{2}}{2} \sin{ \theta } < \frac{r^{2}}{2} \theta < \frac{r^{2} \tan{ \theta } }{2}$ .

Hence:

$\sin{ \theta } < \theta < \tan{ \theta }$ .

The whole idea would also suggest that for small $\theta$ :

$\sin{ \theta } \approx \theta \approx \tan{ \theta }$ , hence:

$\sin{ \theta } \approx \theta, \ and \ \tan{ \theta } \approx \theta$ .

This supports the idea about $\frac{ \sin{ \theta } }{ \theta }$ which was formed before, however, it can now be extended:

$\frac{ \sin{ \theta } }{ \theta } < 1$ .

Consider the identity: $\tan{ \theta } \equiv \frac{ \sin{ \theta } }{ \cos{ \theta }}$ :

$\frac{ \sin{ \theta }}{ \cos{ \theta }} > \theta$

Hence, over the domain of $0 < \theta < \frac{ \pi }{2}$ (as the sector had this domain):

$\frac{ \sin{ \theta }}{ \theta } > \cos{ \theta }$

Hence:

$\cos{ \theta } < \frac{ \sin{ \theta }}{ \theta } < 1$ .

(This can be observed in the graphs of these functions over the aforementioned domain. (Indeed an extended domain can be used:

$- \frac{ \pi }{2} < \theta < \frac{ \pi }{2}$

$\cos{ \theta } \equiv \cos{ - \theta }$

Therefore there is no change in the cosine element.

$\frac{ \sin{ - \theta } }{ - \theta } \equiv \frac{ - \sin{ \theta }}{ - \theta } \equiv \frac{ \sin{ \theta }}{ \theta }$ .

Therefore the sine element is not changed.

Overall there is no change in sign, and therefore the inequality holds for this domain.

These ideas can be used to demonstrate the already given property of the sine graph (that it has gradient 1 at the intersection with the x-axis).

The intersection is at (0, 0). Suppose there is some point $\left( \theta , \ \sin{ \theta } \right)$ .

The gradient of the line joining these points is:

$\frac{ \sin{ \theta } }{ \theta }$ .

As the gradient at a point is taken over an infinitesimally small range of "x", the limit can be used:

$\lim _{ \theta \to 0 } \frac{ \sin{ \theta }}{ \theta } = 1$ .

Hence the gradient at the intersection with the x-axis is 1.

Derivatives of sine and cosine functions

Consider the idea presented in Core 2 of the trigonometric functions being based upon a circle, centre the origin, of unit radius.

Let there be some point "A" which is on the circumference of the circle, and forms an angle of $\theta$ anti-clockwise from the x-axis.

A small increase in this angle ( $\delta \theta$ ) would result in a point "B" on the circumference.

Take a line horizontally from "A", and vertically from "B", these lines intersect forming a right-angled triangle depicting the corresponding changes in "x", and "y" of this increase in the angle.

The point "A" has the coordinates $\left( \cos{ \theta } , \ \sin{ \theta } \right)$ , and the point "B" has the coordinates $\left( \cos{ \left( \theta + \delta \theta \right) } , \ \sin{ \left( \theta + \delta \theta \right) } \right)$ .

Now consider the triangle which was described earlier as having the change in "x", and the change in "y" of the change in the angle.

As this circle is of unit radius, the arc length between the points A, and B, is $1 \times \delta \theta = \delta \theta$ .

Let the angle between the horizontal, and the line between A, and B, be $\phi$ .

Hence expressions for $\delta x$ , and $\delta y$ can be formed:

$\delta x = AB \cos{ \phi }$

$\delta y = AB \sin{ \phi }$ .

Now consider:

$\lim _{ \delta \theta \to 0 } \frac{ \delta x }{ \delta \theta } = \frac{ dx}{d \theta }$ , and:

$\lim _{ \delta \theta \to 0 } \frac{ \delta y }{ \delta \theta } = \frac{ dy}{ d \theta }$ .

$\frac{ \delta x }{ \delta \theta } = \frac{ AB \cos{ \phi } }{ \delta \theta } = \cos{ \phi } \times \frac{ AB }{ \delta \theta }$

$\frac{ \delta y }{ \delta \theta } = \frac{ AB \sin{ \phi } }{ \delta \theta } = \sin{ \phi } \times \frac{ AB}{ \delta \theta }$ .

Now it is necessary to consider the idea of a sector of a circle. Let the circle have centre "O", and radius "r"; let there be two points, "A", and "B" which are joined by the circumference, and also a straight line. This sector is like a doubling of the one mentioned when deducing the initial inequalities and limits.

Consider $\frac{ line \ AB }{ arc \ AB }$ .

$\frac{ line \ AB }{ arc \ AB } = \frac{ 2r \sin{ \theta } }{ 2r \theta } = \frac{ \sin{ \theta }}{ \theta }$ .

Hence, as the angle of a sector of a circle tends to 0, the length of the arc tends towards the length of the line joining the corners.

This can be applied to the calculations of $\frac{ \delta x }{ \delta \theta }$ , and $\frac{ \delta y }{ \delta \theta }$ , as $\delta \theta$ is the arc length between "A", and "B", and "AB" is the line between them, hence:

$\lim _{ \delta \theta \to 0 } \frac{ \delta x }{ \delta \theta } = \cos{ \phi }$ , and

$\lim _{ \delta \theta \to 0 } \frac{ \delta y }{ \delta \theta } = \sin{ \phi }$ .

Also, as the change in the angle tends to 0, the points "A", and "B" become a single point, and therefore $\lim _{ \delta \theta \to 0 } \phi = \theta + \frac{ \pi }{2}$ .

Hence:

$\frac{dx}{d \theta } = \cos{ \left( \theta + \frac{ \pi }{2} \right) }$

$\frac{dy}{d \theta } = \sin{ \left( \theta + \frac{ \pi }{2} \right) }$

Hence:

$\frac{dx}{d \theta} = - \sin{ \theta }$

$\frac{dy}{ d \theta} = \cos{ \theta }$

Consider that the change in "y" on this unit circle is the same as the change in the sine of an angle, and the change in "x" is the same as the change in the cosine of this angle:

$\frac{d}{d \theta } \sin{ \theta } = \cos{ \theta }$

$\frac{d}{d \theta } \cos{ \theta } = - \sin{ \theta }$ .

Working with trigonometric functions

The afore-deduced results can be seen on the graphs of the functions. The intersection of the graph of the sine function (in radians) with the x-axis has gradient of 1, and at that point (x = 0), the cosine graph has a value of "1".

Interestingly, these functions will "loop around":

$f(x) = \sin{x}$

$f^{'} (x) = \cos{x}$

$f^{''} (x) = - \sin{x}$

$f^{'''}(x) = - \cos{x}$

$f^{''''}(x) = \sin{x}$

Now that the derivatives of the sine, and cosine functions are known, the tangent function can have it's derivative calculated:

$\tan{ \theta } \equiv \frac{ \sin{ \theta }}{ \cos{ \theta }}$

Hence:

$\frac{d}{d \theta } \tan{ \theta } = \frac{ \left( \cos{ \theta } \times \cos{ \theta } \right) - \left( \sin{ \theta } \times \left( - \sin{ \theta } \right) \right) }{ \cos ^{2} { \theta } } = \frac{1}{ \cos ^{2} { \theta } } = \sec ^{2} { \theta }$ .

Other useful derivatives can be found:

$\frac{d}{d \theta } \sec{ \theta } = \frac{d}{d \theta } \frac{1}{ \cos{ \theta }} = \frac{ - \left( - \sin{ \theta } \right) }{ \cos ^{2} { \theta } } = \sec{ \theta } \tan { \theta }$ .

Some applications

Due to the periodic nature of the trigonometric functions they will be involved in mathematical models for things such as the motion of a pendulum, or the level of water in a tidal system. Questions can be asked regarding different periodic motions, however they are always merely questions of finding the rate of change, or the minima and maxima (note that often there will be a domain attributed to some of the models).

Integrating trigonometric functions

The basic results obtained previously can give certain results for integration:

$\int \cos{x} \,dx = \sin{x} + C$

$\int \sin{x} \,dx = - \int - \sin{x} \,dx = - \cos{x} + C$ .

Also:

$\int \sec ^{2} {x} \,dx = \tan{x} + C$ .

Consider the following:

$\frac{d}{dx} \ln | \sec{x} |$ .

$\sec{x} > 0 \implies | \sec{x} | = \sec{x}$

$\frac{d}{dx} \ln \left( \frac{1}{ \cos{x} } \right) = \left( \frac{1}{ \cos{x} } \right) \times \left( \sec{x} \tan{x} \right) = \tan{x}$

$\sec{x} < 0 \implies | \sec{x} | = ( - \sec{x} )$

Hence:

$\frac{d}{dx} \ln \left( \frac{1}{ - \cos{x} } \right) = \left( \frac{1}{ - \cos{x} } \right) \times \left( - \sec{x} \tan{x} \right) = \tan{x}$

Hence:

$\frac{d}{dx} \ln | \sec{x} | = \tan{x}$

Hence:

$\int \tan{x} \,dx = \ln | \sec{x} | + C$ .

The idea developed in Core 3 to extend integrals can be used:

$\int \sin{nx} \,dx = - \frac{1}{n} \cos{nx} + C$ .

Several identities are useful in the integration of more complicated trigonometric functions, namely:

$2 \sin{A} \cos{A} \equiv \sin{ 2A }$

$2 \cos ^{2} {A} \equiv 1 + \cos{ 2A }$

$2 \sin ^{2} {A} \equiv 1 - \cos{ 2A }$ .

Example

1. Calculate $\int \cos ^{3} {x} \,dx$ .

Begin with trigonometric manipulation:

$\cos ^{3} {x} = \left( 1 - \sin ^{2} {x} \right) \cos{x} = \cos{x} - \cos{x} \sin ^{2} {x}$

Consider:

$\frac{d}{dx} \sin ^{3} {x} = 3 \sin^{2} {x} \cos{x}$

Hence:

$\frac{1}{3} \sin ^{3} {x} + C = \int \sin ^{2} {x} \cos{x} \,dx$

Hence:

$\int \cos ^{3} {x} = \sin{x} - \frac{1}{3} \sin ^{3} {x} + C$ .

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Revision:OCR Core 4 - Differentiating trigonometric functions

Contents

1. Differentiating trigonometric functions

The advantage of using radians

Some inequalities and limits

Derivatives of sine and cosine functions

Working with trigonometric functions

Some applications

Integrating trigonometric functions