The Trigonometric Functions
Radian angle measure
Contents
To study the calculus relating to the trigonometric functions, we must firstly study the radian as a measure of angle. The radian as a unit for angle measure is more suitable when considering the trigonometric functions. We shall now study this measure of angle.
Definition: A radian is the size of the angle subtended at the centre of a circle by an arc of length equal to the radius of the circle. This situation is depicted on the diagram below.
Now, to measure the number of radians in one revolution we simply measure the number of radius lengths within the circumference of the circle. To do so we have,
as the number of radius lengths within the circumference of the circle.
Hence we have that the number of radians within a revolution is . Hence we have that,
Hence we have that,
The following common angle conversions should be known by heart.
Radians  
Degrees 
To convert other angles, we simply form an equation using ratios. It should be obvious that the ratio of any angle in radians over is equal to the ratio of that angle in degrees over . From this, the unknown is left as some pronumerals and the resultant equation is solved to obtain the required value. Consider the below example to illustrate the conversion of angles.
Example 1a) Convert into a radian measure. b) Convert into degree measure. Solution 1a) We simply create an equation using the equality of ratios as explained above.
Hence we have that radians.
b) Again, we simply create a ratio to find the angle in question.
Hence we have that 
Note: When degrees are intended, the degree sign is included. For radians, at times the angle measure is followed by the word “rad”, but most of the time an angle in radians is left without any units following after.
Note: The reason for the use of the radian angle of measure is because functions such as do not make physical sense when the domain is in degrees. Also, the degree unit of measure is based on the angle subtended by the Earth at the sun after one day (the sun is the centre of the circle made by the Earth’s orbit). Obviously mathematics is universal and is not to be restricted only to our solar system, hence making radians a more universal angle measure and more suitable for mathematics.
Arc length and sector area
Within a circle of radius we have that the following formulae are true.
where is measured in radians.
The first formula gives the length of an arc. The second formula gives the area of the sector of the circle. The most important aspect here is that the angle must be measured in radians.
Now at this point, we can extrapolate the second formula to obtain a formula for the area of the minor segment subtended by the angle. Consider the shaded region below which is the minor segment subtended by the angle at the centre.
Now, obviously the area of this minor segment is equal to the area of the sector subtended by the angle minus the area of the isosceles triangle subtended within the sector. That is,
Since we have that the isosceles triangle has two sides, each of length about an angle of size radians. Hence we have that the area of the minor segment is given by the formula:
where is measured in radians. Be sure to convert your calculator into radian mode to ensure that the correct value of is obtained.
Note: Ensure you understand how to convert your calculator from the degree mode of angle measure to the radian mode of angle measure. Use the latter case when calculating trigonometric ratios of angles in radians.
Example 2An arc of a circle with 8 cm subtends an angle of at the centre. Find the length of the arc and the area of the sector. Solution 2So, to find the length of the arc and the area of the sector, we firstly must convert degrees into radians. We have that converts to radians. Now, using the formula for arc length we have,
To find the area of the sector we simply substitute into the formula.
Do not forget to add the appropriate units. 
Example 3In a circle with radius m, find the area of the minor segment subtended by an angle of radians at the centre. Solution 3To find the area of the minor segment, we simply use the formula derived earlier.
Now, we have that m and radians. Substituting into the formula gives,

Example 4A sector of a circle has an area of . If the angle at the centre is radians, find the radius of the circle correct to two significant figures. Solution 4We simply use the formula for the area of a sector to calculate the radius.
Now, substituting the respective values gives,
Hence the radius of the circle is mm correct to two significant figures. 
Graphs of the basic Trigonometric Functions
You should be familiar with the graphs of the trigonometric functions with the measure of the angle being the degree. We shall now investigate these graphs, using radians as our angle measure. Recall that the trigonometric functions are periodic, which implies that for a function there exists such that . We firstly note that the period of a function is the value of in the above identity. That is, it is simply the difference between functional values that results in the functional values repeating themselves. Also, the amplitude of a periodic function is the maximum value of for all within the domain. Below every curve is a description of the most important facts relating to it.
This curve has zeros at the values where is an integer. The period of revolution is . The amplitude of the curve is .
This curve has zeros at where is an integer. The period of revolution is . The amplitude of the curve is 1.
The curve has zeros at where is an integer. The curve has asymptotes at where is an integer. The period of the function is radians.
The curve has no zeros. The period of the curve is . The curve has asymptotes at where is an integer.
The curve has no zeros. The period of the curve is . Asymptotes are at where is an integer.
The zeros of the curve occur at where is an integer. The asymptotes occur at where is integral. The period of the function is .
More difficult graphs
The Advanced Mathematics syllabus may also ask students to sketch variations (translations and stretches) of the two main trigonometric graphs; and .
In general for the curve , the point is translated to the point . The amplitude of the curve is and the period of the curve is .
In general for the curve , the point is translated to the point and the origin is converted to the point . As before the amplitude of the curve is and the period curve is .
These facts should be memorised. There are also other methods of sketching trigonometric curves (such as plotting points) which may complement the methods taught here.
Note: Almost all questions involving sketching will indicate the domain to be sketched.
Consider the below examples that show utilisation of these formulae.
Example 5Sketch the graph of for . Solution 5Here we firstly notice that the domain we are required to sketch the curve for is . Now, the origin is translated to the point . The amplitude of the curve is and so the maximum and the minimum values of the function are and respectively. The period of the curve is and hence the curve repeats itself every radians. Since the four phases of the graph (equilibrium, minima, equilibrium, maxima) all occur within the period of radians, then it follows that we must consider the graph in intervals of from the translated origin. Also, as for any curve sketching question, we find the intercept. At , . Hence the intercept is . Also, to find the intercepts, we can simply solve the equation for . However, since we previously found the minima of the function to be zero, it thus follows that the intercepts will show themselves once the curve has been plotted. So, using the above information and extrapolating the curve gives the diagram, Which is the graph of for lieing between and . 
Note: It is very easy to become confused when solving questions involving trigonometric functions. To help with the sketching of the curve, plotting a few easy points helps. In the above case, noting the value at and will help with sketching the curve.
The next example illustrates the use of the formulae related to the cosine function.
Example 6Sketch the graph of for . Solution 6Firstly, we notice that the point is translated to the point . Secondly, the period of the graph is and the amplitude of the graph is . The equilibrium position is the line . Finding the exact intercepts is not important in this case since finding them will involve approximations. Now, using this information to sketch the graph gives, 
Differential Calculus involving the Trigonometric Functions
We shall now study the differential calculus of the trigonometric functions. We shall not show the first principles derivation of these differentials; rather we shall present them for the student to memorise. The reason for not showing a derivation is due to the requirement of several results outside the scope of the advanced mathematics course.

The three results shown here must be memorised. The differentials of the other trigonometric functions may be obtained using the chain rule and quotient rules.
Applying the chain rule to the above differentials gives the following results.

We shall now apply these results to some example questions.
Example 7Differentiate the following functions with respect to .
a)
b)
c)
d)
e)
f)
Solution 7a)
b)
c)
d)
e)
f)

As exhibited in the above example, all the rules that previously applied for differentiation still apply for differentiation (i.e. the chain rule, the product rule, the quotient rule). You also must be able to use the derivative to find gradients and equations of tangents and normals and also maxima and minima.
Example 8Find the equation of the tangent to at the point where . Solution 8So firstly we differentiate the function to obtain the gradient at .
At , . Hence we have the tangent at is horizontal. Hence we have that the equation of the tangent is . That is the equation is,

Example 9Show that satisfies
Solution 9So we simply differentiate the expression twice and show that it satisfies the relation.
Now, looking at the LHS of the above expression;
Hence we have that the function satisfies the required relation. 
Integral Calculus involving the Trigonometric Functions
As differentiation is the reverse process of integration, we can simply derive the integrals of the trigonometric functions by applying the fundamental theorem of calculus. The following results should be committed to memory.

The following three rules are given in the HSC Mathematics Standard Integrals table. These do not have to be committed to memory.

Now, the more general forms of these rules are;

All formulae for finding volumes, areas and definite integrals apply to the trigonometric functions as well. We shall now investigate some examples of such questions.
Example 10Find the integral of . Solution 10We are required to find Now, since the argument of the sine function is a linear form, then it follows that we may use the rules provided in the Standard Integral table. This gives,
Of course, the answer should be checked by differentiating the RHS and ensuring that the argument of the integrand is obtained. 
Example 11Find the definite integral
Solution 11To find this integral, we simply find an antiderivative and then use the fundamental theorem of calculus as required.
Noting here that . 
Example 12Find the area under the curve between and . Solution 12Firstly, we give a rough sketch of the curve. The amplitude is and the period is . Hence we have the curve; The shaded region is the area required by the curve. Hence we have;

Example 13Differentiate and hence find the value of
Solution 13So firstly we have that,
After applying the chain rule. That is we have that,
Integrating both sides with respect to gives (applying the fundamental theorem of calculus),
That is,
Hence we can now evaluate the definite integral.

Example 14a) Find the primitive of using the identity
b) Hence find the volume formed when the area underneath the curve between and is rotated about the axis, correct to 2 decimal places. Solution 14a) So, to find the primitive of we simply use the identity provided to rewrite in terms of .
b) So, using the formula for volume,
