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Supplementary Angles
Two angles α and β that together form a straight angle, i.e., α + β =π, (or in degrees α + β = 180°)are said to be supplementary.
θ 180°-θ^ θ
180°- θ
θ
Notice in the examples above, the right triangles formed by θ are exactly the same, just flipped horizontally. Since the triangle on the left is in Quadrant II, the cos, sec, tan, and cot will all be negative, but the sin and csc will be positive, just as the triangle on the right. Also, the ratios will all be the same as the triangle on the right.
So we have these properties of Supplementary Angles If θ ≤ 180° sin(θ) = sin(180° - θ) cos (θ) = -cos(180° - θ) tan(θ) = - tan(180° - θ)
2.5 Finding the Value of the Trig Functions Using a Point on the Unit Circle
A unit circle is a circle with its center at the origin (0,0) and a radius of 1. The equation for this circle is x 2 +y 2 = 1.
Example 1 on p. 144 Find the value of sin t , cos t , tan t , csc t , sec t , and cot t for a point P = (a,b) = on a unit circle.
⎞ ⎜⎜⎝
⎛ −
2
,^3 2
( a , b )^1
trad
a
b (^) θ
r=
cos t = a/r = a sin t = b/r = b tan t = a/b cot t = b/a sec t = r/a csc t = r/b
( a , b ) (0,r)
trad
a
b (^) θ
r
cos t = a/1 = a= -½
sin t = b/1 = b=
tan t = b/a =
cot t = a/b =
sec t = 1/a = 1 / -½ = -
csc t = 1/b =
Now let’s look at a circle with center (0,0) and radius r. The equation for this circle is x 2 +y 2 = r 2. Notice for any point (a,b) on the circle, a 2 +b^2 = r 2.
So r = a^2 + b^2
2
3
3
3 3
1 2
3 2
1 −
−
−
3 1
3 2 12
(^3) =− −
− =
3
2 3 3
2 2
3 1 = =
Now you try #7 on p.
Domain and Range of Tan and Cot
(a,b) =(0,1)
π/
What about tan and cot?
tan(θ) = opp/adj = b/a
If a = 0, then tan(θ) is undefined. Therefore the domain of tan does not include angles at points where a=0. Those are all odd multiples of π/ (i.e. π/2, 3π/2, 5π/2, etc..) The range of tan(θ) is all real numbers, because when b is large and a is small, tan(θ) can increase infinitely; and when b is small and a is large, tan(θ) can decrease infinitely.
cot(θ) = opp/adj = b/a
If b = 0, then cot(θ) is undefined. Therefore the domain of cot does not include angles at points where b=0. Those are all integral (integer valued) multiples of π. (i.e. 0, π, 2π, 3π, 4π, etc..) Similarly to tan, the range of cot is all real numbers.
Period of Trig Functions
Review of Coterminal Angles: If θ is an angle measured in degrees, then θ ±360°(k) , where k is any integer, is an angle coterminal with θ****. If θ is an angle measured in radians, then θ ± 2 π k , where k is any integer, is an angle coterminal with θ****. Since coterminal angles are basically equal, all their trigonometry functions are equal. Therefore sin( θ + 2 π ) = sin(θ ) and cos( θ + 2 π ) = cos(θ )
Functions that exhibit this type of behavior are called periodic functions. A function f is periodic if there is a positive number p such that, whenever θ is in the domain of f , so is θ+p, and f(θ+p) = f(θ). Let’s revisit our “ ASTC ” graph:
- -
0
1
-1 0 1
β= πrad = 180° δ= (^0) rad = 0°
α= π/2 (^) rad = 90°
Φ= 3 π/2rad = 270°
I
A ll trig functions >
II
Sin, csc > cos,sec,tan,cot <
III
Tan,cot > 0 sin, cos, sec, csc <
II
Cos,sec > 0 sin, csc, tan,cot <
θ π+θ
Going from 0 to 2 π, sin(θ ) starts out with the value 0, then rises to 1 at π/2, then goes back to 0 at π. At θ> π, sin(θ ) goes from 0 to - at 3π/2, then back to 0 at 2 π. At this point the sin values repeat. The period of the sine function is 2π. (Likewise for cosine).
Going from 0 to 2 π, tan(θ ) starts out with the value 0, then rises to infinity as θ approaches π/2. In Quadrant II, tan(θ ) rises from negative infinity (since the y-coordinates are + and the x-coordinates are -), then goes back to 0 at π. At θ> π, tan (θ ) repeats the same process. The period of the tangent function is π. (Likewise for cotangent). tan( θ + π ) = tan(θ ) and cot( θ + π ) = cot(θ )
(a,b)
(-a,-b)
Example 3 on p.
Find the exact value of d)
37 π −
The tangent function repeats its values for any integral multiple of π. Therefore, let’s add (since the angle is negative) a multiple of π to that will give us a reference angle.
Since tan is an ODD function, we can use the property that tan(-θ) = -tan(θ).
Now you do #29 on p.152 and #65 on p.
π
π 9 4
36
tan π
= ⎛^ − +
tan 4
tan 4
tan π π π π
1 4
tan 4
tan (^) ⎟ =− ⎠
⎞ ⎜ ⎝
⎛ ⎟= − ⎠
⎞ ⎜ ⎝
⎛ −
π π
Homework
p.141 #79,93,
p. 152 #3,9,15,35,47,53,55, 73,
Practice Test (sections 2.1-2.5)
We will go over this in class in
the next lecture.
You will turn it in with the Test on
Feb.21 for 20 pts + 15 points of
your test score (out of a
possible 150).
Chapter Review p. 191-
Project 2
