Transcript Trig Funct of Gen Angles

OF GENERAL ANGLES
Our method of using right triangles only works for
acute angles. Now we will see how we can find the trig
function values of any angle. To do this we'll place
angles on a rectangular coordinate system with the
initial side on the positive x-axis.
reference
angle
 =135°
What is the measure of
this reference angle?
180°- 135° = 45°
HINT: Since it is 360° all the way
around a circle, half way around
(a straight line) is 180°
If  is 135°, we can find the angle
formed by the negative x-axis and
the terminal side of the angle.
This is an acute angle and is
called the reference angle.
Let's make a right triangle by drawing a line perpendicular to
the x-axis joining the terminal side of the angle and the x-axis.
Let's label the sides of the triangle according to a 45-45-90
triangle. (The sides might be multiples of these lengths but
looking as a ratio that won't matter so 1  1  2 will work)
2
1
45°
-1
 =135°
This is a Quadrant II angle. When
you label the sides if you include
any signs on them thinking of x & y
in that quadrant, it will keep the
signs straight on the trig functions.
x values are negative in quadrant II
so put a negative on the 1
Now we are ready to find the 6 trig
functions of 135°
The values of the trig functions of angles and their reference
angles are the same except possibly they may differ by a
negative sign. Putting the negative on the 1 will take care of this
problem.
o
1
2
sin135  

h
2
2
a 1  2
cos135  

h
2
2
Notice the -1 instead of 1 since the
terminal side of the angle is in quadrant II
where x values are negative.
2
1
45°
-1
 =135°
0 1
tan 135  
 1
a 1
We are going to use this method to find
angles that are non acute, finding an acute
reference angle, making a triangle and
seeing which quadrant we are in to help
with the signs.
Let  denote a nonacute angle that lies in a quadrant.
The acute angle formed by the terminal side of  and
either the positive x-axis or the negative x-axis is called
the reference angle for .
Let's use this idea to find the 6 trig functions for 210°
First draw a picture and label  (We know
that 210° will be in Quadrant III)
=210°
 3
30°
-1
2
210°-180°=30°
Now drop a perpendicular line from the
terminal side of the angle to the x-axis
The reference angle will be the angle
formed by the terminal side of the angle
and the x-axis. Can you figure out it's
measure?
Label the sides of the 30-60-90
The reference angle is the triangle and include any negative
signs depending on if x or y values
amount past 180° of 
are negative in the quadrant.
210°
 3
30°
-1
2
You will never put a negative on the
hypotenuse. Sides of triangles are
not negative but we put the negative
sign there to get the signs correct on
the trig functions.
2
 2
csc 210 
1
You should be thinking csc is the reciprocal of sin
and sin is opposite over hypotenuse so csc is
hypotenuse over opposite.
1
3
tan 210 

3
 3
 3
cos 210 
2
Using this same triangle idea, if we are given a point on the
terminal side of a triangle we can figure out the 6 trig
functions of the angle.
Given that the point (5, -12) is on the terminal side of an angle ,
find the exact value of each of the 6 trig functions.
5   12
2
5
2
 h2
h  13
-12
13
(5, -12)
Use the Pythagorean
theorem to find the
hypotenuse
First draw a picture
Now drop a perpendicular
line from the terminal side
to the x-axis
Label the sides of the triangle
including any negatives. You
know the two legs because they
are the x and y values of the point
Given that the point (5, -12) is on the terminal side of an angle ,
find the exact value of each of the 6 trig functions.

5

-12
13
(5, -12)
sin 
cos 
tan 
We'll call the reference angle . The trig
functions of  are the same as  except
they possibly have a negative sign.
Labeling the sides of triangles with
negatives takes care of this problem.
o 12
 h  13
a 5
 
h 13
 o  12
a
5
csc 
sec 
cot 
h 13
 o   12
h 13
 a 5
a 5
o  12
All trig
functions
positive
+

+
sin is +
cos is tan is -
+
_

In quadrant I both the x
and y values are positive
so all trig functions will be
positive
Let's look at the signs of sine,
cosine and tangent in the other
quadrants. Reciprocal functions will
have the same sign as the original
since "flipping" a fraction over
doesn't change its sign.
In quadrant II x is negative
and y is positive.
We can see from this that any value that
requires the adjacent side will then have a
negative sign on it.
In quadrant III, x is
negative and y is negative.
_
_
Hypotenuse is always positive so if
we have either adjacent or opposite
with hypotenuse we'll get a
negative. If we have both opposite
and adjacent the negatives will
cancel

sin is cos is tan is +
+

sin is cos is +
tan is -
In quadrant IV, x is positive
and y is negative .
_
So any functions using opposite
will be negative.
sin is +
cos is tan is -
All trig
functions
positive
sin is cos is tan is +
sin is cos is +
tan is -
To help
remember
these sign
we look at
what trig
functions
are
positive in
each
quadrant.
Here is a mnemonic
to help you
remember.
(start in Quad I and
go counterclockwise)
Students
Take
S
A
T
C
All
Calculus
What about quadrantal angles?
We can take a point on the terminal side of quadrantal
angles and use the x and y values as adjacent and
opposite respectively. We use the x or y value that is not
zero as the hypotenuse as well
well.(but never with a negative).
Try this with 90°
o 1
1
sin 90    1 cosec90   1
1
h 1
1
a 0
cos 90    0 sec 90  undef
0
h 1
dividing by 0 is
o 1
0
undefined so the tan 90 

cot 90   0
tangent of 90° is
a 0
1
(0, 1)
undefined
Let's find the trig functions of 
(-1, 0)
1
o 0
sin     0 cosec   undef
0
h 1
a 1
1
 1
cos     1 sec  
1
h 1
1
o 0
cot    undef
tan     0
0
a 1
Remember
x is adjacent,
y is opposite
and
hypotenuse
here is 1
Coterminal angles are angles that have the same terminal side.
62°, 422° and -298° are all coterminal because graphed,
they'd all look the same and have the same terminal side.
-298°
62°
422°
Since the terminal
side is the same, all of
the trig functions
would be the same so
it's easiest to convert
to the smallest
positive coterminal
angle and compute
trig functions.
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au