Pertanyaan

Tunjukkan bahwa: sin 14 π sin 14 3 π sin 14 5 π sin 14 7 π sin 14 9 π sin 14 11 π sin 14 13 π = 2 − 6

Tunjukkan bahwa:

$sin \frac{π}{14} sin \frac{3 π}{14} sin \frac{5 π}{14} sin \frac{7 π}{14} sin \frac{9 π}{14} sin \frac{11 π}{14} sin \frac{13 π}{14} = 2^{- 6}$

P. Nur

Master Teacher

Jawaban terverifikasi

Pembahasan

Ingat Rumus sudutganda sinus 2 sin α cos α = sin 2 α . Rumus perkalian trigonometri 2 sin α sin β = cos ( α − β ) − cos ( α + β ) Dari soal tersebut dapat kita uraikan sebagai berikut. sin 14 13 π sin 14 π sin 14 11 π sin 14 3 π sin 14 9 π sin 14 5 π = = = = = = = = = = = = = = = 2 1 ( 2 sin 14 13 π sin 14 π ) 2 1 ( cos ( 14 13 π − 14 π ) − cos ( 14 13 π + 14 π ) ) 2 1 ( cos 14 12 π − cos 14 14 π ) 2 1 ( cos 7 6 π − ( − 1 ) ) 2 1 ( cos 7 6 π + 1 ) 2 1 ( 2 sin 14 11 π sin 14 3 π ) 2 1 ( cos ( 14 11 π − 14 3 π ) − cos ( 14 11 π + 14 3 π ) ) 2 1 ( cos 14 8 π − cos 14 14 π ) 2 1 ( cos 7 4 π − ( − 1 ) ) 2 1 ( cos 7 4 π + 1 ) 2 1 ( 2 sin 14 9 π sin 14 5 π ) 2 1 ( cos ( 14 9 π − 14 5 π ) − cos ( 14 9 π + 14 5 π ) ) 2 1 ( cos 14 4 π − cos 14 14 π ) 2 1 ( cos 7 2 π − ( − 1 ) ) 2 1 ( cos 7 2 π + 1 ) Sehingga diperoleh = = = = = = = = = = = = = = = = = = = = sin 14 π sin 14 3 π sin 14 5 π sin 14 7 π sin 14 9 π sin 14 11 π sin 14 13 π sin 14 13 π sin 14 π sin 14 11 π sin 14 3 π sin 14 9 π sin 14 5 π sin 14 7 π 2 1 ( cos 7 6 π + 1 ) 2 1 ( cos 7 4 π + 1 ) 2 1 ( cos 7 2 π + 1 ) sin 2 π 8 1 ( cos 7 6 π + 1 ) ( cos 7 4 π + 1 ) ( cos 7 2 π + 1 ) 1 ingatcos2α=2cosα-1 8 1 ( 2 cos 2 7 3 π − 1 + 1 ) ( 2 cos 2 7 2 π − 1 + 1 ) ( 2 cos 2 7 π − 1 + 1 ) 8 1 ( 2 cos 2 7 3 π ) ( 2 cos 2 7 2 π ) ( 2 cos 2 7 π ) cos 2 7 3 π cos 2 7 2 π cos 2 7 π ( cos 7 3 π cos 7 2 π cos 7 π ) 2 ( cos 7 π cos 7 2 π cos 7 3 π × 2 s i n 7 π 2 s i n 7 π ) 2 ( 2 s i n 7 π 2 s i n 7 π cos 7 π cos 7 2 π cos 7 3 π ) 2 ( 2 s i n 7 π 1 sin 7 2 π cos 7 2 π cos 7 3 π ) 2 ( 2.2 s i n 7 π 1 2. sin 7 2 π cos 7 2 π cos 7 3 π ) 2 ( 2.2 s i n 7 π 1 sin 7 4 π cos 7 3 π ) 2 ( 2.2 s i n ( π − 7 π ) 1 sin ( π − 7 4 π ) cos 7 3 π ) 2 ( 4 s i n 7 6 π 1 sin 7 3 π cos 7 3 π ) 2 ( 2.4 s i n 7 6 π 1 2. sin 7 3 π cos 7 3 π ) 2 ( 2.4 s i n 7 6 π 1 sin 7 6 π ) 2 ( 8 1 ) 2 64 1 2 6 1 2 − 6 Dengan demikian, terbukti bahwa sin 14 π sin 14 3 π sin 14 5 π sin 14 7 π sin 14 9 π sin 14 11 π sin 14 13 π = 2 − 6

Ingat
Rumus sudut ganda sinus
$2 sin α cos α = sin 2 α$ .
Rumus perkalian trigonometri
$2 sin α sin β = cos (α - β) - cos (α + β)$

Dari soal tersebut dapat kita uraikan sebagai berikut.

$sin \frac{13 π}{14} sin \frac{π}{14} sin \frac{11 π}{14} sin \frac{3 π}{14} sin \frac{9 π}{14} sin \frac{5 π}{14} = = = = = = = = = = = = = = = \frac{1}{2} (2 sin \frac{13 π}{14} sin \frac{π}{14}) \frac{1}{2} (cos (\frac{13 π}{14} - \frac{π}{14}) - cos (\frac{13 π}{14} + \frac{π}{14})) \frac{1}{2} (cos \frac{12 π}{14} - cos \frac{14 π}{14}) \frac{1}{2} (cos \frac{6 π}{7} - (- 1)) \frac{1}{2} (cos \frac{6 π}{7} + 1) \frac{1}{2} (2 sin \frac{11 π}{14} sin \frac{3 π}{14}) \frac{1}{2} (cos (\frac{11 π}{14} - \frac{3 π}{14}) - cos (\frac{11 π}{14} + \frac{3 π}{14})) \frac{1}{2} (cos \frac{8 π}{14} - cos \frac{14 π}{14}) \frac{1}{2} (cos \frac{4 π}{7} - (- 1)) \frac{1}{2} (cos \frac{4 π}{7} + 1) \frac{1}{2} (2 sin \frac{9 π}{14} sin \frac{5 π}{14}) \frac{1}{2} (cos (\frac{9 π}{14} - \frac{5 π}{14}) - cos (\frac{9 π}{14} + \frac{5 π}{14})) \frac{1}{2} (cos \frac{4 π}{14} - cos \frac{14 π}{14}) \frac{1}{2} (cos \frac{2 π}{7} - (- 1)) \frac{1}{2} (cos \frac{2 π}{7} + 1)$

Sehingga diperoleh

$= = = = = = = = = = = = = = = = = = = = sin \frac{π}{14} sin \frac{3 π}{14} sin \frac{5 π}{14} sin \frac{7 π}{14} sin \frac{9 π}{14} sin \frac{11 π}{14} sin \frac{13 π}{14} sin \frac{13 π}{14} sin \frac{π}{14} sin \frac{11 π}{14} sin \frac{3 π}{14} sin \frac{9 π}{14} sin \frac{5 π}{14} sin \frac{7 π}{14} \frac{1}{2} (cos \frac{6 π}{7} + 1) \frac{1}{2} (cos \frac{4 π}{7} + 1) \frac{1}{2} (cos \frac{2 π}{7} + 1) sin \frac{π}{2} \frac{1}{8} (cos \frac{6 π}{7} + 1) (cos \frac{4 π}{7} + 1) (cos \frac{2 π}{7} + 1) 1 ingat cos2α=2 cosα -1 \frac{1}{8} (2 cos^{2} \frac{3 π}{7} - 1 + 1) (2 cos^{2} \frac{2 π}{7} - 1 + 1) (2 cos^{2} \frac{π}{7} - 1 + 1) \frac{1}{8} (2 cos^{2} \frac{3 π}{7}) (2 cos^{2} \frac{2 π}{7}) (2 cos^{2} \frac{π}{7}) cos^{2} \frac{3 π}{7} cos^{2} \frac{2 π}{7} cos^{2} \frac{π}{7} (cos \frac{3 π}{7} cos \frac{2 π}{7} cos \frac{π}{7})^{2} (cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{3 π}{7} \times \frac{2 s i n \frac{π}{7}}{2 s i n \frac{π}{7}})^{2} (\frac{2 s i n \frac{π}{7}}{2 s i n \frac{π}{7}} cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{3 π}{7})^{2} (\frac{1}{2 s i n \frac{π}{7}} sin \frac{2 π}{7} cos \frac{2 π}{7} cos \frac{3 π}{7})^{2} (\frac{1}{2.2 s i n \frac{π}{7}} 2. sin \frac{2 π}{7} cos \frac{2 π}{7} cos \frac{3 π}{7})^{2} (\frac{1}{2.2 s i n \frac{π}{7}} sin \frac{4 π}{7} cos \frac{3 π}{7})^{2} (\frac{1}{2.2 s i n ( π - \frac{π}{7} )} sin (π - \frac{4 π}{7}) cos \frac{3 π}{7})^{2} (\frac{1}{4 s i n \frac{6 π}{7}} sin \frac{3 π}{7} cos \frac{3 π}{7})^{2} (\frac{1}{2.4 s i n \frac{6 π}{7}} 2. sin \frac{3 π}{7} cos \frac{3 π}{7})^{2} (\frac{1}{2.4 s i n \frac{6 π}{7}} sin \frac{6 π}{7})^{2} (\frac{1}{8})^{2} \frac{1}{64} \frac{1}{2 ^{6}} 2^{- 6}$

Dengan demikian, terbukti bahwa
$sin \frac{π}{14} sin \frac{3 π}{14} sin \frac{5 π}{14} sin \frac{7 π}{14} sin \frac{9 π}{14} sin \frac{11 π}{14} sin \frac{13 π}{14} = 2^{- 6}$