Pertanyaan

Tunjukkan bahwa: sin 2 0 ∘ sin 4 0 ∘ sin 6 0 ∘ sin 8 0 ∘ = 16 3

Tunjukkan bahwa:

$sin 2 0^{\circ} sin 4 0^{\circ} sin 6 0^{\circ} sin 8 0^{\circ} = \frac{3}{16}$

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S. Solehuzain

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

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Jawaban

terbukti bahwa sin 2 0 ∘ sin 4 0 ∘ sin 6 0 ∘ sin 8 0 ∘ = 16 3 .

terbukti bahwa

sin 2 0^{\circ} sin 4 0^{\circ} sin 6 0^{\circ} sin 8 0^{\circ} = \frac{3}{16}

Pembahasan

Ingat rumus perkalian trigonometri 2 sin α sin β = cos ( α − β ) − cos ( α + β ) . Sehingga diperoleh = = = = = = sin 2 0 ∘ sin 4 0 ∘ sin 6 0 ∘ sin 8 0 ∘ sin 2 0 ∘ . 2 1 ( 2 sin 8 0 ∘ sin 4 0 ∘ ) sin 6 0 ∘ sin 2 0 ∘ . 2 1 ( cos ( 8 0 ∘ − 4 0 ∘ ) − cos ( 8 0 ∘ + 4 0 ∘ ) ) sin 6 0 ∘ sin 2 0 ∘ . 2 1 ( cos ( 4 0 ∘ ) − cos ( 12 0 ∘ ) ) sin 6 0 ∘ sin 2 0 ∘ . 2 1 ( cos ( 4 0 ∘ ) − ( − 2 1 ) ) sin 6 0 ∘ sin 2 0 ∘ . 2 1 ( cos ( 4 0 ∘ ) + 2 1 ) sin 6 0 ∘ ( 2 1 sin 2 0 ∘ cos ( 4 0 ∘ ) + 4 1 ) sin 6 0 ∘ ingat 2 sin α cos β = sin ( α + β ) + sin ( α − β ) Dengan demikian,terbukti bahwa sin 2 0 ∘ sin 4 0 ∘ sin 6 0 ∘ sin 8 0 ∘ = 16 3 .

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

Sehingga diperoleh

$= = = = = = sin 2 0^{\circ} sin 4 0^{\circ} sin 6 0^{\circ} sin 8 0^{\circ} sin 2 0^{\circ} . \frac{1}{2} (2 sin 8 0^{\circ} sin 4 0^{\circ}) sin 6 0^{\circ} sin 2 0^{\circ} . \frac{1}{2} (cos (8 0^{\circ} - 4 0^{\circ}) - cos (8 0^{\circ} + 4 0^{\circ})) sin 6 0^{\circ} sin 2 0^{\circ} . \frac{1}{2} (cos (4 0^{\circ}) - cos (12 0^{\circ})) sin 6 0^{\circ} sin 2 0^{\circ} . \frac{1}{2} (cos (4 0^{\circ}) - (- \frac{1}{2})) sin 6 0^{\circ} sin 2 0^{\circ} . \frac{1}{2} (cos (4 0^{\circ}) + \frac{1}{2}) sin 6 0^{\circ} (\frac{1}{2} sin 2 0^{\circ} cos (4 0^{\circ}) + \frac{1}{4}) sin 6 0^{\circ}$

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

$\begin{array}{l}\left(\frac12\sin20^\circ\cos\left(40^\circ\right)+\frac14\sin20^\circ\right)\;\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&60\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\=\left(\frac14\left(2\;\sin20^\circ\;\cos40^\circ\right)+\frac14\sin20^\circ\right)\;\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&60\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\=\left(\frac14\left(\sin\left(20^\circ+40^\circ\right)+\sin\left(20^\circ-60^\circ\right)\right)+\frac14\sin20^\circ\right)\;\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&60\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\=\left(\frac14\left(\sin\left(60^\circ\right)+\sin\left(-20^\circ\right)\right)+\frac14\sin20^\circ\right)\;\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&60\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\\;\text{ingat}\;\sin\left(-a\right)=-\sin\left(a\right)\\=\left(\frac14\left(\frac12\sqrt3-\sin\left(20^\circ\right)\right)+\frac14\sin20^\circ\right)\;\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&60\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\=\left(\frac18\sqrt3-\frac14\sin\left(20^\circ\right)+\frac14\sin20^\circ\right)\;\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&60\end{array}\begin{array}{rcl}&&\textdegree\end{array}\\=\frac18\sqrt3\;.\begin{array}{rcl}&&\sin\end{array}\begin{array}{rcl}&&60\end{array}\begin{array}{rcl}&&\textdegree\end{array}\begin{array}{rcl}&&\;\end{array}\\\begin{array}{rcl}&=&\end{array}\frac18\sqrt3\;.\frac12\sqrt3\;\\=\frac3{16}\end{array}$

Dengan demikian, terbukti bahwa $sin 2 0^{\circ} sin 4 0^{\circ} sin 6 0^{\circ} sin 8 0^{\circ} = \frac{3}{16}$ .

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