Nilai dari (sin pi/14 sin 3pi/14 sin 9pi/14) . . . .

1/8

Pembahasan

Soal ini meminta untuk menghitung nilai dari ekspresi trigonometri: (sin(π/14) sin(3π/14) sin(9π/14)).

Kita akan menggunakan identitas trigonometri untuk menyederhanakan ekspresi ini.
Perhatikan bahwa 9π/14 = π - 5π/14 dan 3π/14.
Juga, kita bisa melihat hubungan antara sudut-sudut ini.
Misalkan θ = π/14.
Maka ekspresi menjadi sin(θ) sin(3θ) sin(9θ).

Kita tahu bahwa sin(9θ) = sin(π - 5π/14) = sin(5π/14). Ini tidak membantu secara langsung.

Namun, perhatikan hubungan berikut:
sin(9π/14) = sin(π - 5π/14) = sin(5π/14).
Dan 5π/14 = (7π - 2π)/14 = π/2 - π/7. Jadi sin(5π/14) = cos(π/7).

Kita juga punya 3π/14 dan π/14.

Mari kita coba gunakan identitas perkalian sinus: 2 sin A sin B = cos(A - B) - cos(A + B).

Misalkan kita kelompokkan sin(3π/14) sin(9π/14).
2 sin(3π/14) sin(9π/14) = cos(9π/14 - 3π/14) - cos(9π/14 + 3π/14)
= cos(6π/14) - cos(12π/14)
= cos(3π/7) - cos(6π/7)

Sekarang kalikan dengan sin(π/14):
(1/2) * [cos(3π/7) - cos(6π/7)] * sin(π/14)
= (1/2) * [cos(3π/7) sin(π/14) - cos(6π/7) sin(π/14)]

Gunakan lagi identitas: 2 cos A sin B = sin(A + B) - sin(A - B).

2 cos(3π/7) sin(π/14) = sin(3π/7 + π/14) - sin(3π/7 - π/14)
= sin(6π/14 + π/14) - sin(6π/14 - π/14)
= sin(7π/14) - sin(5π/14)
= sin(π/2) - sin(5π/14)
= 1 - sin(5π/14)

2 cos(6π/7) sin(π/14) = sin(6π/7 + π/14) - sin(6π/7 - π/14)
= sin(12π/14 + π/14) - sin(12π/14 - π/14)
= sin(13π/14) - sin(11π/14)

Perhatikan bahwa sin(13π/14) = sin(π - π/14) = sin(π/14)
dan sin(11π/14) = sin(π - 3π/14) = sin(3π/14).

Jadi, 2 cos(6π/7) sin(π/14) = sin(π/14) - sin(3π/14).

Substitusikan kembali:
(1/2) * [(1 - sin(5π/14))/2 - (sin(π/14) - sin(3π/14))/2]
= (1/4) * [1 - sin(5π/14) - sin(π/14) + sin(3π/14)]

Ini masih terlihat rumit. Mari kita coba pendekatan lain.

Perhatikan bahwa 3 * (π/14) = 3π/14 dan 3 * (3π/14) = 9π/14. Ini menunjukkan pola seperti dalam rumus $\sin(3x) = 3\sin(x) - 4\sin^3(x)$.

Mari kita evaluasi sin(9π/14).
9π/14 = (14π - 5π)/14 = π - 5π/14. Jadi sin(9π/14) = sin(5π/14).
Ekspresi menjadi: sin(π/14) sin(3π/14) sin(5π/14).

Kita tahu sin(5π/14) = sin(π/2 - 2π/14) = sin(π/2 - π/7) = cos(π/7).
Kita juga tahu sin(3π/14) = sin(π/2 - 4π/14) = sin(π/2 - 2π/7) = cos(2π/7).
Dan sin(π/14) = sin(π/2 - 6π/14) = sin(π/2 - 3π/7) = cos(3π/7).

Jadi, ekspresi menjadi cos(3π/7) cos(2π/7) cos(π/7).
Ini adalah bentuk yang umum dikenal.

Misalkan P = cos(π/7) cos(2π/7) cos(4π/7).
(Jika kita punya cos(4π/7) bukan cos(3π/7))

Mari kita kembali ke sin(π/14) sin(3π/14) sin(9π/14).

Identitas yang mungkin berguna: sin(x) sin(60-x) sin(60+x) = (1/4) sin(3x).
Ini tidak berlaku di sini.

Identitas lain: sin(x) sin(2x) sin(4x) = (1/4) sin(8x).

Kita punya sin(π/14), sin(3π/14), sin(9π/14).
Perhatikan bahwa 3 * (π/14) = 3π/14 dan 3 * (3π/14) = 9π/14.

Misalkan x = π/14. Kita punya sin(x) sin(3x) sin(9x).

Kita tahu bahwa sin(9x) = sin(9π/14) = sin(π - 5π/14) = sin(5π/14).

Mari kita gunakan identitas: sin(α) sin(β) = (1/2)[cos(α-β) - cos(α+β)].

sin(π/14) sin(3π/14) = (1/2)[cos(2π/14) - cos(4π/14)] = (1/2)[cos(π/7) - cos(2π/7)].

Sekarang kalikan dengan sin(9π/14) = sin(5π/14).

(1/2)[cos(π/7) - cos(2π/7)] * sin(5π/14)
= (1/2) * [cos(π/7)sin(5π/14) - cos(2π/7)sin(5π/14)]

Gunakan identitas: cos(A)sin(B) = (1/2)[sin(A+B) - sin(A-B)].

cos(π/7)sin(5π/14) = (1/2)[sin(π/7 + 5π/14) - sin(π/7 - 5π/14)]
= (1/2)[sin(2π/14 + 5π/14) - sin(2π/14 - 5π/14)]
= (1/2)[sin(7π/14) - sin(-3π/14)]
= (1/2)[sin(π/2) + sin(3π/14)]
= (1/2)[1 + sin(3π/14)]

cos(2π/7)sin(5π/14) = (1/2)[sin(2π/7 + 5π/14) - sin(2π/7 - 5π/14)]
= (1/2)[sin(4π/14 + 5π/14) - sin(4π/14 - 5π/14)]
= (1/2)[sin(9π/14) - sin(-π/14)]
= (1/2)[sin(9π/14) + sin(π/14)]

Substitusikan kembali:
(1/2) * [(1/2)(1 + sin(3π/14)) - (1/2)(sin(9π/14) + sin(π/14))]
= (1/4) * [1 + sin(3π/14) - sin(9π/14) - sin(π/14)]

Kita tahu sin(9π/14) = sin(5π/14). Jadi:
= (1/4) * [1 + sin(3π/14) - sin(5π/14) - sin(π/14)]

Ini masih belum menghasilkan nilai numerik yang sederhana.

Mari kita coba identitas yang melibatkan cosinus:
cos(π/7) cos(2π/7) cos(4π/7) = 1/8.

Kita punya sin(π/14) sin(3π/14) sin(9π/14).

Sederhanakan sin(9π/14) = sin(π - 5π/14) = sin(5π/14).
Jadi kita punya sin(π/14) sin(3π/14) sin(5π/14).

Ubah ke cosinus:
sin(π/14) = cos(π/2 - π/14) = cos(6π/14) = cos(3π/7).
sin(3π/14) = cos(π/2 - 3π/14) = cos(4π/14) = cos(2π/7).
sin(5π/14) = cos(π/2 - 5π/14) = cos(2π/14) = cos(π/7).

Jadi ekspresi kita adalah cos(3π/7) cos(2π/7) cos(π/7).

Kita tahu identitas: cos(x) cos(2x) cos(4x) = sin(8x) / (8 sin(x)).

Jika kita ambil x = π/7:
cos(π/7) cos(2π/7) cos(4π/7) = sin(8π/7) / (8 sin(π/7)).
Karena sin(8π/7) = sin(π + π/7) = -sin(π/7).
Maka cos(π/7) cos(2π/7) cos(4π/7) = -sin(π/7) / (8 sin(π/7)) = -1/8.

Namun, ekspresi kita adalah cos(π/7) cos(2π/7) cos(3π/7).
Perhatikan bahwa cos(3π/7) = cos(π - 4π/7) = -cos(4π/7).

Jadi, cos(π/7) cos(2π/7) cos(3π/7) = cos(π/7) cos(2π/7) [-cos(4π/7)]
= - [cos(π/7) cos(2π/7) cos(4π/7)]
= - (-1/8)
= 1/8.

Jadi, nilai dari (sin π/14 sin 3π/14 sin 9π/14) adalah 1/8.