James Gray

University of Edinburgh

Applicable Mathematics (Foundation) af0

3: Trigonometry

3.1 sin, cos and tan

`theta` is an angle between `0^circ` and `90^circ`.  Draw a right-angled triangle with `theta` as one of the angles.  Any two such triangles will be similar so the ratio of the sides of the triangle will be the same.

3.1 sin, cos and tan

Can define:

Often remembered by: soh-cah-toa

Note

sin, cos and tan are abbreviations of sine, cosine and tangent. 

3.2 Special Cases

Need to know these triangles for exact values of sin, cos and tan with the angles `30^circ`, `45^circ` and `60^circ`.

Example

Q: A surveyor stands 25m from a  building and measures the angle to the top from the horizontal to be `30^circ`.  How high is the building?

A:

Exercises

3.3 Radians

An angle has size `theta` radians if the arc it subtends on the circumference of a circle of radius `1` has length `theta` units.

Circumference of circle of radius `1` is `2pi`.  So we have

In general

So:

Degrees `360^circ` `180^circ` `90^circ` `60^circ` `45^circ` `30^circ`
Radians `2pi` `pi` `pi/2` `pi/3` `pi/4` `pi/6`

Notes

  1. No symbol means angle is in radians e.g.
    • `sin 10` means sine of 10 radians
    • `sin 10^circ` means sine of 10 degrees
  2. In calculus (i.e. the mf0 course) angles are usually measured in radians.

Exercises

3.4 Trigonometry for any angle

Draw a circle radius 1.  The point P which lies on the circle has coordinates `(cos theta, sin theta)`.

We define `tan theta = (sin theta)/(cos theta)`

Example

Q: What is the exact value of `sin 300^circ`?

A: The coordinates of the point `P` are `(cos 300^circ , sin 300^circ)` so we need to find the values for these.  Draw in a triangle between the `x`-axis and the point `P`. 

We deduce that the angle `theta = (360 - 300)^{circ} = 60^{circ}`. Using our standard triangles we find that `x = 1/2` and `y=sqrt(3)/2`.

Hence the coordinates of `P` are `(1/2, - sqrt(3)/2)`.  Thus we have found that

Note

N.B. Anti-clockwise is defined to be positive as this is the way we think of the Earth rotating (from the persperctive of the Northern hemisphere).

Right-hand Rule: Hold you right hand out in front of you with your thumb up. Your thumb represents the axis of rotation with north up and your curved fingers indicate the direction of rotation of our Planet.

Method: associated acute angle

To simplify the process we use the method of the associated acute angle.  This is the angle that the line makes with the `x`-axis in the shaded part of the diagram below on the left.  We then use that angle and correct the sign using the diagram on the right.

Example

Q: What are the exact values of `sin - 150^circ` and `tan -150^circ`?

A: Associated accute angle is `180^circ - 150^circ = 30^circ`.

`sin 30^circ = 1/2` and `tan 30^circ = 1/sqrt(3).

`-150^circ` is in the 3rd quadrant so sin is negative and tan is positive.

Hence `sin - 150^circ = - 1/2` and `tan -150^circ=1/sqrt(3)`.

Exercises

3.4 Solving Trig Equations

Example

Q:Solve `2sin theta - 1=0` for `0 le theta < 2pi`.

A: Rearrange the equation to get `sin theta = 1/2`.  So `sin theta` is positive, and therefore `theta` lies in the 1st and 2nd quadrants. 

Associated acute angle is `sin^(-1) 1/2 = pi/6`.

In the 1st quadrant `theta = pi/6`.

In the 2nd quadrant `theta = pi - pi/6 = (5pi)/6`.

So the solutions in the range `0 le theta < 2pi` are `theta = pi/6, (5pi)/6`.

Example

Q: Solve `1=1/2 - sin 4x` for `0^circ le x le 150^circ`.

A: Rearrange the equation to get `sin 4x = - 1/2`.  Since `sin 4x` is negative there will be solutions in the 3rd and 4th quadrants.

The associated acute angle is `sin^(-1) = 30^circ`.

In the 3rd quadrant `4x = 180^circ + 30^circ = 210^circ`.

In the 4th quadrant `4x = 360^circ - 30^circ = 330^circ`.

Since `0^circ le x le 150^circ` we are looking for values of `4x` in the range `0^circ le 4x le 600^circ`.

So `4x = 210^circ, 330^circ, 570^circ`.

Therefore `x = 52.5^circ, 82.5^circ, 142.5^circ`.

3.5 Trig with non-right angled triangles

Can only use soh-cah-toa with right-angled triangles.

But for non-right-angled triangles we have:

Cosine rule
`a^2 = b^2 + c^2 - 2bc cos A`
Sine rule
`a/(sin A) = b/(sin B) = c/(sin C)`

Example

Q: What is the size of the angle `A` in the diagram?

A: Use the cosine rule with `a=5`, `b=8` and `c=7`.  So

Therefore `A=38.2^circ qquad` (1 d.p.)

Exercises

Previous Exercises

3.6 Graphs of trig functions