Cos 20 Cos 40 Cos 80

Cos 20 Cos 40 Cos 80.

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– Sehabis mempelajari materi “rumus jumlah dan beda sudut lega trigonometri” dan materi “rumus hasil mana tahu antara dua kerangka trigonometri” serta rumus trigonometri yang lainnya, pada artikel ini kita akan coba mengomongkan tentang
Penerapan Rumus Trigonometri sreg Soal-tanya Bagian 1. Tanya-soal yang mengikutsertakan rumus-rumus trigonometri ini biasanya kita jumpai puas tanya UJian Nasional, pertanyaan seleksi ikut perguruan tinggi baik negeri maupun swasta seperti SBMPTN, UM UGM, SIMAK UI, dan lain-lainnya. Situasi mendasar yang harus kita perhatikan adalah ketelitian baik dalam menggunakan rumusnya maupun dalam melakukan penjabaran dan perhitungannya. Berbarengan semata-mata kita pelajari sejumlah contoh soal berikut ini.

1). Tentukan nilai mulai sejak buram

$ \sin 20^\circ \sin 40^\circ \sin 80^\circ $

?

Penuntasan :

Suka-suka tiga pendirian nan akan kita sajikan dalam menyelesaikan soal nomor 1 :

Cara I :

*). Rumus Dasar nan kita gunakan ialah 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, skor $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.

Cara II :

*). Rumus Asal yang kita gunakan adalah rumus multiplikasi 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)] $

*). Membereskan 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, biji $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.

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Cara III :

*). Rumus Dasar yang kita gunakan adalah rumus perbanyakan 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 tanya :

$ \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 pendirian yang akan kita sajikan internal mengamankan soal nomor 2 :

Cara I :

*). Rumus Dasar yang kita gunakan yakni rumus perkalian arti 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 cak bertanya :

$ \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 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 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 $.

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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)] $

*). Menguasai 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, ponten $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8} . \, \heartsuit $.

Kaidah IV :

*). Rumus Dasar yang kita gunakan ialah “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 $

*). Mengatasi tanya ,

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{(kelihatannya } \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{(bakal } \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 terbit
$ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ $
?

(Tanya UN Ilmu hitung IPA masa 2007)

Penyelesaian :

Soal ini dapat diselesaikan dengan bineka cara, diantaranya :

Pendirian I :

*). Rumus radiks yang digunakan :

$ \cos A + \cos B = 2\cos \frac{(A+B)}{2} \cos \frac{(A-B)}{2} $

$ \cos (180^\circ – A ) = – \cos A $

Biji $ \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 $.

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Kaidah II :

*). Rumus bawah 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, angka $ \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} $

*). Memintasi tanya :

$ \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, skor $ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.

4). Tentukan nilai dari $ \csc 10^\circ – \sqrt{3} \sec 10^\circ $?

(pertanyaan 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 bersumber $ \csc 10^\circ – \sqrt{3} \sec 10^\circ = 4 . \, \heartsuit $.

       Demikian pembahasan materi
Penerapan Rumus Trigonometri pada Tanya-soal Bagian 1
dan teladan-contohnya. Silahkan juga baca materi lain nan berkaitan dengan trigonometri.

Cos 20 Cos 40 Cos 80

Source: https://www.konsep-matematika.com/2017/02/penerapan-rumus-trigonometri-pada-soal-soal-bagian-1.html

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