1. Diketahui sin A = 5/13 dan cos B = 3/5 , A dan B di kuadran I Tentukan a. Sin 2A + cos 2 B b. Sin (A + B) c. Tan (A – B)

Materi : SUDUT SUDUT GANDA TRIGONOMETRI
Kelas : XI SMA MIPA
Mapel : MATEMATIKA PEMINATAN

sin A = 5/13
cos A = ?
tan A = ?

sin = depan / miring
sin = 5 / 13

Gunakan Theorema Phytagoras !
》Anggap :
Miring = AC
Depan = AB
Samping = BC

AC^2 = AB^2 + BC^2
13^2 = 5^2 + BC^2
169 - 25 = BC^2
144 = BC^2
12 = BC

Jadi , BC = samping = 12

cos A = samping / miring
cos A = 12 / 13

tan A = depan / samping
tan A = 5 / 12

cos B = 3/5
sin B = ?
tan B = ?

cos B = 3 / 5
cos B = samping / miring

Gunakan Theorema Phytagoras !
AC^2 = BC^2 + AB^2
5^2 = 3^2 + AB^2
25 - 9 = AB^2
16 = AB^2
4 = AB

Depan = 4

sin B = 4 / 5
tan B = 4 / 3

*Mencari Sin 2A :
Sin 2A = 2 sin A . cos A
Sin 2A = 2 . 5/13 . 12/13
Sin 2A = 120/169

*Mencari cos 2B :
cos 2B = 1 - 2sin^2 B
cos 2B = 1 - 2(4/5)^2
cos 2B = 25/25 - 2(16/25)
cos 2B = 25/25 - 32/25
cos 2B = -7/25

a. sin 2A + cos 2A
》120/169 - 7/25
》3000/4225 - 1183/4225
》1817/4225

B. sin (A + B)
》sin A . cos B + cos A . sin B
》5/13 . 3/5 + 12/13 . 4/5
》15/65 + 48/65
》63/65

C. tan (A - B)
》(tan A - tan B ) / (1 + tan A . tan B)
》(5/12 - 4/3 ) / ( 1 + 5/12 . 4/3)
》(5/12 - 16/12) / ( 36/36 + 20/36)
》(-11/12) / (56/36)
》-11/12 × 36/56
》-11/1 × 3/56
》-33/56

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