Download Unit Circle * Learning Outcomes

1
Trig Functions – Learning Outcomes
 Solve problems about trig functions in right-angled
triangles.
 Solve problems using Pythagoras’ theorem.
 Solve problems about trig functions in all quadrants of a
unit circle.
 Solve problems about trig functions of angles greater
than 360o and less than 0o.
pg 260-263
2
Use Trig Functions (Right-Angled Triangles)
 Recall in a right-angled triangle, we name three sides
based on the angle of interest:
 “Hypotenuse” is always the longest side.
 “Opposite” is literally opposite the angle of interest.
 “Adjacent” is literally adjacent to the angle of interest.
pg 260-263
3
Use Trig Functions (RAT)
 For a particular angle, the ratio between the side lengths
is the same for every triangle, regardless of size.
 The ratio of the opposite to the hypotenuse is called 𝑠𝑖𝑛𝑒:
 sin 𝐴 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
 The ratio of the adjacent to the hypotenuse is called
𝑐𝑜𝑠𝑖𝑛𝑒:
 cos 𝐴 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
 The ratio of the opposite to the adjacent is called
𝑡𝑎𝑛𝑔𝑒𝑛𝑡:
 tan 𝐴 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
pg 260-263
4
Use Trig Functions (RAT)
 These are given on page 13 of the Formula and Tables
Book, but are presented without the words “opposite”,
“adjacent”, or “hypotenuse”.
pg 260-263
5
Use Trig Functions (RAT)
 e.g. Calculate sin(37), cos(37), and tan(37) using the
triangle below:
pg 260-263
6
Use Trig Functions (RAT)
 e.g. Use a calculator to calculate each of the following
and fill in the blanks.
1. sin 60 = 0.866, so the opposite is 0.866 times as long as
the hypotenuse.
2. cos 28 =
long as the
, so the
3. tan 76 =
long as the
, so the
4. sin 62 =
long as the
, so the
.
.
.
is
times as
is
times as
is
times as
pg 260-263
7
Use Trig Functions (RAT)
1. Draw a triangle 𝐴𝐵𝐶 so that 𝐴𝐶 = 100, ∠𝐴𝐵𝐶 = 90𝑜 ,
and ∠𝐶𝐴𝐵 = 50𝑜 . Find |𝐴𝐵| and |𝐵𝐶|.
2. Draw a triangle 𝐷𝐸𝐹 so that 𝐷𝐸 = 7, ∠𝐷𝐸𝐹 = 90𝑜 , and
∠𝐹𝐷𝐸 = 25𝑜 . Find |𝐷𝐹| and |𝐸𝐹|.
3. Draw a triangle 𝐿𝑀𝑁 so that 𝐿𝑀 = 62, ∠𝐿𝑀𝑁 = 37𝑜 ,
and ∠𝑀𝑁𝐿 = 90𝑜 . Find |𝐿𝑁| and |𝑀𝑁|.
pg 260-263
8
Use Trig Functions (RAT)
 While the trig functions apply to angles and return the
ratio of side lengths, there are inverse trig functions that
apply to ratios and return angles.
 If sin(𝐴) =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
,
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
then 𝐴 =
sin−1
 If cos 𝐴 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
,
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
then 𝐴 =
cos −1
 If tan 𝐴 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
,
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
then 𝐴 =
tan−1
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
 On a calculator, these are typed using e.g. SHIFT+sin
 sin−1 𝐵 is pronounced “inverse sine of B” or “sine minus
one of B”, likewise for cos −1 𝐵 and tan−1 𝐵.
pg 260-263
9
Use Trig Functions (RAT)
 e.g. Given the triangle shown, find 𝜃:
 tan(𝜃) =
50
100
 ⇒ 𝜃 = tan−1
 ⇒ 𝜃 ≈ 26.6𝑜
1
2
pg 260-263
10
Use Trig Functions (RAT)
1. Draw a triangle 𝐴𝐵𝐶 so that 𝐴𝐵 = 9, 𝐵𝐶 = 4, and
∠𝐶𝐵𝐴 = 90𝑜 . Find |∠𝐵𝐴𝐶|.
2. Draw a triangle 𝐷𝐸𝐹 so that 𝐷𝐹 = 13, 𝐸𝐹 = 7.5, and
∠𝐷𝐸𝐹 = 90𝑜 . Find |∠𝐹𝐷𝐸|.
3. Draw a triangle 𝐿𝑀𝑁 so that 𝐿𝑁 = 2.2, 𝐿𝑀 = 1.7, and
∠𝐿𝑀𝑁 = 90𝑜 . Find |∠𝑁𝐿𝑀|.
pg 260-263
11
Use Trig Functions (RAT)
 In real world problems, two addition terms are important.
 When looking up, the angle from the horizontal is called
the angle of elevation.
 When looking down, the angle from the horizontal is
called the angle of depression.
 Additionally, a clinometer is a device used to measure
these angles.
pg 260-263
12
Use Trig Functions (RAT)
1. From the top of a light house 60 meters high with its
base at the sea level, the angle of depression of a boat
is 15 degrees. What is the distance of boat from the foot
of the light house?
2. The angle of elevation of the top of an incomplete
vertical pillar at a horizontal distance of 100 m from its
base is 45 degrees. If the angle of elevation of the top
of the complete pillar at the same point is to be 60
degrees, then the height of the incomplete pillar is to
be increased by how much ?
3. A 10 meter long ladder rests against a vertical wall so
that the distance between the foot of the ladder and
the wall is 2 meter. Find the angle the ladder makes
with the wall and height above the ground at which
the upper end of the ladder touches the wall.
pg 262
13
Use Pythagoras’ Theorem
 Pythagoras’ Theorem states that in a right-angled
triangle, the square of the hypotenuse is equal to the
sum of the squares of the other two sides.
pg 262
14
Use Pythagoras’ Theorem
 Find the missing side length in each of the following
triangles.
1.
2.
3.
pg 264-268
15
Solve Quadrant Problems
 The right-angled triangle definitions of sine, cosine, and
tangent limit the possible angles between 0 and 90
degrees.
 To generalise the trig functions, we redefine them based
on the unit circle.
 The unit circle is a circle on the coordinate plane with
centre (0, 0) and radius 1.
pg 264-268
16
Solve Quadrant Problems
 Measuring the angle
anti-clockwise from the
positive x-axis:
 cos 𝐴 is the x-coordinate
where the radius meets
the circumference,
 sin 𝐴 is the y-coordinate
where the radius meets
the circumference.
pg 264-268
17
Solve Quadrant Problems
 Using the unit circle, complete this table.
𝑨 (degrees)
sin 𝐴
0
90
180
cos 𝐴
tan 𝐴
 Recall that angles measure
anti-clockwise from the
positive x-axis.
 cos 𝐴 is the x-coordinate
 sin 𝐴 is the y-coordinate
 tan 𝐴 =
sin 𝐴
cos 𝐴
270
360
pg 264-268
18
Solve Quadrant Problems
 Draw a unit circle
including axes.
 Draw radiuses at 60o,
120o, 240o, and 300o.
 Find cos 𝐴 for each of the
angles drawn.
 Find sin 𝐴 for each of the
angles drawn.
pg 264-268
19
Solve Quadrant Problems
 Each angle can be
reframed as the smallest
angle made with the xaxis, called the reference
angle.
 For 60o, 120o, 240o, and
300o, this angle is 60o.
 Since cos 60 = 0.5, cosine
of each of the other
angles is either 0.5 or -0.5
depending on what side
of the circle it’s on.
pg 264-268
20
Solve Quadrant Problems
 Recall that cos 𝐴
represents the xcoordinate of the
circumference.
 Thus, it is positive in the
first and fourth
quadrants of the circle.
 Similarly, it is negative in
the second and third
quadrants of the circle.
pg 264-268
21
Solve Quadrant Problems
 Similarly, sin 60 = 0.866, so
the sine of each of the
other angles is either
0.866 or -0.866.
pg 264-268
22
Solve Quadrant Problems
 Recall that sin 𝐴
represents the ycoordinate of the
circumference.
 Thus, it is positive in the
first and second
quadrants of the circle.
 Similarly, it is negative in
the third and fourth
quadrants of the circle.
pg 264-268
23
Solve Quadrant Problems
 Recall that tan 𝐴
sin 𝐴
represents
.
cos 𝐴
 Thus, it is positive in the
first and third quadrants
of the circle
 Similarly, it is negative in
the second and fourth
quadrants of the circle.
 (You can think of it as the
slope of the radius).
pg 264-268
24
Solve Quadrant Problems
 If cos 𝐴 =
 If sin 𝐵 =
1
,
2
1
− ,
2
find two values of 𝐴 if 0𝑜 ≤ 𝐴 ≤ 360𝑜 .
find two values of 𝐵 if 0𝑜 ≤ 𝐵 ≤ 360𝑜 .
 If tan 𝐶 = 3, find two values of 𝐶 if 0𝑜 ≤ 𝐶 ≤ 360𝑜 .
pg 268-269
25
Solve > 360o and < 0o Problems
 Sine, cosine, and tangent
are periodic.
 Each function repeats
itself every 360o.
 e.g. cos 400 =
cos(400 − 360) = cos 140.
 In general,
 cos 𝐴 = cos 𝐴 + 360𝑛
 sin 𝐴 = sin(𝐴 + 360𝑛)
 tan 𝐴 = tan(𝐴 + 360𝑛)
pg 268-269
26
Solve > 360o and < 0o Problems
 Likewise, if we measure
negative angles (going
clockwise instead of anticlockwise), the result is
equivalent to some anticlockwise angle.
 In general,
 cos 𝐴 = cos 𝐴 + 360𝑛
 sin 𝐴 = sin(𝐴 + 360𝑛)
 tan 𝐴 = tan(𝐴 + 360𝑛)
 with negative 𝑛
pg 268-269
27
Solve > 360o and < 0o Problems
 Find:
 tan 495𝑜
 sin 840𝑜
 cos(−120𝑜 )
pg 268-269
28
Solve > 360o and < 0o Problems
 e.g. If cos 3𝐴 = 0.5, find all values of 𝐴 if 0𝑜 ≤ 𝐴 ≤ 360𝑜 .
 cos −1 0.5 = 60𝑜 , so our reference angle is 60𝑜 .
 Cosine is positive in the first and fourth quadrants, so:
𝑛
𝟑𝑨 = 𝟔𝟎 + 𝟑𝟔𝟎𝒏
𝟑𝑨 = 𝟑𝟎𝟎 + 𝟑𝟔𝟎𝒏
0
3𝐴 = 60
⇒ 𝐴 = 20𝑜
3𝐴 = 60 + 360
⇒ 3𝐴 = 420
⇒ 𝐴 = 140𝑜
3𝐴 = 60 + 720
⇒ 3𝐴 = 780
⇒ 𝐴 = 260𝑜
3𝐴 = 300
⇒ 𝐴 = 100𝑜
3𝐴 = 300 + 360
⇒ 3𝐴 = 660
⇒ 𝐴 = 220𝑜
3𝐴 = 300 + 720
⇒ 3𝐴 = 1020
⇒ 𝐴 = 340𝑜
1
2
pg 268-269
29
Solve > 360o and < 0o Problems
 If cos 2𝐴 =
 If sin 3𝐵 =
1
,
2
1
− ,
2
find two values of 𝐴 if 0𝑜 ≤ 𝐴 ≤ 360𝑜 .
find two values of 𝐵 if 0𝑜 ≤ 𝐵 ≤ 360𝑜 .
 If tan 4𝐶 = 3, find two values of 𝐶 if 0𝑜 ≤ 𝐶 ≤ 360𝑜 .