2008  Rasmus ehf
and Jhann sak

Trig functions

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Lesson 2      The Unit Circle    


We begin by looking at a right angled triangle where the hypotenuse has a length of 1 unit.

Opposite side = sin v

Adjacent side  = cos v

In a right angled triangle

   sin v = opposite side/hypotenuse

and

   cos v = adjacent side/hypotenuse.

If the hypotenuse in a triangle has length 1 then it follows that

sin v = opposite side and cos v = adjacent side.

We now consider a circle drawn in a coordinate system.

A circle with a radius of 1 unit and its centre  in (0, 0) is called the Unit circle.

                     The Unit circle

If we draw a radius that makes an angle of v with the positive arm of the x axis and drop a perpendicular as the diagram shows we get a  right angled triangle with sides of length  cosv and sin v. This means that the coordinates of the point where the radius intersects the circle must be (cos v, sin v).

Now well look at tan v a similar way.

opposite side = tan v

Adjacent side = 1

This time we choose the adjacent side to be of length 1 unit.

   tan v = opposite side/adjacent side.

The adjacent side = 1 therefore  tan v = opposite side.

We can add this to the  diagram of the unit circle.

To see the connection between sin, cos and tan we do the following calculations:

                   opposite side
tan v =
                    adjacent side

              opposite side/hypotenuse
tan v = 
                  adjacent side/hypotenuse

Dividing the numerator and denominator by the hypotenuse.

The calculations we have shown so far apply to an angle drawn in the unit circle, measured from the x axis and lying in the first quadrant of the coordinate system. If we call the point where the radius r cuts the circle P, and rotate the radius OP anticlockwise round the circle from the x axis we say that the angle v is a positive rotation. If OP is rotated clockwise from the x axis we talk about a negative rotation. ( this is simply a definition that  has been agreed on). So if P moves along the circle up from the x axis we have a positive value for the rotation v. If P moves along the circle and down from the x axis we have a negative value for v.
( see the diagram).

We now generalise the definition of sine, cosine and tangent as follows:


If the radius OP in the unit circle is rotated  v from the positive x-axis then

     cos v = x-coordinate of  P

      sin v = y-coordinate of  P

This definition implies that the trig functions can be positive or negative depending on in which quadrant of the coordinate system the point P lies.

As the point P moves with a positive rotation round the circle it reaches the same position every 360 . This means that the values of sine and cosine are repeated every 360. Tangent  repeats its value every 180 as we will see in example 4.

Example 1

Find the positive rotation that takes us to the same position as −200

To do this we simply have to add on 360.

   −200 + 360 = 160

Example 2

Find an angle on the interval 0 v < 360 equivalent to 1100.

   We need to subtract 360 several times until we reach the interval required.

      1 circle:  1∙ 360 = 360

      2 circles:   2∙ 360 = 720

      3 circles:   3∙ 360 = 1080

   This is sufficient.

      1100 − 1080 = 20

Example 3

We will now see how Pythagoras can be use to calculate some exact values for the trig functions.

First we look at 30. The triangle in the diagram is half of an equilateral triangle and therefore we know that the opposite side is .

sin 30 = = 0,5

cos 2 30 + ()2 = 12      

cos 2 30= 1 − =

0.866

Pythagoras rule

                                                                             

                                                                             0.577

Next we look at the angle 45. The triangle is isosceles, we call the sides a.

a2 + a2 = 1

2a2 = 1

a2 =

Now the angle 60.

Again we have a triangle with 30, 60 and 90 so we can use the same calculations as for 30. Notice that sine and cosine have exchanged values.

The values of the trig functions for 90 can easily be seen from the  unit circle. OP, the arm of the angle lies on the y axis so the x-coordinate is 0 and the y coordinate is  1.

cos 90 = 0

 sin 90 = 1

tan 90 = 1/0 does not exist.

Angles that are in other quadrants can be found by comparing them with angles in the first quadrant.

 

120 lies in the second quadrant  (to the left of the positive y axis) and therefore the cos 120 is negative, but the sin 120 is  positive..

We can continue in this way.

135 (90+ 45)  or ( 180− 45) can be found from 45, only the signs change.

Example 4

Find the values of sine, cosine and tan for 225.

225 is in the third quadrant and can be calculated from 45. (45 + 180 = 225). We begin by drawing a diagram.

We calculate in the same way as in example 3, both sine and cosine are negative but tan is positive
(− divided by −).

Notice that tan 225is the same as tan 45. In other words the value of tan v repeats itself every 180.

Example 5

Find all the solutions of the equation tan x = 2, then write down which of these solutions are on the interval 0  x < 360.

    Use the inverse tan function, tan −1, on your calculator.

       tan −1(2) ≈ 63.44

    We know that the values of tan repeat themselves every 180, so we can write a formula for all the solutions by adding k∙180 where k represents any whole number. The complete solution is therefore:

       x ≈ 63.44 + k∙180          k is a whole number.

    Two of these solutions lie  on the interval 0  x < 360. They are x ≈ 63.44 and x ≈ 63.44 + 180 ≈ 243.44

Example 6

Find the angles where sine takes the value 0.5 .

We first solve the equation sin v = 0.5 using a calculator and the inverse sine function sin −1 . sin −1(0.5) = 30. Now draw a diagram of the unit circle.

               

Draw the angle v as usual starting from the positive x axis and turning anticlockwise. The y coordinate of the point on the circle (height above the x axis ) is sin v.
We are given this value 0.5 (sin v = 0.5).

If we draw a horizontal line through this point (the dotted line in the diagram) we can see that there is a second point on the circle which has the same y coordinate and therefore the same value of sine. Both the shaded triangles in the diagram are congruent ( exactly the same). Measuring the rotation of this second radius from the positive x axis we get the second answer for v:

v= 180 − 30 = 150.

From this example we get the following rule:

sin v = sin (180-v)

We use this rule when we need to solve equations of the type

                    sin v = a

Unfortunately calculators only give us one solution, the smallest angle that satisfies the equation measured from the positive x axis. We find the second answer by subtracting this answer from 180. To find all possible solutions we need to add k∙360 to both solutions.
( k can be any whole number ).

Example 7

Find all the solutions of the equation sin v = −0.6.

Draw a diagram of the unit circle to see what we need to do.

   The calculator gives one solution sin −1(−0.6) ≈ −36.9.

   We get the second answer by subracting from 180.

      180 − (−36.9) ≈ 216.9

    Add 360 to −36.9 to get rid of the negative.

       −36.9 + 360 ≈ 323.1

The complete soution is:   v1216.9 + k∙360 og v2323.1 + k∙360

( Its a good idea to use your calculators to check the answers.
sin 216.9 ≈ −0.6 and sin 323.1 ≈ −0.6 )

Example 8

Solve the equation cos v = 0.7.

Draw the unit circle to see what solutions there are.

In this example we are given the x coordinate ( cos v= 0.7) of the point on the circle, that is the distance from the y axis. A perpendicular line ( dotted line) through the point where the radius intersects the circle has been drawn. This perpendicular line through 0.7 on the x axis cuts the circle in two points telling us there are two angles where cos v= 0.7.

Using the calculator we get cos −1(0.7) ≈ 45.57. We can see that the second angle is −45.57 . We find the  positive angle equivalent to this by adding 360.

Solutions:  v145.57 + k∙360

             v2 ≈ −45.57 + 360 + k∙360 ≈ 314.43 + k∙360

From this example we can see the following rule.

cos v = cos(-v)

This is the rule we use to solve equations of the type

cos v = a

The calculator gives one answer, the second answer is the negative of the first answer. We can always convert negative answers to positive by adding 360. Finally we find all solutions by adding k∙360 to each answer.

Example 9

Solve the inequality cos v < −0.7 on the interval 0  v < 360.

First we solve the equation cos v = −0.7.

The calculator gives cos −1(−0.7) ≈ 134.43 . The second solution is therefore −134.43  or −134.43 + 360 ≈ 225.57. Now we need to draw the unit circle.

We can see that cos v is less than −0.7 if  v is in the shaded area on the diagram.

The solution is therefore 134 < v< 226.


Try Quiz 2 on Trig functions.
Remember to use the checklist to keep track of your work.