trigonometry - elsd.ssru.ac.th
TRANSCRIPT
Trigonometry Teacher: Sunisa Pengmanee (SDSSRU) Grade 9 1. Trigonometric Ratio (Reference from TRIGONOMETRY, MICHAEL CORRAL)
1.1 Trigonometric Functions of an Acute Angle
Consider a right triangle △ ABC, with the right angle at C and with lengths a, b, and c, as in the figure on the right. For the acute angle A, call the leg BC its opposite side, and call the leg AC its adjacent side. Recall that the hypotenuse of the triangle is the side AB. The ratios of sides of a right triangle occur often enough in practical applications to warrant their own names, so we define the six trigonometric functions of A as follows:
Table 1.1: The six trigonometric functions of A
ฉาก
tE
ชด
ขาม① ก
② าฉาก
③ ขาม⇒
1.1 cosec ฉาก_
ขาม
2. 1 %
3. 1ชด¥ม
We will usually use the abbreviated names of the functions. Notice from Table 1.1 that the pairs sin A and cosec A (csc A), cos A and sec A, and tan A and cot A are reciprocals:
Example 1.1: For the right triangle △ ABC shown on the right, find the values of all six trigonometric functions of the acute angles A and B.
Solution: The hypotenuse of △ ABC has length 5. For angle A, the opposite side BC has length 3 and the adjacent side AC has length 4. Thus:
sin A = = cos A = = tan A = =
cosec A = = sec A = = cot A = = For angle B, the opposite side AC has length 4 and the adjacent side BC has length 3. Thus:
sin B = = cos B = = tan B = =
cosec B = = sec B = = cot B = = Notice in Example 1.1 that we did not specify the units for the lengths. This raises the possibility that our answers depended on a triangle of a specific physical size. For example, suppose that two different students are reading this textbook: one in the United States and one in Germany. The American student thinks that the lengths 3, 4, and 5 in Example 1.5 are measured in inches, while the German student thinks that they are measured in centimeters. Since 1 in ≈ 2.54 cm, the
students are using triangles of different physical sizes (see Figure 1.1.1 below, not drawn to scale).
opposite hypotenuse
3 5
adjacent hypotenuse
4 5
opposite adjacent
3 4
hypotenuse opposite
5 3
hypotenuse adjacent
5 4
adjacent opposite
4 3
opposite hypotenuse
4 5
adjacent hypotenuse
3 5
opposite adjacent
4 3
hypotenuse opposite
5 4
hypotenuse adjacent
5 3
adjacent opposite
3 4
t I I
sin
sin A ใน 5in Bหณ €× ↳ ①[อห
b vป③ ¥
แบบฝกหด 1 1. จงเตมชอยอลงบนรปสามเหลยมมมฉากทกาหนดใหตอไปน ตามตาแหนงมมหลก
1) C 2) X 3) R
ฉาก ขาม
A ชด B Y Z P Q
4) 5) A 6) C
F
D E C B A B
7) 8) 9) R
P Q
10) C 11) F 12)
Q R P R P Q
A B D E
a
60D
30D 60D
Practice 1 hyp
. opp.
adj.
1. Define the three side of a right-angled triangle below with a mark angle. Label the hypotenuse, adjacent side and opposite side of the triangle.
k ต•
ฒ
c.
A B
หน7opp . ทฑ
opp
adj adj
""" adj หท adj opp↳P
opp opp adj
Practice A
Exercises 1: Find the values of all six trigonometric functions of angles A and B in the right triangle △ ABC 1. a = 5, b = 12, c = 13
sin A = .............. cos A = .............. tan A = .............. cosec A = .............. sec A = .............. cot A = .............. sin B = .............. cos B = .............. tan B = .............. cosec B = .............. sec B = .............. cot B = .............. 2. a = 8, b = 15, c = 17
sin A = .............. cos A = .............. tan A = .............. cosec A = .............. sec A = .............. cot A = .............. sin B = .............. cos B = .............. tan B = .............. cosec B = .............. sec B = .............. cot B = .............. 3. a = 7, b = 24, c = 25
sin A = .............. cos A = .............. tan A = .............. cosec A = .............. sec A = .............. cot A = .............. sin B = .............. cos B = .............. tan B = .............. cosec B = .............. sec B = .............. cot B = .............. 4. a = 20, b = 21, c = 29 sin A = .............. cos A = .............. tan A = .............. cosec A = .............. sec A = .............. cot A = .............. sin B = .............. cos B = .............. tan B = .............. cosec B = .............. sec B = .............. cot B = .............. 5. a = 9, b = 40, c = 41 sin A = .............. cos A = .............. tan A = .............. cosec A = .............. sec A = .............. cot A = .............. sin B = .............. cos B = .............. tan B = .............. cosec B = .............. sec B = .............. cot B = ..............
2
mmmฑl 5in B ะ { cos B = { tan A = {๐ sina.EE#-tanA ะ 95
B¥ ศ %
ณใน ฑ ¥] 135
13_
13_
5_
12 5 12
C)A
12 C
⑤ ¥ ¥ ร⇒ B
17 17_
IIEน 5 15 8
17I ะ 15_
8×50 17 17 8
เ± ± ะ15 8 15 d
C)A 15
¥ 24_
± Btam 25 25 24
+๐,2¥ E ¥ 25 2424_
± ⇐
25 25 7
2 25- ±
24 7 24 A ๆC
B⑧ ② ห ห
99+50 % ฑ % 29 20
ม 2°- 21
2929 29
29②อ
มา 5° 21 A 21C
259 40 1 B
tyh 4T 4041_
41 ④9 IO 9 4 1 940 9- 4°
4T 41 T4+ ¥ ¥อ40
A 40 C
6. a = 1, b = 2, c =5 sin A = .............. cos A = .............. tan A = .............. cosec A = .............. sec A = .............. cot A = .............. sin B = .............. cos B = .............. tan B = .............. cosec B = .............. sec B = .............. cot B = .............. 7. a = 1, b = 3 c = ? By the Pythagorean Theorem, the length c of the hypotenuse is given by sin A = .............. cos A = .............. tan A = .............. cosec A = .............. sec A = .............. cot A = .............. sin B = .............. cos B = .............. tan B = .............. cosec B = .............. sec B = .............. cot B = .............. 8. a = 2, b = 5 c = ? By the Pythagorean Theorem, the length c of the hypotenuse is given by sin A = .............. cos A = .............. tan A = .............. cosec A = .............. sec A = .............. cot A = .............. sin B = .............. cos B = .............. tan B = .............. cosec B = .............. sec B = .............. cot B = .............. 9. a = 5, b = ? c = 6 By the Pythagorean Theorem, the length b of the hypotenuse is given by sin A = .............. cos A = .............. tan A = .............. cosec A = .............. sec A = .............. cot A = .............. sin B = .............. cos B = .............. tan B = .............. cosec B = .............. sec B = .............. cot B = ..............
<
0
0
0
b2= cำ 92
คำ =36-25B
b2= q
b =3
E Z I6 b 3 6566
_ ZF 3 5Z 5 Z6 I 5
§ ¥ §A b =3
d
3
3. ถา 5sin13
A = จงหาคาของ cos , tanA A
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4. กาหนดให sin 0.6A = จงหาคาของ cos A และ tan A
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5. กาหนดให tan 0.75θ = จงหาคาของ sin , cos , cot , secθ θ θ θ และ cosecθ
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6. ถา 13 sin 5A = จงหาคาของ cos , tanA A
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If , find the value of cos A, tan A
If
If , find the value of and
If 13 , find the value of cos A, tan A
m mmmmos
mm
mmmmm-ttota-tnsroa-T-RTRTnsro-o0-R-tnto-rtr-stoo.tt/RTttttttsoRx.sam mmmmmm
*"
cำaำ b2 -① Sin A = E ขศก
+5 bว = ย- a.2- ③
Emท๐ B b2
= ย - a.2
ผ = อ- นำ - ③ cosA = { %= 132-52 นก
"ca5 = 169-25 cos A = t +9nA = นะ บน
Yb r b2 = 144A b = 12C b = 12 tan A = #
kitty
+5๐ sin A-- อ . เB
cos A = ¥=
% ci 5 3 ะย tan A ะ Z
=3_
X r4
5 A ⑤ 4 C
หว๐ tan อ = อ .75B sin 0 = ย = § coseco =ผ = ว
= 5ๆam ÷ 25100 ÷25 ย a=3 eos0 =ฐ= 4g sec0 =§ = I
tant = 34
F # r tan อ =3_= 9 cot0 ะ1=4
_
= ขามA b = 4
C4 b a 3tano = E-
b ชดย↳๐
13 sin A = 5 b2= ย - a2 Cos A = ¥ =¥
×5๐ 13 × sinA = 5 = 132- 52
sin A = 5_
= 169-25 = }13 52 = 144B
b = 12 tan A = £ = ขาม13 b ด
5=
5)วA b = 12
C
5
แบบฝกหด 2
1. ถา 3tan4
A = จงหาคาของ sin cos cotA A A+ −
C จากรปจะไดดาน AC = 5
3 4 4sin cos cot5 5 3
A A A+ − = + −
9 12 2015
+ −=
115
=
2. ถา 17sec15
A = จงหาคาของ
(1) 2 2sin cosA A+
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(2) 2 21 tan secA A+ −
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(3) 2 21 cot cosA ec A+ −
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5 3 A
A 4 B
Practice 2If , find the value of sin A + cos A - cot A
From picture,
If , find the value of ....
m mmmmmommmm
ฑ mm
16
B
Sin A = tnCSCA
COSA = tsecASecA =
1g Cos A
=# tan A = tnCOSA 17 cot A
B 2
17 g
ย = อ- b2 siทำ- + cos"A = (ภ +¥)C a
ย = 172-152= 289-225 = 64 + 225
gnnrn
X r a = 64 289 * siทำ +องA = 1A C
ธ 5 a ะ 8 SinAtcos A = 28cg = 1
289
ะ 1 +§2-
(กม = 289-289าn_
→ 2ฐ = 289_
225 225
น5 ะ 1 + 64-289 = อ ¥ 1 ttan A = sec"A
×225 E-¥ 225
นะ 225+64_
-289 = ๐
225
* 1 + cot 2A = cos.ec?A
= 1 + ( ศ )2- (ก2
= 289 → 2
¥ = 2¥64
= 1 + 225 - 289 = 0 ⇒ 1+ cot2A = cosec2Aneneen
-
64 64 64
# 64L 64+225-289 = 0=nnnes
64
6 3. กาหนดให 5sin 3A = จงหาคาของ
(1) cos sin tanA A A⋅ ⋅
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(2) 2
2
cos1 cot
ec AA+
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4. ถา 2 tan 1θ = จงหาคาของ 10sin 7cos4sin 3cos
θ θθ θ++
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5. ถา 5tan12
θ = เมอ 0 90θ< <D D แลวจงหาคาของ 4sin 3cos4cos 4sin
θ θθ θ+−
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Define , find the value of ....
If , find
If when , find the value
mm mm
m mm
m m mmm
¥ ะ E B ๗H
sin A = Z C 5 3 a=ht 5
7 rA
b.-4C
sin A ะ นะ ยะ ศ CosA × 5in A × tan A
COSA ะ กะ # ะ 45=
§×
45 x }tan A ะ ศ =
g= ๆ =
µ
61- =
นย§#a
= ง | cot A=# =¥
tan A ะ Z4
2
112 =# = 1
9+16-9* cosec
"A
=
¥ קำ+แมน
= 1 0rcoseยAะา+ cot2A
2 tan อ = 1 → 2 tan A = |tan 0 ะ 1
มB
AB2= ว +22 ะ (เอ × ¥ ) +17× ศ ) =
2ศ.no
fy 1 ABวะ 1+4 14 × g) + (3× E)
= § ¥AB = Fb
0\rc sinอะ¥ -
=
¥ ק = ¥ = ๆA 2cos0 = 2
A
BAB? 52+122
= 25+144 = (4 ×ศ ) + (3 × %) =20-361.mn13
AB2-- 169 ( 4 × %) - (4 × %) 48-20
"5 AB ะ 13 # 2
13-
sin อะ ± = *e) เ3 5
× EmA 12
Ccoso = ท = 2
13
7 6. ถา 5sin 4A = จงหาคาของ
(1) cos sec tan cotA A A A+
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(2) 2 2sin cos
sin cos cos secA A
A ecA A A+
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7. ถา 3cot4
A = จงหาคาของ 5sin 3cossin 2cos
A AA A−+
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8. ถา 2tan3
A = จงหาคาของ ( )( )
cos sec cossin sec sin
A A AA co A A
−−
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If , find the value of....
If , find
If , find
k mmm
แ mm
ห า
y18
sin A-_ 45
Bcosm} ใน = =
5 4¥§ ฑ4¥%)= 2
A ง " (อ# + ห ^ #)¥
ะ
(g)"
1 ถ =
¥5 £5
¥¥× 3
ง ,= 2
5.25= t
cotA =3_
4
→ tan A-- §B
54
= ( 5× ±5) - (3×35) = 1
¥×§)r ¥ + ( 2 × } )
A3
C
=2÷ _ § = I
sin A = ±-
เอง¥ + ±5COSA =
Z5
13 A Bำม+ ำ = 3in4ns§ (ญ - }) 13×3
= 4 t 9 _
aren
-
2 × 9ร" 2 AB"= เ3 } ( ข -ก ) 2×13
.
) AB = ขง = ฏ4×ษ2A 3 C
=
ๆ ( เ3µ9ns )/3ก6
sin A = ก CosA=§_3ps=
Coseca = ๆ seca=ๆ%( ง#) ศ =
¥
Example 1.2: Find the values of all six trigonometric functions of 45◦ Solution: By the Pythagorean Theorem, the length c of the hypotenuse is given by
Thus, using the angle A we get:
sin 45◦ = = cos 45◦ = = tan 45◦ = = = 1
cosec 45◦ = = sec 45◦ = = cot 45◦ = = = 1
Example 1.3: Find the values of all six trigonometric functions of 60◦ Solution: By the Pythagorean Theorem, the length c of the hypotenuse is given by
Thus, using the angle A we get:
sin 60◦ = = cos 60◦ = = tan 60◦ = = = 3
cosec 60◦ = = sec 60◦ = = = 2 cot 60◦ = = Notice that, as a bonus, we get the values of all six trigonometric functions of 30◦, by using angle
B = 30◦ in the same triangle △ ABC above:
sin 30◦ = = cos 30◦ = = tan 30◦ = = = 1
cosec 30◦ = = = 2 sec 30◦ = = cot 30◦ = = = 3
opposite hypotenuse
1 2
adjacent hypotenuse
2 1
opposite adjacent
1 1
hypotenuse opposite
2 1
hypotenuse adjacent
1 2
adjacent opposite
1 1
sin 45◦ = cos 45◦ =
opposite hypotenuse
3 2
adjacent hypotenuse
2 1
opposite adjacent
1 3
hypotenuse opposite
2 3
hypotenuse adjacent
1 2
adjacent opposite
3 1
opposite hypotenuse
1 2
adjacent hypotenuse
2 3
opposite adjacent
3 1
hypotenuse opposite
2 1
hypotenuse adjacent
3 2
adjacent opposite
1 3 คน
Example 1.4: A is an acute angle such that sin A = . Find the values of the other trigonometric functions of A. Solution: By the Pythagorean Theorem, the length c of the hypotenuse is given by
Thus, using the angle A we get:
sin 90◦ = = cos 90◦ = = tan 90◦ = =
cosec 90◦ = = sec 90◦ = = cot 90◦ = = You may have noticed the connections between the sine and cosine, secant and cosecant, and tangent and cotangent of the complementary angles in Examples 1.5 and 1.7. Generalizing those examples gives us the following theorem: Theorem 1.2. Co-function Theorem: If A and B are the complementary acute angles in a right triangle
△ ABC, then the following relations hold: sin A = cos B sec A = cosec B tan A = cot B sin B = cos A sec B = cosec A tan B = cot A We say that the pairs of functions {sin, cos}, {sec, cosec}, and {tan, cot} are co-functions. So sine and cosine are co-functions, secant and cosecant are co-functions, and tangent and cotangent are co-functions. That is how the functions cosine, cosecant, and cotangent got the “co” in their names. The Co-function Theorem says that any trigonometric function of an acute angle is equal to its co-function of the complementary angle.
opposite hypotenuse
2 3
adjacent hypotenuse
3 5
opposite adjacent
5 3
hypotenuse opposite
3 2
hypotenuse adjacent
5 3
adjacent opposite
2 5
2 3
5
Memorization Trick!
Trigonometric Hand Trick This is an easy way to remember the values of common values of trigonometric functions in the first quadrant. It’s a lengthy explanation, but once you know this by heart, you can use this trick
for all four quadrants. All you need is your non-dominant hand.
Step 1: Hold out your non-dominant hand.
Step 2: “Assign” the following values to your fingers.
If your non-dominant hand is your left hand…
If your non-dominant hand is your right hand…
Step 3: Find a trig problem.
For example: cos (π/6)
Step 4: Hold down the finger assigned for that angle.
For example: Hold down your ring finger for π/6.
Step 5: Know the following formulas.
sin 𝜃 = √𝑏𝑜𝑡𝑡𝑜𝑚 𝑓𝑖𝑛𝑔𝑒𝑟𝑠
2 cos 𝜃 =
√𝑡𝑜𝑝 𝑓𝑖𝑛𝑔𝑒𝑟𝑠2
tan 𝜃 = √𝑏𝑜𝑡𝑡𝑜𝑚 𝑓𝑖𝑛𝑔𝑒𝑟𝑠
√𝑡𝑜𝑝 𝑓𝑖𝑛𝑔𝑒𝑟𝑠
“Bottom fingers” refer to how many fingers are “below” the finger you’ve held down. “Top fingers” refer to how many fingers “above” the finger you’ve held down. Your thumb counts.
Step 6: Calculate the values for your trig expression using the appropriate formula.
For example: When you hold down your ring finger, there is 1 finger below your ring finder (your pinkie), and there are 3 fingers above your ring finger (your thumb, your index finger, and
your middle finger.) Therefore, cos (π/6) = √32
. If you need sine, sin (π/6) = √12
= 12.
π/2
π/3
π/4
0
π/6
π/2 π/3
π/4
0
π/6
6
การหาคาตรโกณมต sin , cos และ tan โดยใชโมเดลฝามอ
พบนวโปง
นว 0q
ดานซายเหลอ 0 นว และ ดานขวาเหลอ 4 นว
sin = √
=
= 0
cos = √
=
= 1
tan = ซายขวา
= √ √
=
= 0
พบนวช
นว 30q
ดานซายเหลอ 1 นว และ ดานขวาเหลอ 3 นว
sin = √
=
= 0.5
cos = √
=
= 0.8660
tan = ซายขวา
= √ √
= √
=
= 0.5774
พบนวกลาง
นว 45q
ดานซายเหลอ 2 นว และ ดานขวาเหลอ 2 นว
sin = √
= √
× √ √
= √
= √
=
= 0.7071
cos = √
= √
× √ √
= √
= √
=
= 0.7071
tan = ซายขวา
= √ √
= 1
0
30
45
Trigonometric Hand Trick
RL
RL
RL
tmmmtssgrapoorangnoroaooooo+
มา
MMMMMtmwss.am
-+ฒ
mmmmmm•
.mtmmmmmhtts
ฒ•
าn-←
7
⇒ฐ 8ออ-
2
siin อ ¥ ×
2- tan =§
sifน {~°5
0
-
20
พบนวนาง
นว 60q
ดานซายเหลอ 3 นว และ ดานขวาเหลอ 1 นว
sin = √
=
= 0.8660
cos = √
=
= 0.5
tan = ซายขวา
= √ √
= √
= √ = 1.7321
พบนวกอย
นว 90q
ดานซายเหลอ 4 นว และ ดานขวาเหลอ 0 นว
sin = √
=
= 1
cos = √
=
= 0
tan = ซายขวา
= √ √
=
= หาคาไมได
จากการค านวณผานโมเดลฝามอเขยนอตราสวนตรโกณมตของคา sin, cos และ tan ลงในตารางไดดงตอไปน
องศา มม 0q 30q 45q 60q 90q
sin 0
= 0.5 √ = √
= 0.7071 √
= 0.8660 1
cos 1 √
= 0.8660 √ = √
= 0.7071
= 0.5 0
tan 0 √
= 0.5774 1 √ = 1.7321 หาคาไมได
60
90
RL
RL not defined
csc
sec
cot
""ญ" อ ."
ภญต 00 ไมนยาม
๏
9
แบบฝกหด 3 1. จากรปจงหาอตราสวนตรโกณมตของมมตอไปน
(1) sin 30D = 12
(2) cos 30ec D = ………………. (3) cos30D = ……………….
(4) sec30D = ………………. (5) tan 30D = ………………. (6) cot 30D = ……………….
(7) sin 60D = ………………. (8) cos 60ec D = ………………. (9) cos60D = ……………….
(10) sec60D = ………………. (11) tan 60D = ………………. (12) cot 60D = ……………….
2. จากรปจงหาอตราสวนตรโกณมตของมมตอไปน
(1) sin 45D = ………………. (2) cos 45ec D = ………………. (3) cos 45D = ……………….
(4) sec45D = ………………. (5) tan 45D = ………………. (6) cot 45D = ……………….
3. จงหาคา sin 60 cos30−D D
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4. จงหาคาของ sin 45 cos 45 sin 30 cos60−D D D D
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60D
30D 2
3
1
1
1 2
45D
Practice 2
Practice 3Find the trigonometric ratios from the picture below.
Find the trigonometric ratios from the picture below.
Find
Calculate
mmammoasatanmmtmmpommmttasนา
ttthtthrt.TT/fTftTMf'RTRf'Rtrgng.go-stTMe 20ย
sin A = E sinB = 5_
B c c
sin A=ฬก C05A = น COSB = £c c
COSA ะ ชด c a
- tana = q tanB-_ นจาก
Abc b a
tan A-. ขาม_
ชด ๆ = 2โ2
215ง ¥ =งงม 1T ง 2
ว = 2 ¥ โ3 งB sin = E
C
a cos = ยA
b ctan = ย
b
sin เอะ ข3_
2
Cos แ°
= ๆ }= SIท 6อ - cos 3อ
= ร - ๆปท = 0
ยน sina÷ ⇐ ยญฏ={ฬษCOS 45
°
= E.2
sin 3อ°
= E
= ม-
#2
= 1
cos 6อะ E Tt2
10 5. จงหาคาของ 2 2tan 60 tan 30−D D
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6. จงหาคาของ 2 2 2sin 60 cos 30 tan 45+ −D D D
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7. จงหาคาของ 2 2 2tan 60 4cos 45 3sec 30+ +D D D
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8. จงหาคาของ 2 2 2 24 1 13tan 30 cos 30 sec 45 sin 603 2 3
+ − −D D D D
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Calculate
Calculate
Calculate
Calculate
mm
wmn
mm
mm
21
④
2 2
tan เอ - (กณmญ | ห -i= ๆ×3- F×3
ะ
g- } -
- g
นายสน
sin เอา = E ะ (ถ + ( ฑ - (นำ2 J"" "
"
= ¥2 | = ¥ #
- งtan 45- = ข = E- } = } - }รด
= 0
- COS 30°
tan 60ะฐญญt %) + ( 3 ×↳)'
)ะ 3 + (*×# + [3×#= 3+2 + 4 = 9
tan 30 = E
r3.im#lHmi)-Hx()C0S30 = งวะยง × ;) + ( ↳ ×*4)
% หญ ] = ะ + า - า - ะ4
sin เอ = I = 1-t
2 4
= 4# = #
11 9. จงหาคาของ 2 2 22cos 45 4sin 45 3cos 60ec+ +D D D
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10. จงหาคาของ cos 60 sec30 cot 60 cos0ec D D D D
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11. จงหาคาของ sin 45 tan 45cos 45
+D
DD
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12. จงหาคาของ 2 2 2 2sin 30 sin 60 tan 60 cot 45+ + +D D D D
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Calculate
Calculate
Calculate
Calculate
mm
mm
m
mm
@22
→sin เอ
:ห°
.. กณ = แ ×↳)
'
1 + ( 4 × (ฑ 1 + (ง ×( :)ำ2
sin เอ = E = # × g) + (*× E) + (5×#Cosec เอ = ญ สม f
B = 1 + 2+4 = 7
sin เอะ E cosec แ°
= ม = cosec แ°
sec 3อ cot 6อ cosอ2 ข3
COS 3อะ ข3_
sec 3อ = ม
=
ศ × ศ × ต × 1
2 B=
tan เอ°
= E cot เอ°
ะ E %¥ขา
= ฐ3 = 4งC050
°
= ฐ = ๆ = 1 ว ←3= 4ศ
tan 45°
= SIn_45.COS 45
°
= sin45.mnCOS45
อ
+9ท 45°
ะ tan 45°
+ tan 45°
→ = 2tan 45°
= § + ข = 2 (E)กด
= 1 + 1 = 2= 2×1 = 2
ptan 45°
sin 3อ = E :(ก + (ญ +(ฑ + (ฑ2
sin แ°
= ขว2 # + } + 3 + 1
tan เอ°
= E = 4_
+4ขา 4
tan 45°
= ม = 1 +4
ข2Cot 45
°
ะ rg= 5
รด
8
9. ในรปสามเหลยม ABC ม l 90B = D ให 5sin13
C = จงหาคาของ tan cotsec cos
C AC ecA
++
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10. กาหนดให 3tan7
A = จงหาคาของ 2 2
2sin cos cos1 sin sin cos
A A AA A A
−− + −
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11. ถา 5 tan 4θ = จงหาคาของ 5sin 3cossin 2cos
θ θθ θ
−+
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12. ถา sec 4θ = จงหาคาของ cotθ
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Define triangle , and , find
Define , calculate
If , calculate
If , calculate
Bmoattn • ก มคsนน
แ นเ
m mm
m mm
q19
ขามฉTก
A tanc = I % + นะห132=52+ c.52
12 1 2
13 ห 5 [132 = 132-52 cosc = 12
=
ม=¥ ×¥
-
12t r CTำ = 169 - 2 5
13 = #C 12 B EB
"= 144 gin Aะ 12
26
CB = 12 tan A-- ห13 =±5 ขาม 13
B A 132 = Aย+B c.2-
ด= §
= 2ๆ+32 ( 2 × ¥ × ศ 8) - % = 42-7ข58
งาน8 3 A Pำ= 49+9 58oneeA 13
วะ 58 ( เ - น +(นำ - (นำ 58-3โ8+9-495_ne
a % rd งนำศโ = 4÷ - ¥% | = 7 §ร-ำ× 5สอง×58 ×ข58-0
f- 18-3โ58
!¥×ณ ¥8""ศ = ฑส.nl = §-
= ๆ (6-โ/58)
-42-7โ58 =
3C b-/คง
18-3โ587 (6- ข58) 3 ( 6- F)8)
sec① = 4_
Bc? 42 ำ1
= 16-1
Cos 0 = t 13ำ = 15B 4
A
#? *อtอ =
↳→= ¥
เC tan อะ ขา
4 7. B กาหนด ∆ ABC ใหดงรปจงหา
(1) lsin BAD
(2) lsin ABD
(3) lcos BCD
(4) lcos CBD
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8. ให ABC เปนรปสามเหลยมดานเทา ดงรป จงหา
(1) lsin ABD
(2) lcos BCD
(3) ltan CBD
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9. จงหาคาอตราสวนตรโกณมตจากตาราง
(1) sin 5D = ………………. (2) sin18D = ………………. (3) sin 36D = ……………….
(4) sin 45D = ………………. (5) sin 54D = ………………. (6) sin 72D= ……………….
(7) cos10D = ………………. (8) cos 20D = ………………. (9) cos 24D = ……………….
(10) cos37D = ………………. (11) cos75D = ………………. (12) cos80D = ……………….
(13) tan 9D = ………………. (14) tan12D = ………………. (15) tan 28D= ……………….
(16) tan 42D = ………………. (17) tan 56D = ………………. (18) tan 76D= ……………….
A D C
12 15
14
B
A 2 D C
4
Define , find....
Define is the Equilateral triangle, find....
Find the trigonometric ratios below by using trigonometric table.
Trigonometric tableScan Here!!
mm mm
mm mmmmmmm
mmmmmmmmnsa
13Em④axSใdY + เอ๐ 43µวa po
15
ไCไ A)5 D
b
BC
×
14-9ะ 5วน qa 12
2 งD 9 C
b
① DC2=13อ - BD2
② AB2= AB+131ำ 1) Sin BกD= [ = 12
= 152-122 = 52+122 c 13
= 225-144 = 25+144 2) sin A B^D = ¥ = ศDย = 81 A 132=169DC = 9 AB ะ 13 3) COS BอD= ฐ ะ ¥
4) Cos CID = ยะ ¥B
4 x.2ง ผ = <ำ b
2
xpn a4
\ a= 42-22
น
2 = 16-4
0 แ 4 A วาง ย = 12 → a. =ม31- 4 -1 B
1) sin B⑤ = ร =ทIะฐ 2ง+ 4
sin AB^D = ฐ = ม =
g าf
cD 2
2)_COS 13อD= ม = ท
3) tan 4วD%
= ง
tan R ะ นะ }Sin P = จะ ±5
COKCP =§sin R ะ % = } §ญ3
¥ = } ×÷ = ศ
¥× ณ1- % ะ 4# ะ 24
A SIN(A) COS(A) Tan(A) A SIN(A) COS(A) Tan(A)0 0.0000 1.0000 0.0000 45 0.7071 0.7071 1.00001 0.0175 0.9998 0.0175 46 0.7193 0.6947 1.03552 0.0349 0.9994 0.0349 47 0.7314 0.6820 1.07243 0.0523 0.9986 0.0524 48 0.7431 0.6691 1.11064 0.0698 0.9976 0.0699 49 0.7547 0.6561 1.15045 0.0872 0.9962 0.0875 50 0.7660 0.6428 1.19186 0.1045 0.9945 0.1051 51 0.7771 0.6293 1.23497 0.1219 0.9925 0.1228 52 0.7880 0.6157 1.27998 0.1392 0.9903 0.1405 53 0.7986 0.6018 1.32709 0.1564 0.9877 0.1584 54 0.8090 0.5878 1.376410 0.1736 0.9848 0.1763 55 0.8192 0.5736 1.428111 0.1908 0.9816 0.1944 56 0.8290 0.5592 1.482612 0.2079 0.9781 0.2126 57 0.8387 0.5446 1.539913 0.2250 0.9744 0.2309 58 0.8480 0.5299 1.600314 0.2419 0.9703 0.2493 59 0.8572 0.5150 1.664315 0.2588 0.9659 0.2679 60 0.8660 0.5000 1.732116 0.2756 0.9613 0.2867 61 0.8746 0.4848 1.804017 0.2924 0.9563 0.3057 62 0.8829 0.4695 1.880718 0.3090 0.9511 0.3249 63 0.8910 0.4540 1.962619 0.3256 0.9455 0.3443 64 0.8988 0.4384 2.050320 0.3420 0.9397 0.3640 65 0.9063 0.4226 2.144521 0.3584 0.9336 0.3839 66 0.9135 0.4067 2.246022 0.3746 0.9272 0.4040 67 0.9205 0.3907 2.355923 0.3907 0.9205 0.4245 68 0.9272 0.3746 2.475124 0.4067 0.9135 0.4452 69 0.9336 0.3584 2.605125 0.4226 0.9063 0.4663 70 0.9397 0.3420 2.747526 0.4384 0.8988 0.4877 71 0.9455 0.3256 2.904227 0.4540 0.8910 0.5095 72 0.9511 0.3090 3.077728 0.4695 0.8829 0.5317 73 0.9563 0.2924 3.270929 0.4848 0.8746 0.5543 74 0.9613 0.2756 3.487430 0.5000 0.8660 0.5774 75 0.9659 0.2588 3.732131 0.5150 0.8572 0.6009 76 0.9703 0.2419 4.010832 0.5299 0.8480 0.6249 77 0.9744 0.2250 4.331533 0.5446 0.8387 0.6494 78 0.9781 0.2079 4.704634 0.5592 0.8290 0.6745 79 0.9816 0.1908 5.144635 0.5736 0.8192 0.7002 80 0.9848 0.1736 5.671336 0.5878 0.8090 0.7265 81 0.9877 0.1564 6.313837 0.6018 0.7986 0.7536 82 0.9903 0.1392 7.115438 0.6157 0.7880 0.7813 83 0.9925 0.1219 8.144339 0.6293 0.7771 0.8098 84 0.9945 0.1045 9.514440 0.6428 0.7660 0.8391 85 0.9962 0.0872 11.430141 0.6561 0.7547 0.8693 86 0.9976 0.0698 14.300742 0.6691 0.7431 0.9004 87 0.9986 0.0523 19.081143 0.6820 0.7314 0.9325 88 0.9994 0.0349 28.636344 0.6947 0.7193 0.9657 89 0.9998 0.0175 57.290045 0.7071 0.7071 1.0000 90 1.0000 0.0000 ∞
Trigonometry Table
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