Download Section 9.4 Trigonometric Functions of any Angle

Section 9.4 Trigonometric Functions of any Angle
So far we have only really looked at trigonometric functions of acute (less than 90º)
angles. We would like to be able to find the trigonometric functions of any angle.
To do this follow these steps:
1.
2.
3.
4.
Draw the angle in standard position on the coordinate axes
Draw a reference triangle and find the reference angle
Label the reference triangle
Write down the answer
OR use your calculator.
Note: Your calculator will only give you decimal approximations but on the test your
answers must be exact.
What is a reference triangle and reference angle?
If you draw the angle θ in the standard postion its reference angle is the acute angle θ’
formed by the terminal side of θ and the horizontal axis.
The reference triangle is the triangle which is formed by drawing a perpendicular line
from the terminal side of your angle θ in standard position to the horizontal axis.
Ex: Here is a reference angle and triangle in the 2nd quadrant.
Reference Angle θ’
Original Angle θ
θ
θ’
Reference Triangle
The size of the reference angle in the second quadrant will be 180 – θ or π – θ depending
on whether the angle is given in degrees or radians respectively.
Here is one in the 4th quadrant:
Reference Angle θ’
Original Angle θ
θ
θ’
Reference Triangle
The size of the reference angle in the fourth quadrant will be 360 – θ or 2π – θ depending
on whether the angle is given in degrees or radians respectively.
Ex 1: Find the reference triangles and angles:
a) θ=309º
The reference angle will be 360-309 = 51º
Reference Angle θ’
Original Angle θ
309º
51º
Reference Triangle
b) θ = −
7π
4
The reference angle will be 2π −
7π π
=
4
4
Reference Angle θ’
−
7π
4
π
4
Reference Triangle
Original Angle θ
10π
3
This angle is larger than one full revolution so to find it in standard position we need to
find a coterminal angle that is smaller than 2π (One time around the circle). To do this
we simply subtract 2π untill our angle is less than 2π.
c) θ =
10π
10π 6π 4π
− 2π =
−
=
3
3
3
3
So to find the reference angle we will start with the coterminal angle
to get:
4π
π
−π =
3
3
Reference Angle θ’
4π
and subtract π
3
Original Angle θ
π
3
Reference Triangle
Now we will use these reference angles to find the values of some trigonometric
functions.
Ex 2. Find the values of the six trigonometric functions for θ = 150º
1. Draw the angle in standard position
2. Draw the reference triangle and angle
3. Label the triangle. Here we will label using the standard 30 – 60 - 90 triangle
2
150
1
30
- 3
NOTE: Since one side of the reference triangle is on the negative x-axis
that side is labeled as − 3 . This is VERY IMPORTANT. You will
notice that this makes the cosine, secant, tangent and cotangent
negative.
4. Now we can find the 6 trigonometric functions by reading them off the reference
triangle:
sin(150) =
opp 1
=
hyp 2
csc(150) =
hyp 2
= =2
opp 1
cos(150) =
3
adj − 3
=
=−
hyp
2
2
sec(150) =
hyp
2
2
=
=−
adj − 3
3
tan(150) =
opp
1
1
=
=−
adj − 3
3
cot(150) =
adj − 3
=
=− 3
opp
1
Ex 3: Find the values of the six trigonometric functions for θ = −
π
4
Note that the angle is negative.
1 Draw the angle in standard position
2 Draw the reference triangle and angle
3 Label the triangle. Here we will label using the standard 30 – 60 - 90 triangle
1
π
4
-1
2
NOTE: Since one side of the reference triangle is in the negative y
direction that side is labeled as -1. This is VERY IMPORTANT. You
will notice that this makes the sine, cosecant, tangent and cotangent
negative.
4. Now we can find the 6 trigonometric functions by reading them off the reference
triangle:
π
opp − 1
1
=
=−
sin(− ) =
hyp
4
2
2
2
π
hyp
csc(− ) =
=
=− 2
opp − 1
4
π
adj
1
=
cos(− ) =
hyp
4
2
π
hyp
sec(− ) =
= 2
adj
4
π
opp − 1
tan(− ) =
=
= −1
adj
4
1
cot(150) =
1
adj
=
= −1
opp − 1
4
and θ in quadrant III. Find the values of the six
5
trigonometric functions for θ.
Ex 4: Suppose cos θ = −
The first thing we need to do is to draw a reference triangle. We know we are in quadrant
4 adj
III and we know that cos θ = − =
so we know two sides of the triangle already.
5 opp
We will use the Pythagorean theorem to find the last side. Here θ’ is the reference angle.
θ
-4
θ’
-y=-3
5
Solve for y: (−4) 2 + y 2 = 5 2 so y = 3.
NOTE: Since one side of the reference triangle is in the negative y
direction and the other is in the negative x direction that both of those
sides are negative. This is VERY IMPORTANT. You will notice that
this makes the sine, cosecant, cosine and secant negative.
Now we can find the 6 trigonometric functions by reading them off the reference triangle:
sin(θ ) =
3
opp − 3
=
=−
hyp
5
5
csc(θ ) =
1
5
5
=
=−
sin(θ ) − 3
3
cos(θ ) =
4
adj − 4
=
=−
hyp
5
5
sec(θ ) =
1
5
5
=
=−
cos(θ ) − 4
4
tan(θ ) =
opp − 3 3
=
=
adj − 4 4
cot(θ ) =
1
−4 4
=
=
tan(θ ) − 3 3
What this has shown us is that we can determine the sign of the trigonometric functions
by which quadrant the terminal side in.
Quadrant II
sin(θ): +
cos(θ): tan(θ): -
Quadrant I
sin(θ): +
cos(θ): +
tan(θ): +
Quadrant IV
sin(θ): cos(θ): +
tan(θ): -
Quadrant III
sin(θ): cos(θ): tan(θ): +
Ex 5: Suppose csc θ = 4 and cot θ > 0 . Find the values of the six trigonometric
functions for θ
1
= csc θ = 4 the sine is positive so θ is in quadrant I or II.
sin θ
Since the cot θ > 0 then the tangent is positive so θ is in quadrants I or III .
Since the
The overlap of these two regions is quadrant I so we can draw our triangle:
4 hyp
csc θ = =
1 opp
4
1
θ
x= 15
To solve for x we use the Pythagorean theorem: (1) 2 + x 2 = 4 2 so x = 15 .
1
opp 1
sin(θ ) =
=
csc(θ ) =
=4
hyp 4
sin(θ )
cos(θ ) =
15
adj
=
hyp
4
sec(θ ) =
1
4
=
cos(θ )
15
tan(θ ) =
opp
1
=
adj
15
cot(θ ) =
1
= 15
tan(θ )