Jawaban Uji Kompetensi 8 3

Oleh Lilik Nur Fadlilah

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Transkrip Jawaban Uji Kompetensi 8 3

JAWABAN UJI KOMPETENSI 8.3
1. SINUS, COSINUS, TANGEN, SECAN, COSEC, COTANGEN
a.
y

r = √ 52 +62

A (5,6)

6

= √ 25+36

r=?

=√ 61
0

5

x

y

b.

P (-3,8)
r = √ (−3)2 +82
= √ 9+64

r=?

8

= √ 73
-3

0

x

c.
y

r = √ (−3)2 +(−5)2
x

(-5,-3) r = ?

=√ 34

-5

T

=√ 9+25

-3

d.

x
r = √ 22+(−8)2

r=?

= √ 4 +64
= √ 68

-8
y

2

Q (2,-8)

JAWAB:

6
√61 6 √ 61
x
=
√61 √61 61
5
√61 5 √ 61
Cos A =
x
=
√61 √61 61
6
Tan A = 5
61
Sec A = √
5

a) Sin A =

61
Cosec A = √
6

5

Cot A = 6
8
√73 8 √ 73
x
=
√73 √73 73
−3
√73 −3 √ 73
2 Cos P =
x
=
√ 73 √73 73
8
−8
3 Tan P = −3 = 3
73
4 Cosec P = √
8
−√ 73
5 Sec P =
3
−3
6 Cot P = 8

b) 1 Sin P =

−5
√34 −5 √ 34
x
=
√ 34 √34 34
−3
√34 −3 √ 34
Cot T =
x
=
√34 √34 34
−5 5
Tan T = −3 = 3
34 − 34
Sec T = √ = √
−3
3
34 − 34
Cosec T = √ = √
−5
5
−3 3
Cot T = −5 = 5

c) Sin T =

d) Sin Q =

2
√68 2 √ 68 2 √ 68
x
=
= 34
√68 √68 68

−8
√ 68 −8 √ 68 −2
x
=
= 17
√68 √ 68 68
2
1
Tan Q = −8 =- 4
68 − 68
Sec Q = √ = √
−8
8
68
Cosec Q = √
2
−2
Cot Q = 2 = -4

Cos Q =

√ 68

2. A. P (5,12)
y
12

P (5,12)
r = √ 52 +122
= √ 25+144

12

= √ 164
= 13
5

Sin P =

12
13

Cos P =

5
13

Tan P =

x

12
5

B. Q (-5.2,7.2)
r = √ (7,2)2+(−5,2)2

Q (-5.2,7.2)

7,2

7,2

= √ 51,84+27,04
= √ 78,88
= √ 79

-5,2

Sin Q =

7,2
√79 7,2 √ 79
x
=
√79 √79 79

Cos Q =

−5,2 √79 −5,2
x
= 79 √ 79
√ 79 √79
−7,2

Tan Q = 5,2
C. R (-5,-2)

5√ 29
(-5,-2)
R

-2

-5

Sin R =

−2 √ 29 −2
x
=
√ 29
√29 √ 29 29

Cos R =

−5 √ 29 −5
√ 29
x
=
√29 √29 29

Tan R =

−2 2
=
−5 5

D. T (3.5,-7.75)

3,5
r = √ (3,5)2 +(−7,75)2

r=?

= √ 12,25+60,0625
= √ 72,3125
-7,75

Sin T =
Cos T =

−7,75 √ 72 −7,75
x
= 72 √ 72
√ 72 √72
3 ,5 √ 72 3 ,5
√ 72
x
=
√72 √ 72 72

T

= √ 72

−7,75

Tan T = 3,5

3. a. Karena pada kuadran IV nilai tanda untuk Sin x dan Cos x bertanda negatif (-),
1
yang mana Sec =
cos x
b. Tidak karena antara nilai Sinus dan Cosinus nilainya berkebalikan
c.
4.
Sin α > 0
Cos α >0








Sin α < 0

Cos α < 0

Tanα < 0

Sinα > 0

Sin α > 0 → 00 < α < 900 & 900 < α < 1800
Cos α > 0 → 00 < α < 900 & 2700 < α < 3600
Sin α < 0 → 1800 < α < 3600
Cos α < 0 → 900 < α < 2700
Tan α < 0 → 900 < α < 1800 & 2700 < α < 3600
Sin α > 0 → 00 < α < 900 & 900 < α < 1800
−8

5. Tan x = 15 Sin α > 0
sin α

8

r = √ 64+225

y

= cos x = 15 = x

= √ 289
= 17

−15

 Cos α = 17

17

 Sec α = −15
17

 Cosec α = 8

17
8
Cosec α
17
 Cotan α = −15 = −15
8
8 −15
 ( sin α ) (cos α) = 17 ∙ 17
120
= 289 = 0,4

Kuadran II & I
Kuadran IV & I
Kuadran III & IV
Kuadran II & III
Kuadran II & IV
Kuadran II & 1

−15

 Cotan α = 8

π
π
≤ β ≤ 3 ; nilai Cotan tidak terdefinisi
2
2
0
Misal : β = 180
1
Ambil , Cotan β =
tan β
1
= tan 180 °
1
= 0 → tak terdefinisi
a. Sin β = Sin 1800

6. Diket :

=1
b. Cos β = Cos 1800
=0
sin β

sin 180 °

1

c. tan β +1 = tan(80° +1) = 0+1 = 1
2
cos β
tan β−1
2
= cos β x tan β−1
2(tan β−1)
=
cos β
2(tan 180 °−1)
=
cos 180 °
2(0−1)
=
0
2(0−1) −2
=
= 0 → tak terdefinisi
0
7. A. Cos x ∙ Cosec x ∙ Tan x
1
sin x
= Cos x ∙ sin x ∙ cos x
cos x sin x 1
= sin x ∙ cos x = 1 = 1
B. Cos x ∙ Cotan x ∙ Sin x
1
= Cos x ∙ tan x + Sin x
1
= Cos x ∙ sin x + Sin x
cos x
2 Sec β
d. tan β−1 =

2

cos x
=
+ Sin x
sin x
2
cos x+ sin x
=
sin x
1
= sin x

= Cosec x
− 3 −x
8. Cos β = √ = r
4
x = -√ 3
r =4
y = √ 4 2−¿ ¿ )2
= √ 16−3
=√ 13


 Sec β = cos β =
x
= 3 √3
√ 3 √3
−4

1

3

−4

− 3
 Sin β = √

4
−√ 3
√ 3 √3 1 √ 3 9
 Tan β =
=
x
=
−√ 3
√ 3 √3 3

 Cos 15 = 60-45
1

1

= 2 - 2 √2

45-30
1

1 1
= 2 - 2 ∙ √2
= √2

a.

2

Sec β−tan β
+ Sec β
tan β

−4
4
−¿ ¿ + (- )
√3
√3
−4 13 4
−12−13 √3−12
− −
3
3 √3
√3
√3
=
√ 13
√13
√3
√3
−24−13 √ 3 √ 3
=
x
3 √3
√3
= ¿¿

=

−24 √ 3−39 √ 39
x
3 √39
√ 39
−24 √ 117−39 √ 39
=
117

=

1

= 2 √2 - 2 √3

2

2

Sec β + Sec β−1
Sec 2 β+ tan 2 β
b.
=
2
2
2
2sin β +2 cos β 2(sin ¿ ¿ 2 β+ cos β) ¿

=

2 Sec 2 β−1
2−1

2
−4
3) −1
2 Sec 2 β−1 2(

=
=
3
2−1
2
2
−4
2(
3) −1

= 3
2

−16
2(
3) −1
9
=
2

−16
2(
2.1) −1
3
=
2
−32 3

35 1
35
= 3 3 =-3 x 2=- 6
2

9. Jika α = 2040 , Hitunglah:
−1
√3 −1 √ 3
sin α
sin 240
2
2
a.
= -2√ 3
2 =
2 =
2 =
1
(cos α )
(cos 240)
−1
(
)
4
2
1 1
+ √3
2 2



α =
240 = −1 1
x 1 1 ❑
cos α + sin( )
cos 240+sin(
)
+ √3
4
4
+ √3
2 2
2 2
tan α

b.

√3

tan 240

1
1
1
3
√3+ ∙ 3
√3+
2
2
2
2
= 1 1 ❑ = 1 3❑
− + ∙3
− +
4 4
4 4

1
3
1
3
√ 3+
√3+
2
2
2
2
=
=
2
1
4
2
1

3

= 2 (2 √3 + 2)
= √3 + 3

c. 2 sin α - cos (2 α) = 2 sin 240 - cos 480
1

1

= 2 ∙ - 2 √ 3 - (- 2 ¿
1

= - √3 + 2
d. Sin2 α + Cos2 α + √ 3
= Sin2 240 + Cos2 240 + √ 3
−1
−1 2
2
= ( 2 √ 3¿ ¿ + ( 2 ¿ + √ 3
1
1
= 4 ∙ 3 +4 + √3
3

1

= 4 + 4 +√ 3
4

= 4 + √3
= 1 + √3
sin A

sin A

10. a. 1+ cos A + 1−cos A =
=

=

sin A ( 1−cos A )+ sin A (1 cos A)
( 1+cos A ) (1−cos A )

sin A ( 1−cos A )+ sin A (1 cos A)
2

1−cos A +cos A−C 0 s A
sin A ( 1−cos A )+ sin A (1 cos A)
2

1−C 0 s A

sin A – sin A cos A+sin A+sin A cos A
2 sin A
=
2
2
1−C 0 s A
1−C 0 s A
2 sin A
=
2
2
2
C 0 s A+ sin A−C 0 s A
2sin A
= 2
sin A
2
= sin A

=

= 2 Cosec A
b . (Sin B + Cos B)2 + (Sin B – Cos B)2
= Sin2 B + 2 Sin B Cos B + Cos2 B + Sin2 A – 2 Sin B Coc B +
Cos2 B
= 2 Sin2 B + 2 Cos2 B

= 2 (Sin2 B + Cos2 B)
= 2 (1)
=2
c. (Cosec A – Cotan A) (1 + Cos A)
= Cosec A + Cosec A Cos A – Cotan A Cos A
1

1

1

1

cos A

cos A

1

cos A

cos2 A

= sin A + sin A Cos A - tan A ∙ Cos A
= sin A + sin A - sin A ∙ Cos A
= sin A + sin A sin A
=

sin2 A+ cos2 A +cos A−cos A
sin A

2
2
cos A
sin A+ cos A sin A
=
=
+ sin A = Sin A + Cotan A
sin A
sin A

11. Dggnmj,k

12.y = 3 Cos 2x – 2
a). y maks. Dicapai bila Cos x = 1
Cos 2x = 2 Cos2 x – 1
= 2 ∙ 1-1
=1
y min. Dicapai bila Cos x = -1
Cos 2x = 2 Cos2x – 1
= 2(-1)2 – 1
= 2-1
=1
b). y = 5 Sin x + Cos 2x
y maks. dicapai bila Sin x = 1

,

y maks = 3∙ 1 - 2
= 3-2
=1
y min= 3 Cos 2x-2
= 3∙ 1 - 2
= 3-2
=1
Cos x = 1 → Cos 2x = 2 Cos2 x-1

= 2 ∙ 12 – 1 = 1
y maks. = 5 ∙ 1 + 1 = 6
y min dicapai bila Sin x = - 1 Cos x = - 1 → Cos 2x = 2 Cos2 x – 1 = 2 (-1)2 – 1
=1

c).
7

Sinx+cos x

d).y = Sinx−cos x x Sinx+cos x
= 7¿¿ =

7(Sinx+Cosx)
7( Sinx+Cosx)
=
2
sin x +sin 2 x−1
2sin 2 x−1

y maks. dicapai bila Sin x = 1 ,
y maks =

7.2
7(−1−1)
= 2−1 = 14
2.1−1

y min dicapai bila Sin x = -1 ,
y min =

Cos x = 1

7(−1−1)
= 7(-2)
2.1−1

Cos x = -1

= -14

13.a. √ 2 Sin 2x = tg 2x
sin 2 x

√ 2 Sin 2x = cos 2 x
√ 2 cos 2 x = 1
1

Cos 2x = 2 √ 2
2x = 45°
x = 22,5°
b. (Cos x + Sin x )2 = 12
Cos2x + 2 Cos x Sin x + Sin2x = 1
2 Cos x Sin x + 1 = 1
2 Cos x Sin x
=0
Sin 2x = 0
2x = 0° , 180° , 360° , 540° , 720°
x = 0° , 90° , 180° , 270° , 360°
2
2
c Sin x + Cos x = 1
Cos 2x – Cos2 x + Cos2x = 1
Cos 2x = 1
2x = 0° , 360°
x = 0° , 180°

Judul: Jawaban Uji Kompetensi 8 3

Oleh: Lilik Nur Fadlilah


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