Download Chapter 6: Trigonometric Functions of Angles

Lecture Notes
Algebra II Honors
Chapter 6: Trigonometric Functions of Angles
6.1 Angle Measure
If θ is a central angle (with vertex at the center) of a circle, the measure of the angle in radians is:
s
θ = . The arc length s and the radius r must be measured in the same units, leaving θ
r
dimensionless. The radian or rad or r signifies that the number is the measure of an angle. Since 1
revolution gives an arc length equal to the circumference of the circle, and since circumference =
2π times radius, θ = 1 rev = 360° = 2π radians.
•To convert ° to rad, multiply by the equivalence factor of
Example: 150° ·
π rad 5π
rad
=
180°
6
•To convert rad to °, multiply by
π rad
.
180°
180°
π rad
•Consider an angle to be in standard position in the xy-plane if its vertex is at the origin and its
initial side is along the +x-axis.
•Two angles in standard position are coterminal if their terminal sides coincide.
For example, 60°, 420°, –300° or π/3, 7π/3, –5π/3 are all coterminal.
•Length of arc: s = r · θ if θ is measured in radians and r is the radius of the circle.
•Area of Sector: The sector of a circle (a pie-shaped wedge inside a central angle) has area:
A=
1 2
⋅ r ⋅ θ , where θ is measured in radians and r is the radius of the circle.
2
•Circular Motion: If a point object moves along a circle of radius r, covering arc length s
(subtended by angle θ) in time t, define:
Angular Speed ω =
s
θ
(θ must be in rad) and Linear Speed v =
t
t
Note v = r · ω
1
Lecture Notes
Algebra II Honors
6.2 Trigonometry of Right Triangles
•Here we invoke the definitions of the trig functions of an acute angle θ in terms of the ratios of
sides of a right triangle. Recall SOH-CAH-TOA.
opposite
hypotenuse
csc ( θ ) =
hypotenuse
opposite
adjacent
hypotenuse
cos ( θ ) =
sec ( θ ) =
hypotenuse
adjacent
opposite
adjacent
tan ( θ ) =
cot ( θ ) =
adjacent
opposite
sin ( θ ) =
•To extend these definitions to angles other than acute angles, consider the diagram below.
Draw the angle θ in standard position and drop a vertex from the terminal side of the angle to the
x-axis. Consider the reference angle and reference triangle formed, where the reference angle θ
is the acute angle between the terminal side of θ and the x-axis. Apply the SOH-CAH-TOA
definitions to get the magnitude of the trig functions, and the sign will be correct if the sides are
taken + or – in the usual manner (always take the hypotenuse as +). This definition is a bit
difficult to apply for quadrantal angles (0, π/2, π, 3π/2); an alternative is in the next section.
•Make use of the special right triangles 30-60-90 and 45-45-90.
2
Lecture Notes
Algebra II Honors
•Example: Find the six trig functions of 5π/6 rad or 150°.
The reference triangle is shown above, with θ = π/6 or 30°.
2
1
30
–√3
The trig functions of 5π/6 rad are the same, when adjusted for sign, as the functions of π/6.
Use the reference triangle and SOH-CAH-TOA to write the trig functions.
sin(5π/6) = 1/2
tan(5π/6) = –√3/3
csc(5π/6) = 2
cot(5π/6) = –√3
cos(5π/6) = –√3/2 sec(5π/6) = –2√3/3
•Example: Given a right triangle with sides 3, 4, and 5. Find the measures of its angles.
A
B
C
sin(B) = 3/5 = 0.6 Our calculator says Sin–1(0.6) = 36.87° = 0.644 rad.
Since C = 90° = π/2 rad, we have A = 53.13° = 0.927 rad.
3
Lecture Notes
Algebra II Honors
6.3 Trigonometric Functions of Angles
P(X,Y)
r
Y
θ
X
Consider the angle θ in standard position. Pick some point P on its terminal side a distance r from
the origin. Now define:
sin(θ) = Y / r
csc(θ) = r / Y
cos(θ) = X / r
sec(θ) = r / X
tan(θ) = Y / X
cot(θ) = X / Y
This is essentially the scheme used in the last section, but it more directly gives the sign of the
trig function and there is no problem if x or y is zero regarding reference triangles.
•Example: Find the trig functions of π rad or 180°.
Pick any point on the –x-axis as P. Why not choose P(–1, 0) to bring back fond memories of the
unit circle? Then with x = –1, y = 0 and r = 1:
sin(π) = 0
csc(π) = undef
cos(π) = –1
sec(π) = –1
tan(π) = 0
cot(π) = undef
•Areas of Triangles
Recall the area of the triangle is (1/2 )· base · height. It is shown in the text, using the definition of
sine and the fact that sin(180° – θ) = sin(θ):
Area = (1/2) · side1 · side2 · (sin(included angle)) = (1/2) · a · b · sin(θ)
4
Lecture Notes
Algebra II Honors
•Trig Identities
csc(θ) = 1 / sin(θ)
sec(θ) = 1 / cos(θ)
cot(θ) = cos(θ) / sin(θ)
tan(θ) = sin(θ) / cos(θ)
sin2(θ) + cos2(θ) = 1
1 + tan2(θ) = sec2(θ)
cot(θ) = 1 / tan(θ)
1 + cot2(θ) = csc2(θ)
•Example: Find the values of the other trig functions of θ if given sin(θ) = 3/5, θ in II.
Using identities, we see csc(θ) = 5/3. Note the other four trig functions are negative in II.
⎛3⎞
⎝5⎠
2
sin2(θ) + cos2(θ) = 1 → cos(θ) = ± 1 − ⎜ ⎟ = ±
16
4
= ± and we choose –4/5 in II.
25
5
Thus sec(θ) = –5/4.
Using tan(θ) = sin(θ) / cos(θ) gives tan(θ) = –3/4 and thus cot(θ) = –4/3.
•Be sure to have memorized All Students Take Calculus and know:
All trig functions are positive in I
Sine and its reciprocal are positive in II
Tan and its reciprocal are positive in III
Cos and its reciprocal are positive in IV
5
Lecture Notes
Algebra II Honors
6.4 Law of Sines
a
sin ( A )
sin ( A )
a
=
=
b
c
=
sin ( B ) sin ( C )
sin ( B )
b
=
sin ( C )
c
Recall the area of a triangle given two sides and the included angle is half the product of those
two sides and the sine of the included angle:
Area = (1/2) · b · c · sin(A) = (1/2) · a · c · sin(B) =(1/2) · a · b · sin(C)
Divide by (1/2) · a · b · c to find:
sin ( A )
a
=
sin ( B )
b
=
sin ( C )
c
•Use the Law of Sines to solve triangles if given AAS, ASA, or (ambiguously) SSA.
•Example: Solve the triangle pictured above where A = 30°, C = 65°, b = 10.
We are asked to find B, a, and c. Note the given in this case are ASA, so we use the law of sines.
Since the angles add to 180°, B = 85°.
b
a
10
a
10
a
=
→
=
→
=
sin ( B ) sin ( A )
sin ( 85° ) sin ( 30° )
0.9962 0.5000
Solving the proportion by cross-multiplying, (10)(0.5000) = (0.9962)(a) or a = 5.02
•Example: Solve the triangle pictured above if a = 28, b = 15, A = 110°
We are asked to find c, B, and C. Note this case is SSA. There may be 0, 1, or 2 solutions!
If you are not given the largest side or angle, try to solve for that first. (Remember, the largest
side is opposite the largest angle.)
sin ( B )
b
=
sin ( A )
a
→
sin ( B )
15
=
sin (110° )
28
→ sin ( B ) =
15 ⋅ ( 0.9397 )
28
= 0.5034
Thus B = 30.23° or 149.77°. (Recall sine is positive in I or II.) Here the latter answer cannot fit in
the given triangle, but it could lead in other cases to multiple solutions. If a sine is ever found to
be > 1, that is a clue no solution is possible.
Now C = 180° – 110° – 30.23° = 39.77°.
( 28 )( 0.6397 )
c
a
c
28
c
28
=
→
=
→
=
→c=
sin ( C ) sin ( A )
sin ( 39.77° ) sin (110° )
0.6397 0.9397
0.9397
or c = 19.06. Thus B = 30.23°, C = 39.77°, c = 19.06 is the only solution in this case.
•Note a / sin(A) etc is the diameter of the circle that circumscribes ABC.
6
Lecture Notes
Algebra II Honors
6.5 Law of Cosines
c 2 = a 2 + b 2 − 2 ⋅ a ⋅ b ⋅ cos ( C )
b 2 = a 2 + c 2 − 2 ⋅ a ⋅ c ⋅ cos ( B )
a 2 = b 2 + c 2 − 2 ⋅ b ⋅ c ⋅ cos ( A )
The law of cosines is used to solve triangles when given SSS or SAS.
Proof: Place a triangle on a coordinate system with C(0, 0), B(a, 0) and A(b · cos(C), b · sin(C)).
Using the distance formula for side c (the distance between A and B):
( b ⋅ cos ( C ) − a ) + ( b ⋅ sin ( C ) − 0 )
= ( b ⋅ cos ( C ) − a ) + ( b ⋅ sin ( C ) − 0 )
2
c=
c2
2
2
2
c 2 = b 2 ⋅ cos 2 ( C ) − 2 ⋅ a ⋅ b ⋅ cos ( C ) + a 2 + b 2 ⋅ sin 2 ( C )
(
)
c 2 = a 2 + b 2 ⋅ cos 2 ( C ) + sin 2 ( C ) − 2 ⋅ a ⋅ b ⋅ cos ( C )
c 2 = a 2 + b 2 ⋅ (1) − 2 ⋅ a ⋅ b ⋅ cos ( C )
•Example: Two boats leave port at the same time. One boat travels at 30 knots in the direction N
50° E, and the other boat travels at 26 knots in the direction S 70° E. How far apart are the boats
after one hour?
In navigation problems, the direction is often given as a bearing, an acute angle measure from due
north or due south. N 50° E points 50° to the east of due north. (An alternative is to give a
heading from 0° to 360° where 0° is due north and the angle increases CW; due east = 90°, etc.)
The diagram is basically that shown above. Side c is the distance 30 nautical miles that the first
boat travels in one hour, and side b is 26 nautical miles. We are asked to find the length of side a.
Given the bearings in the problem we find A = 180° – 50° – 70° = 120°. Thus we have SAS, and
the law of cosines is written: a 2 = b 2 + c 2 − 2 ⋅ b ⋅ c ⋅ cos ( A )
a 2 = 302 + 262 − 2 ⋅ 30 ⋅ 26 ⋅ cos ( 60° ) = 900 + 676 − 780 = 796
Thus a = the distance between the boats after one hour = 28.2 nautical miles.
•Using the law of cosines, one may prove (see text) Heron's Formula for the area of a triangle.
Heron's Formula: Given the sides a, b, c of a triangle, define s (the semiperimeter) as
Then the area of the triangle A =
a+b+c
.
2
s ( s − a )( s − b )( s − c )
7