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– Sesudah mempelajari materi “rumus total dan selisih kacamata pada trigonometri” dan materi “rumus hasil barangkali antara dua bentuk trigonometri” serta rumus trigonometri yang lainnya, puas artikel ini kita akan coba ceratai mengenai
Penerapan Rumus Trigonometri pada Soal-soal Penggalan 1. Soal-tanya nan melibatkan rumus-rumus trigonometri ini kebanyakan kita jumpai pada tanya Eksamen Nasional, pertanyaan seleksi timbrung perguruan tinggi baik provinsi maupun swasta sebagaimana SBMPTN, UM UGM, SIMAK UI, dan tidak-lainnya. Keadaan mendasar nan 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 biji semenjak susuk
$ \sin 20^\circ \sin 40^\circ \sin 80^\circ $
?
Perampungan :
Ada tiga mandu nan akan kita sajikan dalam menyelesaikan cak bertanya nomor 1 :
Cara I :
*). Rumus Dasar nan 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 $
*). Mengendalikan pertanyaan :
$ \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, skor $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.
Mandu II :
*). Rumus Pangkal yang kita gunakan ialah 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 pertanyaan :
$ \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} $
kaprikornus, nilai $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.
Cara III :
*). Rumus Dasar yang kita gunakan yaitu rumus multiplikasi arti 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} $
makara, nilai $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.
2). Tentukan nilai dari tulangtulangan
$ \cos 20^\circ \cos 40^\circ \cos 80^\circ $
?
Penuntasan :
Ada empat kaidah yang akan kita sajikan intern mengamankan pertanyaan nomor 2 :
Kaidah I :
*). Rumus Bawah 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 $
*). Membereskan tanya :
$ \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 $.
Prinsip II :
*). Rumus Bawah nan kita gunakan adalah rumus perkalian maslahat 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 $
*). Tanggulang 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, ponten $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8} . \, \heartsuit $.
Cara III :
*). Rumus Pangkal yang kita gunakan adalah rumus perkalian kepentingan 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 $.
Mandu IV :
*). Rumus Dasar yang kita gunakan yaitu “Rumus Trigonometri untuk Sudut Ganda” :
$ \sin 2A = 2\sin A \cos A \rightarrow \sin A \cos A = \frac{1}{2} \sin 2A $
Rumus Tak :
$ \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{(kerjakan } \sin 20^\circ ) \\ P & = \frac{1}{8} \end{align} $
makara, nilai $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = P = \frac{1}{8} . \, \heartsuit $.
3). Tentukan poin pecah
$ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ $
?
(Cak bertanya UN Matematika IPA tahun 2007)
Perampungan :
Soal ini boleh dikerjakan dengan berbagai cara, diantaranya :
Pendirian 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 $
Kredit $ \cos 160^\circ = \cos (180^\circ – 20^\circ ) = – \cos 20^\circ $
*). Menguasai 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} $
kaprikornus, nilai $ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.
Kaidah 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} $
kaprikornus, kredit $ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.
Mandu III :
*). Rumus dasar yang digunakan :
$ \cos A + \cos B = 2\cos \frac{(A+B)}{2} \cos \frac{(A-B)}{2} $
*). Menyelesaikan cak bertanya :
$ \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} $
bintang sartan, ponten $ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.
4). Tentukan nilai mulai sejak $ \csc 10^\circ – \sqrt{3} \sec 10^\circ $?
(soal SIMAK UI tahun 2013 Matematika IPA kode 133)
Penyelesaian :
*). Rumus dasar nan 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 terbit $ \csc 10^\circ – \sqrt{3} \sec 10^\circ = 4 . \, \heartsuit $.
Demikian pembahasan materi
Penerapan Rumus Trigonometri pada Pertanyaan-pertanyaan Bagian 1
dan sempurna-contohnya. Silahkan pula baca materi bukan yang berkaitan dengan trigonometri.
