Pertanyaan

Tunjukkan bahwa: sin 1 0 ∘ sin 3 0 ∘ sin 5 0 ∘ sin 7 0 ∘ = 16 1

Tunjukkan bahwa:

$sin 1 0^{\circ} sin 3 0^{\circ} sin 5 0^{\circ} sin 7 0^{\circ} = \frac{1}{16}$

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S. Solehuzain

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Jawaban

terbukti bahwa sin 1 0 ∘ sin 3 0 ∘ sin 5 0 ∘ sin 7 0 ∘ = 16 1 .

terbukti bahwa

sin 1 0^{\circ} sin 3 0^{\circ} sin 5 0^{\circ} sin 7 0^{\circ} = \frac{1}{16}

Pembahasan

Ingat rumus perkalian trigonometri 2 sin α sin β = cos ( α − β ) − cos ( α + β ) . Sehingga diperoleh = = = = = = = sin 1 0 ∘ sin 3 0 ∘ sin 5 0 ∘ sin 7 0 ∘ 2 1 sin 1 0 ∘ sin 5 0 ∘ sin 7 0 ∘ 2 1 . 2 1 ( 2 sin 1 0 ∘ sin 5 0 ∘ ) 4 1 ( cos ( 1 0 ∘ − 5 0 ∘ ) − cos ( 1 0 ∘ + 5 0 ∘ ) ) sin 7 0 ∘ 4 1 ( cos ( − 4 0 ∘ ) − cos ( 6 0 ∘ ) ) sin 7 0 ∘ 4 1 ( cos ( 4 0 ∘ ) − 2 1 ) sin 7 0 ∘ ( 4 1 cos ( 4 0 ∘ ) − 8 1 ) sin 7 0 ∘ 4 1 cos 4 0 ∘ sin 7 0 ∘ − 8 1 sin 7 0 ∘ ingat 2 cos α sin β = sin ( α + β ) − sin ( α − β ) Dengan demikian, terbukti bahwa sin 1 0 ∘ sin 3 0 ∘ sin 5 0 ∘ sin 7 0 ∘ = 16 1 .

Ingat rumus perkalian trigonometri
$2 sin α sin β = cos (α - β) - cos (α + β)$ .

Sehingga diperoleh

$= = = = = = = sin 1 0^{\circ} sin 3 0^{\circ} sin 5 0^{\circ} sin 7 0^{\circ} \frac{1}{2} sin 1 0^{\circ} sin 5 0^{\circ} sin 7 0^{\circ} \frac{1}{2} . \frac{1}{2} (2 sin 1 0^{\circ} sin 5 0^{\circ}) \frac{1}{4} (cos (1 0^{\circ} - 5 0^{\circ}) - cos (1 0^{\circ} + 5 0^{\circ})) sin 7 0^{\circ} \frac{1}{4} (cos (- 4 0^{\circ}) - cos (6 0^{\circ})) sin 7 0^{\circ} \frac{1}{4} (cos (4 0^{\circ}) - \frac{1}{2}) sin 7 0^{\circ} (\frac{1}{4} cos (4 0^{\circ}) - \frac{1}{8}) sin 7 0^{\circ} \frac{1}{4} cos 4 0^{\circ} sin 7 0^{\circ} - \frac{1}{8} sin 7 0^{\circ}$

ingat $2 cos α sin β = sin (α + β) - sin (α - β)$

$\begin{array}{l}\begin{array}{rcl}&&\frac14\end{array}\begin{array}{rcl}&&\cos\end{array}\begin{array}{rcl}&&40^\circ\end{array}\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}\begin{array}{rcl}&&\textdegree\end{array}\begin{array}{rcl}&&-\end{array}\begin{array}{rcl}&&\frac18\end{array}\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\=\frac18\left(2\;\cos\begin{array}{rcl}&&40^\circ\end{array}\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}\begin{array}{rcl}&&\textdegree\end{array}\right)+^\circ\begin{array}{rcl}&&-\end{array}\begin{array}{rcl}&&\frac18\end{array}\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\=\frac18\left(\sin\left(40^\circ+70^\circ\right)-\sin\left(40^\circ-70^\circ\right)\right)-\begin{array}{rcl}&&\frac18\end{array}\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\=\frac18\left(\sin\left(110^\circ\right)-\sin\left(-30^\circ\right)\right)-\frac18\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\\;\text{ingat}\;\sin\left(-a\right)=-\sin\left(a\right)\\=\frac18\left(\sin\left(110^\circ\right)-\left(-\sin\left(30^\circ\right)\right)\right)-\frac18\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\=\frac18\left(\sin\left(110^\circ\right)+\frac12\right)-\frac18\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}^\circ\\=\frac18\sin\left(180^\circ-70^\circ\right)+\frac1{16}-\frac18\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}^\circ\\=\frac18\sin\left(70^\circ\right)+\frac1{16}-\frac18\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&70\end{array}^\circ\\=\frac1{16}\\\end{array}$

Dengan demikian, terbukti bahwa $sin 1 0^{\circ} sin 3 0^{\circ} sin 5 0^{\circ} sin 7 0^{\circ} = \frac{1}{16}$ .

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