Your revision and your exam period is stressful time for most students! This App contains practical and powerful stress-busting strategies to keep you calm and composed so that you deliver your best work in the exam.
Graphs of trigonometric functions
The graphs of the three major functions are very important and you need to learn the characteristics of each.
The sine function
- This graph is continuous (there are no breaks).
- The range is -1 ≤ sin θ ≤ +1.
- The shape of the graph from θ = 0 to θ = 2π is repeated every 2π radians.
- This is called a periodic or cyclic function and the width of the repeating pattern that is measured on the horizontal axis, is called the period. The sine wave has a period of 2π, a maximum value of +1, and a minimum value of -1.
- The greatest value of the sine wave is called the amplitude.
The cosine function
- This graph is continuous.
- The range is -1 ≤ cos θ ≤ +1.
- It has a period of 2π.
- The shape is the same as the sine wave but displaced a distance of π ⁄ 2 to the left on the horizontal axis. This is called a phase shift.
The tan function
The tan function is found using:
It therefore follows that tan θ = 0, when sin θ = 0, and tan θ is undefined when cos θ = 0.
1. This graph is continuous, but is undefined when
2. The range of values for tan θ is unlimited.
3. It has a period of π.
All of the three functions periodically repeat their values and the simplest way to learn this, is to make sure that you understand the general rules below which use 'n' to represent any integer (i.e. any whole number, both positive and negative).
(Remember: nπ means "every 180 degrees", and 2nπ means "every 360 degrees".)
Sin curves |
---|
sin θ = 0 when θ = nπ |
sin θ = 1 when θ = 2nπ + π ⁄ 2 |
sin θ = −1 when θ = 2nπ − π ⁄ 2 |
Cos curves |
---|
cos θ = 0 when θ = 2nπ |
cos θ = 1 when θ = (2n + 1) π ⁄ 2 |
cos θ = −1 when θ = (2n + 1) π |
Tan curves |
---|
tan θ = 0 when θ = nπ |
tan θ = ± ∞ when θ = (2n + 1) π ⁄ 2 |
A trigonometric equation contains at least one trigonometric function, and when asked to solve the equation we must find the angle(s) for which it is valid.
We are normally required to find particular values of θ in a given interval.
Example:
Solve the equation cos θ = 0, for −π ≤ θ ≤ +π.
The finite solution set is θ = − π ⁄ 2 and π ⁄ 2.
There are two methods to find the solution of a trigonometric equation:
- Use the graph of the trigonometric functions.
- Use the four quadrants of the coordinate grid.
The first step in both cases is to find the principal value, (or PV of θ which is the value you get from the calculator).
Principal values for sin θ
Any equation for sin θ = S for the domain
has one solution in this interval called the principal value of θ.
It is in the first or fourth quadrant.
The range is shown in the diagram.
Principal values for cos θ
Any equation cos θ = C for the domain [0, π], has one solution in this interval called the principal value of θ.
It is in the first or second quadrant.
The range is shown in the diagram.
Principal values for tan θ
All the possible values for tan θ = T occur in the interval
The one solution in this interval called the principal value of θ.
It is in the first or fourth quadrant.
The range is shown in the diagram.
Each trig. function has two solutions in a 360^{0} or 2π interval. The first solution is the principal value, the other solution is called the secondary value, (SV), and lies in a different quadrant.
This can be found by drawing the graph or using the four quadrants of the coordinate grid as follows.
Solve tan θ = 2, −π ≤ θ ≤ +π
The first solution, (or principal value) is found using the calculator.
θ = 1.11 (3 sf)
The second solution, (or secondary value), is found using the fact that tan is also positive in the third quadrant (it repeats every θ radians).
θ = π + 1.11 = 4.25 (3 sf)
General Solutions of Trigonometric Equations
A general solution refers to all angles that satisfy the equation. So, it is an infinite set of angles. This is the same as before but we have to remember the period of the graph to list the rest of the solutions.
As sin and cos repeat every 360^{0} or 2π radians we:
- Find the two solutions in the initial range, (e.g. −π ≤ θ ≤ +π)
- Add 360n or 2nπ to both of these
As tan repeats every 180^{0} or π radians we:
- Find the first solution (PV) initial range, (e.g. −π ≤ θ ≤ +π)
- Add 180n or nπ
This can be summarised as:
For sin θ = S (where |S| ≤ 1), the general solution is
θ = PV + 2nπ | or | θ = PV + 360^{0} |
θ = SV + 2nπ | θ = SV + 360^{0} |
For cos θ = C, the general solution is
θ = ± PV + 2nπ | or | θ = ± PV + 360^{0} |
For tan θ = T, the general solution is
θ = PV + nπ | or | θ = PV + 180^{0} |
The graphs of trigonometric functions of compound angles
The graph of the function sin cθ where c is a constant, is a sine wave with a period of 2π ⁄ c. The frequency is c times that of sin θ. This is shown in the diagram below:
This rule is also true for cos θ, and tan θ.
This means that when solving trigonometric equations with a multiple of θ, there will be a different number of solutions in a 360^{0} range. In these situations find the two initial solutions, make the general set of solutions, and then rearrange to find θ.
Example:
Solve cos (3θ + 45) = −0.5
(3θ + 45) = −120 (from calculator) and,
(3θ + 45) = 120 (cos is negative in the second and third quadrants)
Therefore, (3θ + 45) = −120 ± 360n and (3θ + 45) = 120 ± 360n.
Therefore, θ = −55 ± 120n , and θ = 25 ± 120n
A final hint. Watch out for trigonometric equations that are quadratics.
Example:
2 sin^{2} θ + sin θ − 1 = 0
This has to be factorised and then solved.
(2 sin θ − 1) (sin θ + 1) = 0, where 0 ≤ θ ≤ 360
sin θ = 0.5 or sin θ = -1 and solve as before to get,
θ = 30, 150, or 270. (See if you can get these solutions.)