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TRIGONOMETRY FUNCTIONS

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OF GENERAL ANGLES TRIGONOMETRY FUNCTIONS Our method of using right triangles only works for acute angles. Now we will see how we can find the trig function values of ... – PowerPoint PPT presentation

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Title: TRIGONOMETRY FUNCTIONS


1
TRIGONOMETRY FUNCTIONS
OF GENERAL ANGLES
2
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.
HINT Since it is 360 all the way around a
circle, half way around (a straight line) is 180
?
? 135
referenceangle
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.
What is the measure of this reference angle?
180- 135 45
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.
3
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 will work)
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
? 135
1
45
-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.
4
Notice the -1 instead of 1 since the terminal
side of the angle is in quadrant II where x
values are negative.
? 135
1
45
-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.
5
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)
Now drop a perpendicular line from the terminal
side of the angle to the x-axis
?210
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?
30
-1
2
210-18030
Label the sides of the 30-60-90 triangle and
include any negative signs depending on if x or y
values are negative in the quadrant.
The reference angle is the amount past 180 of ?
6
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.
210
30
-1
2
You should be thinking csc is the reciprocal of
sin and sin is opposite over hypotenuse so csc is
hypotenuse over opposite.
7
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.
First draw a picture
5
Now drop a perpendicular line from the terminal
side to the x-axis
-12
13
(5, -12)
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
Use the Pythagorean theorem to find the hypotenuse
8
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
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.
?
-12
13
(5, -12)
9
In quadrant I both the x and y values are
positive so all trig functions will be positive
All trig functions 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.

sin is cos is -tan is -
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.
10
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
In quadrant IV, x is positive and y is negative .

So any functions using opposite will be negative.
sin is -cos is tan is -
11
To help remember these sign we look at what trig
functions are positive in each quadrant.
All trig functions positive
sin is cos is -tan is -
A
S
T
C
sin is -cos is tan is -
sin is -cos is -tan is
Here is a mnemonic to help you remember. (start
in Quad I and go counterclockwise)
Students
All
Take
Calculus
12
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.
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 (but never with a negative).
Try this with 90
(0, 1)
dividing by 0 is undefined so the tangent of 90
is undefined
13
Let's find the trig functions of ?
(-1, 0)
Remember x is adjacent, y is opposite and
hypotenuse here is 1
14
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.
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.
62
-298
422
15
Acknowledgement I wish to thank Shawna Haider
from Salt Lake Community College, Utah USA for
her hard work in creating this PowerPoint. www.sl
cc.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
Stephens School Carramar www.ststephens.wa.edu.
au
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