The document discusses trigonometric ratios and functions. It provides information on:
1) Trigonometric ratios in the four quadrants and for angles greater than 360 degrees, including how the ratios change based on the quadrant or whether the angle is an even or odd multiple of 90 degrees.
2) How to solve trigonometric equations involving sin, cos, and tan, including using identities and periodicity.
3) How to draw and analyze simple graphs of the trigonometric functions sin x, cos x, and tan x, including finding intercepts and maximum/minimum values.
This is a basic intoductory unit on trigonometry meant for high school students in geometry. It is aligned to the Common Core States Standards covering right triangular geometry.
This is a basic intoductory unit on trigonometry meant for high school students in geometry. It is aligned to the Common Core States Standards covering right triangular geometry.
It is a ppt on Trigonometry for th students of class 10 .
The basic concepts of trigonometry are provided here with examples Hope that that you like it .!! Thankyou ..!! :)
The PowerPoint presentation is on the "BASICS OF TRIGONOMETRY".
It includes the --
1) Definition of Trigonometry,
2) History of Trigonometry and its Etymology,
3) Angles of a Right Triangle,
4) About different Trigonometric Ratios,
5) Some useful Mnemonics to remember the Trig. ratios,
6) Theorem, which states that --
"Trigonometric Ratios are same for the same angles"
7) Trigonometric Ratios for some specific/ standard angles.
This is a school standard presentation for class 10 students .
It will be very helpful to you all.
Hope you all like this .
And pass your exams with flying colors
It is a ppt on Trigonometry for th students of class 10 .
The basic concepts of trigonometry are provided here with examples Hope that that you like it .!! Thankyou ..!! :)
The PowerPoint presentation is on the "BASICS OF TRIGONOMETRY".
It includes the --
1) Definition of Trigonometry,
2) History of Trigonometry and its Etymology,
3) Angles of a Right Triangle,
4) About different Trigonometric Ratios,
5) Some useful Mnemonics to remember the Trig. ratios,
6) Theorem, which states that --
"Trigonometric Ratios are same for the same angles"
7) Trigonometric Ratios for some specific/ standard angles.
This is a school standard presentation for class 10 students .
It will be very helpful to you all.
Hope you all like this .
And pass your exams with flying colors
Discusses trigonometric functions, graphing, and defining using the Unit Circle. Explains how to convert from radians to degrees, and vice versa. Describes how to calculate arc lengths in given circles.
This is a simple PowerPoint on the properties of Sine and Cosine functions. It was created for a student teaching lesson that I had in the past. Feel free to use and modify! :-)
The data is present below the pictures so as to edit it as per your needs. I wanted to use good fonts and this was the only way i could do it as the fonts would not be available on your computer.
Hasil presentasi TIK Kelompok V:
Oleh:
Febriansyah ramadhan
Muchammad hawari ichwan
Anggit metha
Kelas X Aksel SMAN 1 Jember
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Fungsi dan Cara Penggunaan Terminal (CLI, Gedit, dan VI Editor)Fadhel Hizham
Hasil presentasi TIK Kelompok IV: Oleh: Aulia Rafikasari Fadhel Akhmad Hizham Radis Selfiana Sari Kelas X Aksel SMAN 1 Jember Yang slide 179, 180, 181, 186, 188, 193, 195, 197, 200, 203, 206, 208, 209, dan 210 itu tidak ada gambarnya. Gunakan Libre Office untuk membuka file tersebut
Hasil presentasi TIK Kelompok III:
Oleh:
Annisa Aulia M.S.
Diinar Athika F.
Rosa Pinanda F.
Kelas X Aksel – SMAN 1 Jember
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Hasil presentasi TIK Kelompok I:
Oleh:
HAYATI PUSPAMURTI
NOVAIL ALIF MUHARROM H
ROCHIMA DEWI MAHARANI
Kelas X aksel SMAN 1 Jember
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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