- Kerjakan Dikertas Lembar Kumpul Pada Guru Piket
- Selanjutnya hasil Kerjaan pada poin 1 anda salin kembali dikolom komentar/buku tamu dengan waktu yang sama 90 menit
- Jika lewat dari waktu 90 menit tugas anda tidak diterima dan dianggap remedial.
TUGAS MATEMATIKA KELAS XI IPA & XI IPS
Tanggal 19 – 22 November 2012
|
Hari/tanggal
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Jam ke-
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Kelas / Ruang
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Tugas
|
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Senin,
19
November 2012
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4
- 5
|
XI IPA 2 / 4
|
Buatlah Resume dibuku
catatan Rumus Trigonometri untuk
jumlah dan Selisih Dua sudut + Rumus Trigonometri Sudut Ganda, selanjutnya
buatlah masing-masing rumus tersebut dengan dua buah contoh soal
|
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6
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XI
IPS 1 / 9
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Kerjakan
dibuku latihan halaman :
Lat.
9 Hal. 136, No. 1a, 1c
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|
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9
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XI
IPS 2 /3
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Kerjakan
dibuku latihan halaman :
Lat.
9 Hal. 136, No. 1a, 1c
|
|
|
10
|
XI
IPA 1 / 4
|
Buatlah
Resume dibuku catatan Rumus
Trigonometri untuk jumlah dan Selisih Dua sudut + Rumus Trigonometri Sudut
Ganda, selanjutnya buatlah masing-masing rumus tersebut dengan dua buah
contoh soal
|
|
|
12
|
XI
IPA 2 / 4
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Kerjakan
dibuku latihan halaman :
Lat.
1 Hal 116, No. 1a, 5, 6
Lat.
2 Hal 118, No. 1d, 5
Lat.
3 Hal 121, No. 1a, 2
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|
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Selasa,
20
November 2012
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3
- 4
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XI
IPS 2 / 4
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Buatlah
Resume dibuku catatan Apa itu Fungsi Invers dan Invers dari Fungsi
Komposisis, selanjutnya buatlah masing-masing rumus tersebut dengan dua buah
contoh soal
|
|
7
- 8
|
XI
IPS 1/ 4
|
Buatlah
Resume dibuku catatan Apa itu Fungsi Invers dan Invers dari Fungsi
Komposisis, selanjutnya buatlah masing-masing rumus tersebut dengan dua buah
contoh soal
|
|
|
Ekstra
KIR
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Fisika
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Diskusi
karya tulis Kelas X dan XI dengan bimbingan kelas XII
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|
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Rabu,
21
November 2012
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7
|
XI
IPS 2 / 7
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Kerjakan
latihan dan dikumpul halaman :
Lat.
11 Hal 143, No. 1a,f 2c dan 4d
Lat.
12 Hal 146, No. 1b, 3b, 4b
|
|
|
8
9-10
|
XI
IPA 1 / 2
|
Kerjakan
dibuku latihan halaman :
Lat.
1 Hal 116, No. 1a, 5, 6
Lat.
2 Hal 118, No. 1d, 5
Lat.
3 Hal 121, No. 1a, 2
Buatlah
Resume dibuku catatan Rumus Perkalian
Trigonometri + Rumus Penjumlahan dan Pengurangan, selanjutnya buatlah
masing-masing rumus tersebut dengan dua buah contoh soal
|
|
|
11
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XI
IPS 1 / 10
|
Kerjakan
latihan dan dikumpul halaman :
Buatlah
lima soal beserta jawabannya tentang materi fungsi invers
|
|
|
12
|
XI
IPA 2 / 2
|
Buatlah
Resume dibuku catatan Rumus Perkalian
Trigonometri + Rumus Penjumlahan dan Pengurangan, selanjutnya buatlah
masing-masing rumus tersebut dengan dua buah contoh soal
|
|
Kamis,
22
November 2012
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1
- 2
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XI
IPS 1 / 4
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Kerjakan
latihan dan dikumpul halaman :
Lat.
11 Hal 143, No. 1a,f 2c dan 4d
Lat.
12 Hal 146, No. 1b, 3b, 4b:
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|
|
5
– 6
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XI
IPA 1 / 2
|
Latihan
Ulangan Trigonometri dan dikumpul, soalnya bisa down load di blog
mediaharja.blogspot.com + jawabannya masing2 siswa juga diposting komentar blog ini.
|
|
|
7
– 8
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XI
IPA 2 / 1
|
Latihan
Ulangan Trigonometri dan dikumpul, soalnya bisa down load di blog
mediaharja.blogspot.com + jawabannya masing2 siswa juga diposting komentar blog ini.
|
|
Juma’at
23
November 2012
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9
- 10
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XI
IPS 2
|
Kerjakan
latihan dan dikumpul halaman :
Buatlah
lima soal beserta jawabannya tentang materi invers dari fungsi komposisi
|
Pangkalan
Balai, November 2012
Media
Harja, S.Pd
NIP
19830807 200801 1 010
Jawab 2 (Okta nindita priambodo)
BalasHapusP (a,b) = (-3,3) dan R = 4
Persamaan = (x-a)¬¬2 + (y-b)2 = R2¬¬¬
= (x-(-3))2 + (y-3)2 = 42
= (x+3)2 + (y-3)2 = 16
Jawab 3 (Subhan)
X2 + y2 + 4x + 6y – 12 = 0 melalui titik (1,7)
a = 4 b=6 c= -12
pusat (-2, -3)
r = √(〖(-2)〗^2+ 〖(-3)〗^2 )- (-12)
r = √(4+9+12 )
r = √25
r = 5
Jawab 4 (Muhammad Iqbal Dwi Pernama)
X2 + y2 + 6x + 4y – 12 = 0 melalui titik (7,1)
Persamaan
x1x + y1y + 1/2A +(x + x1)+ 1/2B +(y + y1)+C=0
7x + y + 3(x + 7)+ 2(y + 1)-12=0
7x + y + 3x + 21 + 2y + 2-12=0
10x + 3y + 11 = 0
Jawab 5(Arief Hidayatullah dan Mulpi)
Y = m x ± r √(1+m)2
m x ± 6,3 √(1+(6,3))2
m x ± 6,3 √(1+39,69)
m x ± 6,3 √40,69
(5 ×180 )/6 = 150
(xI-a) (x-a) + (yI-b) (y-b) = r2
(-3-(-4)) (x-(-4)) + (4-(3)) (y-(3)) = 40
1x + 4 + 1y – 3 + 40
1x + 1y – 39 = 0
maaf pak ..
BalasHapusnomor 1 tidak bisa d pos disini ..
jadi kami kirim d email bapak yang mediaharja@gmail.com
maaf pak sebelumnya
Nama :Cyndiana Pawestri
BalasHapusKelas : XI IPA 2
a.) Sin 750 = …?
Sin 750 = sin (450+ 300)
sin (450+ 300) = sin 45.cos 30 + cos 45. Sin 30
= 1/2 √2 . 1/2 √3 + 1/2 √2 . 1/2
= 1/4 √6 + 1/4 √2
= 1/4 √2 (√3 + 1)
Cos 15 =…?
cos (60 - 45) = cos 60.cos 45 + sin 60.sin 45
= 1/2 . 1/2 √2 + 1/2 √3. 1/2 √2
= 1/4 √2 + 1/4 √6
= 1/4 √2 (1 + √3)
b.) diketahui : cos (A+B) = 2/5
Cos A.cos B= 3/4
SinA . sin B = ..?
Jawab :
cos (A+B) = Cos A.cos B - SinA . sin B
2/5 = 3/4 - SinA . sin B
SinA . sin B = 3/4 - 2/5
= 7/20
tan A.tan B =( SinA .sin B)/(Cos A.cos B)
= (7/20)/(3/4)
= 28/60=7/15
c.) diketahui : sin A =4/5
cos B =24/25
a dan b adalah sudut lancip.
Sin B =√(1-〖cos〗^(2 ) )B
=√(1-576/625)
=√(49/625)= 7/25
Cos A=√(1-〖sin〗^(2 ) A)
= √(1-16/25)
= √(9/25) = 3/5
Tan A = (sin A)/(Cos A) = (4/5)/(3/5) = 4/3
Tan B = (Sin B)/(cos B)=(7/25)/(24/25)= 7/24
cos (A+B) = Cos A.cos B - SinA . sin B
= 3/5 . 24/25 - 4/5 7/25
= 72/125- 28/125= 44/125
tan (A –B)= tan〖A-tanB 〗/(1+tanA.tanB)
= (4/3-7/24)/(1+ 4/3.7/25) = ((32-7)/24)/(1+28/72) = (25/24)/(72/72+28/72)
= (25/24)/(100/72) =(25/24)/(25/18) =450/600 = 3/4
diketahui :
A= sudut lancip
sinA = 4/5
sin 2A =2 sinA.cosA
= 2 .4/5.3/5
=2.12/25
=24/25
cos 2A =..?
cos A = √(1-〖sin〗^(2 ) A)
= √(1-16/25)
= √(9/25) = 3/5
cos 2A = 1 – 2 〖sin〗^(2 ) A
=1- 2. 16/25
=1- 32/25
= 25/25 - 32/25
= 7/25
tan 2A=…?
Tan A = (sin A)/(Cos A) = (4/5)/(3/5) = 4/3
Tan 2A = (2 tanA)/(1- 〖tan〗^a ) =(2.4/3)/(1-16/9)=(8/3)/(-7/9)=72/(-21)= -24/7
a. sin π/12=sin 180/12=sin15
sin 15 = sin (45 -30)
= sin 45.cos 30 –cos 45. Sin 30
=1/2 √2. 1/2 √3- 1/2 √2. 1/2
= 1/4 √6-1/4 √2
= 1/4 √2 (√3 - 1)
cos 71/2 =√((1+cos4.7 1/2)/4) =√((1+cos30)/4)
=√((1+1/2 √3)/4)
=1/2 √(1+1/2 √3)
tan 1121/2 =√((1-cos225)/(1+cos225 )) =√((1-(-1/2 √2))/(1+(-1/2 √2))) =√((2+√2)/(2-√2))
4.) a. 2 cos 75. Cos 15 = ….?
Jawab : 2 cos 75. Cos 15 = cos (75 + 15) + cos (75 - 15)
= cos 90 + cos 60
= 0 + 1/2
= 1/2
b. 4 sin 3x cos x =2[ sin (3x +x) +sin (3x -x)]
=2.( sin 4x + sin 2x)
c. 4 sin 521/2 sin 71/2 = 2 [cos (521/2 – 71/2) – cos (521/2 +7 1/2)
= 2( cos 45 – cos 60)
= 2 (1/2 √2 - 1/2 )
= √2 – 1
5. a. sin 75 + sin 15 = 2 sin 1/2 (75 +15). Cos 1/2 (75 – 15)
= 2 sin 45.cos 30
= 2. 1/2 √2. 1/2 √3
= 2. 1/4 √6
= 1/2 √6
b. cos 105 – cos 15 = -2 sin1/2 (105 + 15).sin1/2 (105 – 15)
=-2. sin 60.sin 45
= -2. 1/2 √3 . 1/2 √2
= -2. 1/4 √6
= - 1/2 √6
c. sin 6x- sin 2x = 2 cos1/2(6x + 2x). sin 1/2 . (6x- 2x)
= 2. Cos 4x. sin 2x
6. diketahui : cos (A+B) = 2/5
Cos A.cos B= 3/4
SinA . sin B = ..?
Jawab :
cos (A+B) = Cos A.cos B - SinA . sin B
2/5 = 3/4 - SinA . sin B
SinA . sin B = 3/4 - 2/5
= 7/20
tan A.tan B =( SinA .sin B)/(Cos A.cos B)
= (7/20)/(3/4)
= 28/60=7/15
Nama Santi Kartika
BalasHapusXI IPA 2
a.) Sin 750 = …?
Sin 750 = sin (450+ 300)
sin (450+ 300) = sin 45.cos 30 + cos 45. Sin 30
= 1/2 √2 . 1/2 √3 + 1/2 √2 . 1/2
= 1/4 √6 + 1/4 √2
= 1/4 √2 (√3 + 1)
Cos 15 =…?
cos (60 - 45) = cos 60.cos 45 + sin 60.sin 45
= 1/2 . 1/2 √2 + 1/2 √3. 1/2 √2
= 1/4 √2 + 1/4 √6
= 1/4 √2 (1 + √3)
b.) diketahui : cos (A+B) = 2/5
Cos A.cos B= 3/4
SinA . sin B = ..?
Jawab :
cos (A+B) = Cos A.cos B - SinA . sin B
2/5 = 3/4 - SinA . sin B
SinA . sin B = 3/4 - 2/5
= 7/20
tan A.tan B =( SinA .sin B)/(Cos A.cos B)
= (7/20)/(3/4)
= 28/60=7/15
c.) diketahui : sin A =4/5
cos B =24/25
a dan b adalah sudut lancip.
Sin B =√(1-〖cos〗^(2 ) )B
=√(1-576/625)
=√(49/625)= 7/25
Cos A=√(1-〖sin〗^(2 ) A)
= √(1-16/25)
= √(9/25) = 3/5
Tan A = (sin A)/(Cos A) = (4/5)/(3/5) = 4/3
Tan B = (Sin B)/(cos B)=(7/25)/(24/25)= 7/24
cos (A+B) = Cos A.cos B - SinA . sin B
= 3/5 . 24/25 - 4/5 7/25
= 72/125- 28/125= 44/125
tan (A –B)= tan〖A-tanB 〗/(1+tanA.tanB)
= (4/3-7/24)/(1+ 4/3.7/25) = ((32-7)/24)/(1+28/72) = (25/24)/(72/72+28/72)
= (25/24)/(100/72) =(25/24)/(25/18) =450/600 = 3/4
diketahui :
A= sudut lancip
sinA = 4/5
sin 2A =2 sinA.cosA
= 2 .4/5.3/5
=2.12/25
=24/25
cos 2A =..?
cos A = √(1-〖sin〗^(2 ) A)
= √(1-16/25)
= √(9/25) = 3/5
cos 2A = 1 – 2 〖sin〗^(2 ) A
=1- 2. 16/25
=1- 32/25
= 25/25 - 32/25
= 7/25
tan 2A=…?
Tan A = (sin A)/(Cos A) = (4/5)/(3/5) = 4/3
Tan 2A = (2 tanA)/(1- 〖tan〗^a ) =(2.4/3)/(1-16/9)=(8/3)/(-7/9)=72/(-21)= -24/7
a. sin π/12=sin 180/12=sin15
sin 15 = sin (45 -30)
= sin 45.cos 30 –cos 45. Sin 30
=1/2 √2. 1/2 √3- 1/2 √2. 1/2
= 1/4 √6-1/4 √2
= 1/4 √2 (√3 - 1)
cos 71/2 =√((1+cos4.7 1/2)/4) =√((1+cos30)/4)
=√((1+1/2 √3)/4)
=1/2 √(1+1/2 √3)
tan 1121/2 =√((1-cos225)/(1+cos225 )) =√((1-(-1/2 √2))/(1+(-1/2 √2))) =√((2+√2)/(2-√2))
4.) a. 2 cos 75. Cos 15 = ….?
Jawab : 2 cos 75. Cos 15 = cos (75 + 15) + cos (75 - 15)
= cos 90 + cos 60
= 0 + 1/2
= 1/2
b. 4 sin 3x cos x =2[ sin (3x +x) +sin (3x -x)]
=2.( sin 4x + sin 2x)
c. 4 sin 521/2 sin 71/2 = 2 [cos (521/2 – 71/2) – cos (521/2 +7 1/2)
= 2( cos 45 – cos 60)
= 2 (1/2 √2 - 1/2 )
= √2 – 1
5. a. sin 75 + sin 15 = 2 sin 1/2 (75 +15). Cos 1/2 (75 – 15)
= 2 sin 45.cos 30
= 2. 1/2 √2. 1/2 √3
= 2. 1/4 √6
= 1/2 √6
b. cos 105 – cos 15 = -2 sin1/2 (105 + 15).sin1/2 (105 – 15)
=-2. sin 60.sin 45
= -2. 1/2 √3 . 1/2 √2
= -2. 1/4 √6
= - 1/2 √6
c. sin 6x- sin 2x = 2 cos1/2(6x + 2x). sin 1/2 . (6x- 2x)
= 2. Cos 4x. sin 2x
6. diketahui : cos (A+B) = 2/5
Cos A.cos B= 3/4
SinA . sin B = ..?
Jawab :
cos (A+B) = Cos A.cos B - SinA . sin B
2/5 = 3/4 - SinA . sin B
SinA . sin B = 3/4 - 2/5
= 7/20
tan A.tan B =( SinA .sin B)/(Cos A.cos B)
= (7/20)/(3/4)
= 28/60=7/15
nama : Syntha Ariska
BalasHapuskelas : XI IPA 1
tugas trigonometri
a.) Sin 750 = …?
Sin 750 = sin (450+ 300)
sin (450+ 300) = sin 45.cos 30 + cos 45. Sin 30
= 1/2 √2 . 1/2 √3 + 1/2 √2 . 1/2
= 1/4 √6 + 1/4 √2
= 1/4 √2 (√3 + 1)
Cos 15 =…?
cos (60 - 45) = cos 60.cos 45 + sin 60.sin 45
= 1/2 . 1/2 √2 + 1/2 √3. 1/2 √2
= 1/4 √2 + 1/4 √6
= 1/4 √2 (1 + √3)
b.) diketahui : cos (A+B) = 2/5
Cos A.cos B= 3/4
SinA . sin B = ..?
Jawab :
cos (A+B) = Cos A.cos B - SinA . sin B
2/5 = 3/4 - SinA . sin B
SinA . sin B = 3/4 - 2/5
= 7/20
tan A.tan B =( SinA .sin B)/(Cos A.cos B)
= (7/20)/(3/4)
= 28/60=7/15
c.) diketahui : sin A =4/5
cos B =24/25
a dan b adalah sudut lancip.
Sin B =√(1-〖cos〗^(2 ) )B
=√(1-576/625)
=√(49/625)= 7/25
Cos A=√(1-〖sin〗^(2 ) A)
= √(1-16/25)
= √(9/25) = 3/5
Tan A = (sin A)/(Cos A) = (4/5)/(3/5) = 4/3
Tan B = (Sin B)/(cos B)=(7/25)/(24/25)= 7/24
cos (A+B) = Cos A.cos B - SinA . sin B
= 3/5 . 24/25 - 4/5 7/25
= 72/125- 28/125= 44/125
tan (A –B)= tan〖A-tanB 〗/(1+tanA.tanB)
= (4/3-7/24)/(1+ 4/3.7/25) = ((32-7)/24)/(1+28/72) = (25/24)/(72/72+28/72)
= (25/24)/(100/72) =(25/24)/(25/18) =450/600 = 3/4
diketahui :
A= sudut lancip
sinA = 4/5
sin 2A =2 sinA.cosA
= 2 .4/5.3/5
=2.12/25
=24/25
cos 2A =..?
cos A = √(1-〖sin〗^(2 ) A)
= √(1-16/25)
= √(9/25) = 3/5
cos 2A = 1 – 2 〖sin〗^(2 ) A
=1- 2. 16/25
=1- 32/25
= 25/25 - 32/25
= 7/25
tan 2A=…?
Tan A = (sin A)/(Cos A) = (4/5)/(3/5) = 4/3
Tan 2A = (2 tanA)/(1- 〖tan〗^a ) =(2.4/3)/(1-16/9)=(8/3)/(-7/9)=72/(-21)= -24/7
a. sin π/12=sin 180/12=sin15
sin 15 = sin (45 -30)
= sin 45.cos 30 –cos 45. Sin 30
=1/2 √2. 1/2 √3- 1/2 √2. 1/2
= 1/4 √6-1/4 √2
= 1/4 √2 (√3 - 1)
cos 71/2 =√((1+cos4.7 1/2)/4) =√((1+cos30)/4)
=√((1+1/2 √3)/4)
=1/2 √(1+1/2 √3)
tan 1121/2 =√((1-cos225)/(1+cos225 )) =√((1-(-1/2 √2))/(1+(-1/2 √2))) =√((2+√2)/(2-√2))
4.) a. 2 cos 75. Cos 15 = ….?
Jawab : 2 cos 75. Cos 15 = cos (75 + 15) + cos (75 - 15)
= cos 90 + cos 60
= 0 + 1/2
= 1/2
b. 4 sin 3x cos x =2[ sin (3x +x) +sin (3x -x)]
=2.( sin 4x + sin 2x)
c. 4 sin 521/2 sin 71/2 = 2 [cos (521/2 – 71/2) – cos (521/2 +7 1/2)
= 2( cos 45 – cos 60)
= 2 (1/2 √2 - 1/2 )
= √2 – 1
5. a. sin 75 + sin 15 = 2 sin 1/2 (75 +15). Cos 1/2 (75 – 15)
= 2 sin 45.cos 30
= 2. 1/2 √2. 1/2 √3
= 2. 1/4 √6
= 1/2 √6
b. cos 105 – cos 15 = -2 sin1/2 (105 + 15).sin1/2 (105 – 15)
=-2. sin 60.sin 45
= -2. 1/2 √3 . 1/2 √2
= -2. 1/4 √6
= - 1/2 √6
c. sin 6x- sin 2x = 2 cos1/2(6x + 2x). sin 1/2 . (6x- 2x)
= 2. Cos 4x. sin 2x
6. diketahui : cos (A+B) = 2/5
Cos A.cos B= 3/4
SinA . sin B = ..?
Jawab :
cos (A+B) = Cos A.cos B - SinA . sin B
2/5 = 3/4 - SinA . sin B
SinA . sin B = 3/4 - 2/5
= 7/20
tan A.tan B =( SinA .sin B)/(Cos A.cos B)
= (7/20)/(3/4)
= 28/60=7/15