7. Trigonometric Ration on
Quadrants
NOTE:
For quadrant I, all ratio of trigonometric is positive (+)
For quadrant II, only sinus and cosec are positive (+)
For quadrant III, only tan and cotg are positive (+)
For quadrant IV, only cosine and sec are positive (+)
16. Trigonometric Ration of Related
Angle
Angle (n . 90° ± α) for n ϵ real number
For n (even number) consist
Example : sin (180 – α)° = sin (2 . 90 – α)° = sin α
cos (360 – α)° = cos (4 . 90 – α)° = cos α
tan (360 – α)° = tan (4 . 90 – α)° = - tan α
positive (+) or negative (-) based on their quadrant.
17. Trigonometric Ration of Related
Angle
For n (odd number) Function is changed
(complement)
The change are : sin cos
cos sin
tag cotg
Example : sin (90 – α)° = (1 . 90 – α)° = cos α°
cos (270 – α)° = (3 . 90 – α)° = - sin α°
tan (90 – α)° = (1 . 90 – α)° = cotg α°
positive (+) or negative (-) based on their quadrant.
18. Trigonometric Ration of Related
Angle
So, we can conclude that :
90 ± α = change trigonometric function with their complement
180 ± α = the trigonometric function is consist
270 ± α = change trigonometric function with their complement
360 ± α = the trigonometric function is consist
REMEMBER! positive (+) or negative (-) based
on their quadrant.
19. Trigonometric with ANGLE >
360°
K . 90 + α
The Method :
1. Determine K, it’s even or odd. If even the
trigonometry is consistent. If odd, the
trigonometry to be complement of it.
2. K divided by 4, and find the remainder.
3. Then the remainder is the last quadrant that
indicate the quadrant
4. Solve it
28. Trigonometric Equation
A trigonometric equation is any
equation that contains a trigonometric
function.
There are 3 kinds of Trigonometric
Equation:
1. Sinus
2. Cosine
3. Tangent
To solve this problem, we use different formula
for sinus, cosine, and tangent.
29. Problem..
1. Find the value of x that satisfies
trigonometric equation of ,
for 0° ≤ x ≥ 360°
◦ Problem solution: x = A + K.360
x = 30 + K.360
or x = (180-A) + K.360
x = (180-30) +
K.360
x = 150 + K.360
if:
K = 0 x = 30 SS = {30,° 150°}
K = 1 x = 390 (not satisfied)
or K = 0 x = 150
30. Problem..
2. Find the solution set of
, for 0° ≤ x ≥ 360°
Problem Solving 1. K = 0 x = 60
I. K = 1 x = 240
K = 2 x = 420 (NS)
2. K = 0 x = -60 (NS)
K = 1 x = 120
K = 2 x = 300
2.
SS = {60°, 120°,
240°, 300°}
240°,
31. Problem..
3. Find the solution set of
, for 0° ≤ x ≥ 360°
Problem solution
K=0
K =1 (NS)
SS = {
}
32. SINUS
In degrees (°) A is on
A is on
◦ sin x = sin A quadrant I Iand
quadrant and
II, so the sin
II, so the sin
x = A + K . 360 always (+)
always (+)
◦ or x = (180-A) + K . 360
In radian (π)
◦ sin x = sin A
x = A + K . 2π
x = (π-A) + K. 2π
K is integer
(K = + 1, + 2, + 3, + 4, ….)
33. COSINE
In degrees (°)
◦ cos x = cos A
x = A + K . 360
◦ or x = -A + K . 360
In radian (π)
◦ cos x = sin A
x = A + K . 2π
x = -A + K. 2π
K is integer
(K = + 1, + 2, + 3, + 4, ….)
34. TANGENT
In degrees (°)
◦ tan x = sin A
x = A + K . 180
In radian (π)
◦ tan x = sin A
x=A+K.π
K is integer
(K = + 1, + 2, + 3, + 4, ….)
35.
36. Trigonometric Function
Real Number set or its section set.
Trigonometric function value is value
from the function for each given value
x.
e.g: f(x) = sin x
f(x) = cos x
f(x) = tan x
Based on this function, we can make a
simple graph easily.
37. Trigonometric Function
Example:
If function f is defined by f(x) = sin2 x – cos2 x, determine
value x causing function f intersecting X-axis!
• Problem Solving:
function f intersects X-axis b) sin x + cos x = 0
if f(x) = 0 sin x = -cos x
sin2x – cos2x =0 = -1 tan x = -
(sin x – cos x)(sin x + cos 1
x) =0
a) sin x – cos x = 0 x = 135° and 315°
sin x = cos x So, the solution set is:
= 1 tan x = 1 {45°, 135°, 225°, 315°}
can causing function f has
x = 45° and 225° zero value
38. Trigonometric Function Graph
Simple Trigonometric Function Graph
is a geometric description of a
trigonometric function.
This graph can make us easily to
analyze the value of the
function, types of functions, etc.
39. Trigonometric Function Graph
The simple trigonometric function
graph:
1. The graph of function f(x) = sin x, for 0 ≤
x ≥ 360
2. The graph of function f(x) = cos x, for 0
≤ x ≥ 360
3. The graph of function f(x) = tan x, for 0 ≤
x ≥ 360
40. Drawing a Simple Graph
How to make it?
1. Determine the intersect point of x-axis
2. Determine the intersect point of y
3. Find vertex point (max and min)
4. Draw that graph
41. Problem..
If the function of trigonometric is f(x) =
sin 3x, for 0° ≤ x ≥ 360°. Draw the
graph of that function!
i. Find the intersect point of x-axis y = 0
ii. Find the intersect point of y-axis x = 0
iii. Find the vertex point (max-min)
iv. Draw that graph
42. Problem..
i. f(x) = sin 3x
y = sin 3x K=0x=0
K = 1 x = 120
0 = sin 3x K = 2 x = 240
sin 3x= sin 0 K = 3 x = 360
3x = 0 + K.360
K = 0 x = 60
x = 0 + K.120
K = 1 x =
or 3x = (180-0) + 180
K.360
K = 2 x =
3x = 180 + K.360 300
x = 60 + K.120
Intersect point x-axis are
{(0°,0), (60°, 0), (120°, 0), (180°, 0), (240°, 0), (300°,0), (360°, 0)}
43. Problem..
ii. y = sin 3x K = 0 x = 90°
Intersect point K = 1 x =
y = sin 0 y-axis = (0°,0) 210°
K = 2 x =
y=0 330°
Min
iii. Vertex point for y = sin 3x the max point is
Max 1
-1 = sin 3x
for y = sin 3x the max point is 1 sin 3x = sin 270°
1 = sin 3x K = 0 x = 30° 3x = 270° + K.360
K = 1 x = x = 90° + K.120
sin 3x = sin 90° Min = {(90°,-
150°
3x = 90° + K.360 K = 2 x = 1),
(210°, -1),
x = 30° + K.120 270°
(330°, -1)}
Max = {(30°,1),
(150°,1), (270°,
1)}
44. Y = sin 3x Y = 3sin ( x + 30 0 )
1
SHIFT GRAPH
TRIGONOMETRIC
FUNCTIONS
90 0 210 0 330 0
30 0 60 0 1200 1500 1800 240 0270 0300 0 360 0
0
-1
a = sum of wave If sin was replaced by cos
Y = a sin (x+ α) Sin = shape of graph or tan it can change the
α = shifting shape of graph.
45. Problem..
If given trigonometric function f(x) = 2 cos
(x-30), draw the graph of that function!
Problem solution:
i. Find the intersect point of x-axis y = 0
ii. Find the intersect point of y-axis x = 0
iii. Find the vertex point (max-min)
iv. Draw that graph
46. Problem..
i. f(x) = 2 cos (x-30)
y = 2 cos (x-30)
0 = 2 cos (x-30)
0 = cos (x-30)
cos (x-30) = sin 90
x-30 = 90 + K.360 K = 0 x =
120
x = 120 + K.360
or K=0
x-30 = -90 + K.360 x = -60
K = 1 x =
x = -60 + K.360 300
Intersect point x-axis are
{(-60°,0), (120°, 0), (300°, 0)}
47. Problem..
Intersect point
y-axis = (0°, √3)
ii. y = 2cos (x-30)
y = 2cos (-30) *cosmin-cosplus
K = 0 x =
y = 2cos 30 210°
y = 2. ½√3 y = √3 Min
for y = sin 3x the max point is -2
iii. Vertex point -2 = 2cos (x-30)
-1 = cos (x-30)
Max cos (x-30) = cos 180°
for y = 2cos (x-30) x-30 = 180 + K.360
x = 210° + K.360
the max point is 2
K = 0 x = 2 = 2cos (x-30)
Min =
30°
1 = cos (x-30) {(210°, -2),
cos (x-30) = cos 0°
Max =
x-30 = 0 + K.360 {(30°,2)}
x = 30° + K.360
56. Sinus Rules C
b a
A B
D
It can be used if
there’re given 2 angles,
and a side
57. Problem..
In Triangle ABC, given c = 6cm, B =
600 and C = 450. Find the length of b!
• Problem Solving: 0
1
b c 2 3 6
b
1
SinB SinC 2 2
b 6 6 3 2
b
Sin600 Sin450 2 2
b 6 6 6
1 b 3 6
2 3 1 2
2 2
59. Problem..
In Triangle ABC, given a = 6, b = 4
and C = 1200. Find the length of c!
Problem Solution:
c2 = a2 + b2 – 2.a.b.cos C
c2 = (6)2 + (4)2 – 2.(6).(4).cos 1200
c2 = 36 + 16 – 2.(6).(4).( – ½ ) C
c2 = 52 + 24 120 0
b=4 a
c 2 = 76
c = √76 = 2√19 A c= B
6
60.
61. C
Triangle’s Area
b a
A B
2 Sides and Measure 1 of c D
its Angel are Determined.
2 Angles and Measure 1
of its side is Determined.
3 Sides
62. Problem..
In triangle ABC given A = 1500 b = 12
cm and c = 5 cm, so the area of triangle
ABC is..
Problem Solution :
C
Area of ABC = ½ b c sin A
b=12 a
= ½ (12) (5) sin 1500
A 1500 = ½(12) (5) sin (1800 –
300)c= B
5
= ½ (12) (5) sin 300
= ½ (12) (5) ½
63. Problem..
Find the area of ABC Triangle if given
a = 10, b = 12, and c = 14
Problem Solution:
C
b=12 a=10
A
c=14 B
64. Problem..
Find the area of triangle ABC if given a
= 5, A = 60°, C = 75°.
Problem Solution:
C
b 75° a=5
60° 45°
A
c B
