Penerapan Rumus Trigonometri pada Soal-soal Bagian 1

         Blog Koma - Setelah mempelajari materi "rumus jumlah dan selisih sudut pada trigonometri" dan materi "rumus hasil kali antara dua bentuk trigonometri" serta rumus trigonometri yang lainnya, pada artikel ini kita akan coba membahas tentang Penerapan Rumus Trigonometri pada Soal-soal Bagian 1. Soal-soal yang melibatkan rumus-rumus trigonometri ini biasanya kita jumpai pada soal UJian Nasional, soal seleksi masuk perguruan tinggi baik negeri maupun swasta seperti SBMPTN, UM UGM, SIMAK UI, dan lain-lainnya. Hal mendasar yang harus kita perhatikan adalah ketelitian baik dalam menggunakan rumusnya atau dalam melakukan penjabaran dan perhitungannya. Langsung saja kita pelajari beberapa contoh soal berikut ini.

1). Tentukan nilai dari bentuk $ \sin 20^\circ \sin 40^\circ \sin 80^\circ $ ?

Penyelesaian :
Ada tiga cara yang akan kita sajikan dalam menyelesaikan soal nomor 1 :
Cara I :
*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :
$ \sin A . \sin B = -\frac{1}{2} [ \cos (A+B) - \cos (A-B)] $
$ \sin A . \cos B = \frac{1}{2} [ \sin (A+B) + \sin (A-B)] $
$ \sin ( 180^\circ - A) = \sin A $
$ \sin 100^\circ = \sin ( 180^\circ - 80^\circ ) = \sin 80^\circ $
*). Menyelesaikan soal :
$ \begin{align} \sin 20^\circ \sin 40^\circ \sin 80^\circ & = (\sin 20^\circ . \sin 40^\circ ) . \sin 80^\circ \\ & = (\sin 40^\circ . \sin 20^\circ ) . \sin 80^\circ \\ & = \left(-\frac{1}{2} [ \cos (40^\circ + 20^\circ) - \cos (40^\circ - 20^\circ)] \right) . \sin 80^\circ \\ & = \left(-\frac{1}{2} [ \cos 60^\circ - \cos 20^\circ ] \right) . \sin 80^\circ \\ & = \left(-\frac{1}{2} [ \frac{1}{2} - \cos 20^\circ ] \right) . \sin 80^\circ \\ & = \left( - \frac{1}{4} + \frac{1}{2} \cos 20^\circ \right) . \sin 80^\circ \\ & = - \frac{1}{4}\sin 80^\circ + \frac{1}{2} \sin 80^\circ . \cos 20^\circ \\ & = - \frac{1}{4}\sin 80^\circ + \frac{1}{2} (\sin 80^\circ . \cos 20^\circ ) \\ & = - \frac{1}{4}\sin 80^\circ + \frac{1}{2} \times \frac{1}{2} [ \sin (80^\circ + 20^\circ ) + \sin (80^\circ - 20^\circ )] \\ & = - \frac{1}{4}\sin 80^\circ + \frac{1}{4} [ \sin 100^\circ + \sin 60^\circ ] \\ & = - \frac{1}{4}\sin 80^\circ + \frac{1}{4} [ \sin 80^\circ + \sin 60^\circ ] \\ & = - \frac{1}{4}\sin 80^\circ + \frac{1}{4} \sin 80^\circ + \frac{1}{4} \sin 60^\circ \\ & = \frac{1}{4} \sin 60^\circ \\ & = \frac{1}{4} \times \frac{1}{2} \sqrt{3} \\ & = \frac{1}{8} \sqrt{3} \end{align} $
jadi, nilai $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.

Cara II :
*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :
$ \sin A . \sin B = -\frac{1}{2} [ \cos (A+B) - \cos (A-B)] $
$ \cos A . \sin B = \frac{1}{2} [ \sin (A+B) - \sin (A-B)] $
*). Menyelesaikan soal :
$ \begin{align} \sin 20^\circ \sin 40^\circ \sin 80^\circ & = \sin 20^\circ . (\sin 40^\circ . \sin 80^\circ ) \\ & = \sin 20^\circ . (\sin 80^\circ . \sin 40^\circ ) \\ & = \sin 20^\circ . \left( -\frac{1}{2} [ \cos (80^\circ + 40^\circ ) - \cos (80^\circ - 40^\circ )]\right) \\ & = \sin 20^\circ . \left( -\frac{1}{2} [ \cos 120^\circ - \cos 40^\circ ]\right) \\ & = \sin 20^\circ . \left( -\frac{1}{2} [ -\frac{1}{2} - \cos 40^\circ ]\right) \\ & = \sin 20^\circ . \left( \frac{1}{4} + \frac{1}{2} \cos 40^\circ \right) \\ & = \frac{1}{4} \sin 20^\circ + \frac{1}{2} \cos 40^\circ \sin 20^\circ \\ & = \frac{1}{4} \sin 20^\circ + \frac{1}{2} ( \cos 40^\circ \sin 20^\circ ) \\ & = \frac{1}{4} \sin 20^\circ + \frac{1}{2} \times ( \frac{1}{2} [ \sin (40^\circ + 20^\circ ) - \sin ( 40^\circ - 20^\circ )] ) \\ & = \frac{1}{4} \sin 20^\circ + ( \frac{1}{4} [ \sin 60^\circ - \sin 20^\circ ] ) \\ & = \frac{1}{4} \sin 20^\circ + ( \frac{1}{4} [ \frac{1}{2}\sqrt{3} - \sin 20^\circ ] ) \\ & = \frac{1}{4} \sin 20^\circ + \frac{1}{8} \sqrt{3} - \frac{1}{4} \sin 20^\circ \\ & = \frac{1}{8} \sqrt{3} \end{align} $
jadi, nilai $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.

Cara III :
*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :
$ \sin A . \sin B = -\frac{1}{2} [ \cos (A+B) - \cos (A-B)] $
$ \cos A . \sin B = \frac{1}{2} [ \sin (A+B) - \sin (A-B)] $
$ \sin ( 180^\circ - A) = \sin A $
$ \sin 140^\circ = \sin ( 180^\circ - 40^\circ ) = \sin 40^\circ $
*). Menyelesaikan soal :
$ \begin{align} \sin 20^\circ \sin 40^\circ \sin 80^\circ & = \sin 40^\circ . (\sin 80^\circ . \sin 20^\circ ) \\ & = \sin 40^\circ . (\sin 80^\circ . \sin 20^\circ ) \\ & = \sin 40^\circ . \left( -\frac{1}{2} [ \cos (80^\circ + 20^\circ ) - \cos (80^\circ - 20^\circ )] \right) \\ & = \sin 40^\circ . \left( -\frac{1}{2} [ \cos 100^\circ - \cos 60^\circ ] \right) \\ & = \sin 40^\circ . \left( -\frac{1}{2} [ \cos 100^\circ - \frac{1}{2} ] \right) \\ & = \sin 40^\circ . \left( -\frac{1}{2} \cos 100^\circ + \frac{1}{4} \right) \\ & = -\frac{1}{2} \cos 100^\circ \sin 40^\circ + \frac{1}{4} \sin 40^\circ \\ & = -\frac{1}{2} (\cos 100^\circ \sin 40^\circ ) + \frac{1}{4} \sin 40^\circ \\ & = -\frac{1}{2} \times ( \frac{1}{2} [ \sin (100^\circ + 40^\circ ) - \sin (100^\circ - 40^\circ )] ) + \frac{1}{4} \sin 40^\circ \\ & = ( -\frac{1}{4} [ \sin 140^\circ - \sin 60^\circ ] ) + \frac{1}{4} \sin 40^\circ \\ & = ( -\frac{1}{4} [ \sin 140^\circ - \frac{1}{2}\sqrt{3} ] ) + \frac{1}{4} \sin 40^\circ \\ & = -\frac{1}{4} \sin 140^\circ + \frac{1}{8}\sqrt{3} + \frac{1}{4} \sin 40^\circ \\ & = -\frac{1}{4} \sin 40^\circ + \frac{1}{8}\sqrt{3} + \frac{1}{4} \sin 40^\circ \\ & = \frac{1}{8} \sqrt{3} \end{align} $
jadi, nilai $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.

2). Tentukan nilai dari bentuk $ \cos 20^\circ \cos 40^\circ \cos 80^\circ $ ?

Penyelesaian :
Ada empat cara yang akan kita sajikan dalam menyelesaikan soal nomor 2 :
Cara I :
*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :
$ \cos A . \cos B = \frac{1}{2} [ \cos (A+B) + \cos (A-B)] $
$ \cos ( 180^\circ - A) = -\cos A $
$ \cos 100^\circ = \cos ( 180^\circ - 80^\circ ) = -\cos 80^\circ $
*). Menyelesaikan soal :
$ \begin{align} \cos 20^\circ \cos 40^\circ \cos 80^\circ & = (\cos 20^\circ \cos 40^\circ ) \cos 80^\circ \\ & = (\cos 40^\circ \cos 20^\circ ) \cos 80^\circ \\ & = \left( \frac{1}{2} [ \cos (40^\circ + 20^\circ ) + \cos (40^\circ - 20^\circ )] \right) \cos 80^\circ \\ & = \left( \frac{1}{2} [ \cos 60^\circ + \cos 20^\circ ] \right) \cos 80^\circ \\ & = \left( \frac{1}{2} [ \frac{1}{2} + \cos 20^\circ ] \right) \cos 80^\circ \\ & = \left( \frac{1}{4} + \frac{1}{2} \cos 20^\circ \right) \cos 80^\circ \\ & = \frac{1}{4} \cos 80^\circ + \frac{1}{2} \cos 80^\circ \cos 20^\circ \\ & = \frac{1}{4} \cos 80^\circ + \frac{1}{2} ( \cos 80^\circ \cos 20^\circ ) \\ & = \frac{1}{4} \cos 80^\circ + \frac{1}{2} \times ( \frac{1}{2} [ \cos (80^\circ + 20^\circ ) + \cos (80^\circ - 20^\circ )] ) \\ & = \frac{1}{4} \cos 80^\circ + ( \frac{1}{4} [ \cos 100^\circ + \cos 60^\circ ] ) \\ & = \frac{1}{4} \cos 80^\circ + ( \frac{1}{4} [ -\cos 80^\circ + \frac{1}{2} ] ) \\ & = \frac{1}{4} \cos 80^\circ -\frac{1}{4} \cos 80^\circ + \frac{1}{8} \\ & = \frac{1}{8} \end{align} $
jadi, nilai $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8} . \, \heartsuit $.

Cara II :
*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :
$ \cos A . \cos B = \frac{1}{2} [ \cos (A+B) + \cos (A-B)] $
$ \cos ( 180^\circ - A) = -\cos A $
$ \cos 140^\circ = \cos ( 180^\circ - 40^\circ ) = -\cos 40^\circ $
*). Menyelesaikan soal :
$ \begin{align} \cos 20^\circ \cos 40^\circ \cos 80^\circ & = (\cos 80^\circ \cos 20^\circ ) \cos 40^\circ \\ & = (\cos 80^\circ \cos 20^\circ ) \cos 40^\circ \\ & = \left( \frac{1}{2} [ \cos (80^\circ + 20^\circ ) + \cos (80^\circ - 20^\circ )] \right) \cos 40^\circ \\ & = \left( \frac{1}{2} [ \cos 100^\circ + \cos 60^\circ ] \right) \cos 40^\circ \\ & = \left( \frac{1}{2} [ \cos 100^\circ + \frac{1}{2} ] \right) \cos 40^\circ \\ & = \left( \frac{1}{2} \cos 100^\circ + \frac{1}{4} \right) \cos 40^\circ \\ & = \frac{1}{4} \cos 40^\circ + \frac{1}{2} \cos 100^\circ \cos 40^\circ \\ & = \frac{1}{4} \cos 40^\circ + \frac{1}{2} ( \cos 100^\circ \cos 40^\circ ) \\ & = \frac{1}{4} \cos 40^\circ + \frac{1}{2} \times ( \frac{1}{2} [ \cos (100^\circ + 40^\circ ) + \cos (100^\circ - 40^\circ )] ) \\ & = \frac{1}{4} \cos 40^\circ + ( \frac{1}{4} [ \cos 140^\circ + \cos 60^\circ ] ) \\ & = \frac{1}{4} \cos 40^\circ + ( \frac{1}{4} [ -\cos 40^\circ + \frac{1}{2} ] ) \\ & = \frac{1}{4} \cos 40^\circ -\frac{1}{4} \cos 40^\circ + \frac{1}{8} \\ & = \frac{1}{8} \end{align} $
jadi, nilai $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8} . \, \heartsuit $.

Cara III :
*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :
$ \cos A . \cos B = \frac{1}{2} [ \cos (A+B) + \cos (A-B)] $
*). Menyelesaikan soal :
$ \begin{align} \cos 20^\circ \cos 40^\circ \cos 80^\circ & = (\cos 80^\circ \cos 40^\circ ) \cos 20^\circ \\ & = \left( \frac{1}{2} [ \cos (80^\circ + 40^\circ ) + \cos (80^\circ - 40^\circ )] \right) \cos 20^\circ \\ & = \left( \frac{1}{2} [ \cos 120^\circ + \cos 40^\circ ] \right) \cos 20^\circ \\ & = \left( \frac{1}{2} [ -\frac{1}{2} + \cos 40^\circ ] \right) \cos 20^\circ \\ & = \left( -\frac{1}{4} + \frac{1}{2} \cos 40^\circ \right) \cos 20^\circ \\ & = - \frac{1}{4} \cos 20^\circ + \frac{1}{2} \cos 40^\circ \cos 20^\circ \\ & = - \frac{1}{4} \cos 20^\circ + \frac{1}{2} ( \cos 40^\circ \cos 20^\circ ) \\ & = - \frac{1}{4} \cos 20^\circ + \frac{1}{2} \times ( \frac{1}{2} [ \cos (40^\circ + 20^\circ ) + \cos (40^\circ - 20^\circ )] ) \\ & = - \frac{1}{4} \cos 20^\circ + ( \frac{1}{4} [ \cos 60^\circ + \cos 20^\circ ] ) \\ & = - \frac{1}{4} \cos 20^\circ + ( \frac{1}{4} [ \frac{1}{2} + \cos 20^\circ ] ) \\ & = - \frac{1}{4} \cos 20^\circ + \frac{1}{8} + \frac{1}{4} \cos 20^\circ \\ & = \frac{1}{8} \end{align} $
jadi, nilai $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8} . \, \heartsuit $.

Cara IV :
*). Rumus Dasar yang kita gunakan adalah "Rumus Trigonometri untuk Sudut Ganda" :
$ \sin 2A = 2\sin A \cos A \rightarrow \sin A \cos A = \frac{1}{2} \sin 2A $
Rumus Lain :
$ \sin (180^\circ - A) = \sin A $
$ \sin 160^\circ = \sin (180^\circ - 20^\circ ) = \sin 20^\circ $
*). Menyelesaikan soal ,
Kita misalkan hasilnya $ P $ atau $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = P $ :
$ \begin{align} P & = \cos 20^\circ \cos 40^\circ \cos 80^\circ \, \, \, \, \text{(kali } \sin 20^\circ ) \\ P . \sin 20^\circ & = \sin 20^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ \\ & = (\sin 20^\circ \cos 20^\circ ) \cos 40^\circ \cos 80^\circ \\ & = ( \frac{1}{2}\sin 40^\circ ) \cos 40^\circ \cos 80^\circ \\ & = \frac{1}{2}\sin 40^\circ \cos 40^\circ \cos 80^\circ \\ & = \frac{1}{2} ( \sin 40^\circ \cos 40^\circ ) \cos 80^\circ \\ & = \frac{1}{2} \times ( \frac{1}{2} \sin 80^\circ ) \cos 80^\circ \\ & = \frac{1}{4} \sin 80^\circ \cos 80^\circ \\ & = \frac{1}{4} ( \sin 80^\circ \cos 80^\circ ) \\ & = \frac{1}{4} \times ( \frac{1}{2} \sin 160^\circ ) \\ P . \sin 20^\circ & = \frac{1}{8} \sin 20^\circ \, \, \, \, \text{(bagi } \sin 20^\circ ) \\ P & = \frac{1}{8} \end{align} $
jadi, nilai $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = P = \frac{1}{8} . \, \heartsuit $.

3). Tentukan nilai dari $ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ $ ?
(Soal UN Matematika IPA tahun 2007)

Penyelesaian :
Soal ini bisa diselesaikan dengan berbagai cara, diantaranya :
Cara I :
*). Rumus dasar yang digunakan :
$ \cos A + \cos B = 2\cos \frac{(A+B)}{2} \cos \frac{(A-B)}{2} $
$ \cos (180^\circ - A ) = - \cos A $
Nilai $ \cos 160^\circ = \cos (180^\circ - 20^\circ ) = - \cos 20^\circ $
*). Menyelesaikan soal :
$ \begin{align} \cos 40^\circ + \cos 80^\circ + \cos 160^\circ & = ( \cos 80^\circ + \cos 40^\circ ) + \cos 160^\circ \\ & = ( 2\cos \frac{(80^\circ + 40^\circ )}{2} \cos \frac{(80^\circ - 40^\circ )}{2} ) + \cos 160^\circ \\ & = ( 2\cos \frac{(120^\circ )}{2} \cos \frac{(40^\circ )}{2} ) + (- \cos 20^\circ ) \\ & = ( 2\cos 60^\circ \cos 20^\circ ) - \cos 20^\circ \\ & = 2 . \frac{1}{2} \cos 20^\circ - \cos 20^\circ \\ & = \cos 20^\circ - \cos 20^\circ \\ & = 0 \end{align} $
jadi, nilai $ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.

Cara II :
*). Rumus dasar yang digunakan :
$ \cos A + \cos B = 2\cos \frac{(A+B)}{2} \cos \frac{(A-B)}{2} $
*). Menyelesaikan soal :
$ \begin{align} \cos 40^\circ + \cos 80^\circ + \cos 160^\circ & = ( \cos 160^\circ + \cos 80^\circ ) + \cos 40^\circ \\ & = ( 2\cos \frac{(160^\circ + 80^\circ )}{2} \cos \frac{(160^\circ - 80^\circ )}{2} ) + \cos 40^\circ \\ & = ( 2\cos \frac{(240^\circ )}{2} \cos \frac{(80^\circ )}{2} ) + \cos 40^\circ \\ & = ( 2\cos 120^\circ \cos 40^\circ ) + \cos 40^\circ \\ & = 2 . -\frac{1}{2} \cos 40^\circ + \cos 40^\circ \\ & = - \cos 40^\circ + \cos 40^\circ \\ & = 0 \end{align} $
jadi, nilai $ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.

Cara III :
*). Rumus dasar yang digunakan :
$ \cos A + \cos B = 2\cos \frac{(A+B)}{2} \cos \frac{(A-B)}{2} $
*). Menyelesaikan soal :
$ \begin{align} \cos 40^\circ + \cos 80^\circ + \cos 160^\circ & = ( \cos 160^\circ + \cos 40^\circ ) + \cos 80^\circ \\ & = ( 2\cos \frac{(160^\circ + 40^\circ )}{2} \cos \frac{(160^\circ - 40^\circ )}{2} ) + \cos 80^\circ \\ & = ( 2\cos \frac{(200^\circ )}{2} \cos \frac{(120^\circ )}{2} ) + \cos 80^\circ \\ & = ( 2\cos 100^\circ \cos 60^\circ ) + \cos 80^\circ \\ & = 2 . \cos 100^\circ . \frac{1}{2} + \cos 80^\circ \\ & = \cos 100^\circ + \cos 80^\circ \\ & = 2\cos \frac{(100^\circ + 80^\circ )}{2} \cos \frac{(100^\circ - 80^\circ )}{2} \\ & = 2\cos \frac{180^\circ }{2} \cos \frac{20^\circ }{2} \\ & = 2\cos 90^\circ \cos 10^\circ \\ & = 2 \times 0 \times \cos 10^\circ \\ & = 0 \end{align} $
jadi, nilai $ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.

4). Tentukan nilai dari $ \csc 10^\circ - \sqrt{3} \sec 10^\circ $?
(soal SIMAK UI tahun 2013 Matematika IPA kode 133)

Penyelesaian :
*). Rumus dasar yang digunakan :
i). Sudut Rangkap :
$ \sin 2A = 2\sin A \cos A \rightarrow \sin A \cos A = \frac{1}{2} \sin 2A $.
ii). Selilisih sudut : $ \sin (A - B ) = \sin A \cos B - \cos A \sin B $.
iii). Rumus lain :
$ \csc A = \frac{1}{\sin A} \, $ dan $ \sec A = \frac{1}{\cos A } $.
*). Menyelesaikan soal :
$ \begin{align} \csc 10^\circ - \sqrt{3} \sec 10^\circ & = \frac{1}{\sin 10^\circ } - \frac{\sqrt{3} }{\cos 10^\circ } \\ & = \frac{1}{\sin 10^\circ } - \frac{\sqrt{3} }{\cos 10^\circ } \\ & = \frac{\cos 10^\circ }{\sin 10^\circ \cos 10^\circ } - \frac{\sqrt{3} \sin 10^\circ }{\sin 10^\circ \cos 10^\circ } \\ & = \frac{\cos 10^\circ - \sqrt{3} \sin 10^\circ }{\sin 10^\circ \cos 10^\circ } \, \, \, \, \, \, \text{(modifikasi)} \\ & = \frac{ 2 \times ( \frac{1}{2} . \cos 10^\circ - \frac{1}{2} \sqrt{3} . \sin 10^\circ ) }{\sin 10^\circ \cos 10^\circ } \\ & = \frac{ 2 \times ( \sin 30^\circ . \cos 10^\circ - \cos 30^\circ . \sin 10^\circ ) }{\frac{1}{2} . \sin 2 \times 10^\circ } \\ & = \frac{ 2 \sin ( 30^\circ - 10^\circ ) }{\frac{1}{2} . \sin 20^\circ } \\ & = \frac{ 4 \sin ( 20^\circ ) }{ \sin 20^\circ } \\ & = 4 \end{align} $
Jadi, nilai dari $ \csc 10^\circ - \sqrt{3} \sec 10^\circ = 4 . \, \heartsuit $.

       Demikian pembahasan materi Penerapan Rumus Trigonometri pada Soal-soal Bagian 1 dan contoh-contohnya. Silahkan juga baca materi lain yang berkaitan dengan trigonometri.

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