1. Sin 60° x cos 30° + cos 60° × sin 30=

Kategori Soal : Matematika – Trigonometri
Kelas : X (1 SMA)
Pembahasan :
Diketahui 
sin 30° = 1/2
cosec 30° = 2
cos 30° = 1/2√3
sec 30° = 2/√3 = 2/3 √3
tan 30° = sin 30°/cos 30° = (1/2)/(1/2√3) = 1/√3 = 1/3 √3
sin 45° = 1/2 √2
cosec 45° = 2/√2 = √2
cos 45° = 1/2 √2
sec 45° = 2/√2 = √2
tan 45° = 1
sin 60° = 1/2 √3
cosec 60° = 2/√3 = 2/3 √3
cos 60° = 1/2
sec 60° = 2
tan 60° = sin 60°/cos 60° = (1/2 √3)/(1/2) = √3

Mari kita lihat soal tersebut.
sin 60° x cos 30° + cos 60° x sin 30° 
= (1/2 √3 x 1/2√3) + (1/2 x 1/2)
= (1/2 x 1/2 x √3 x √3) + 1/4
= (1/4 x 3) + 1/4
= 3/4 + 1/4
= 4/4
= 1

2 x tan 45° + cos 30° – (sin 60°)²
= 2 x 1 + 1/2 √3 – (1/2 √3)²
= 2 + 1/2 √3 – (1/4 x 3)
= 2 – 3/4 + 1/2 √3
= 8/4 – 3/4 + 1/2 √3
= 5/4 + 1/2 √3

(cos 45°/sec 30°) + cosec 30° =
= [(1/2 √2)/(2/3 √3)] + 2
= [(√2/2)/(2√3/3)] + 2
= [(√2/2) x (3/2√3)] + 2
= (3√2/4√3) + 2
= (3√2/4√3) x (√3/√3) + 2
=(3√2 x√3)/(4√3 x √3) + 2
= (3√6)/(4 x 3) + 2
= √6/4 + 2
= 1/4 √6 + 2

(sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cotan 45°)
= (1/2 + 1 – 2/3 √3)/(2/3 √3 + 1/2 + 1)
= (1/2 + 2/2 – 2/3 √3)/(2/3 √3 + 1/2 + 2/2)
= (3/2 – 2/3 √3)/(2/3 √3 + 3/2)

[2(cos 60°)² + 4(sec 30°)² – (tan 45°)²]/[(sin 30°)² + (cos 30)²]
= [2 x (1/2)² + 4 x (2/3 √3)² – 1²]/[(1/2)² + (1/2 √3)²]
= [2 x 1/4 + 4 x 4/9 x 3 – 1]/[1/4 + 1/4 x 3]
= [1/2 + 16/3 – 1]/[1/4 + 3/4]
= (6/12 + 64/12 – 12/12)/(4/4)
= [58/12]/1
= 58/12
= 4 10/12
= 4 5/6

Semangat!