Sin 20 Sin 40 Sin 80

Sin 20 Sin 40 Sin 80.

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– Setelah mempelajari materi “rumus jumlah dan tikai sudut lega trigonometri” dan materi “rumus hasil kali antara dua bentuk trigonometri” serta rumus trigonometri yang lainnya, plong artikel ini kita akan coba membahas akan halnya
Penerapan Rumus Trigonometri pada Pertanyaan-soal Bagian 1. Soal-pertanyaan nan melibatkan rumus-rumus trigonometri ini galibnya kita jumpai pada soal Eksamen Nasional, soal pemilahan ikut perguruan tataran baik negeri maupun swasta sebagai halnya SBMPTN, UM UGM, SIMAK UI, dan enggak-lainnya. Keadaan mendasar yang harus kita perhatikan adalah ketelitian baik dalam menggunakan rumusnya maupun internal melakukan penjabaran dan perhitungannya. Berbarengan semata-mata kita pelajari beberapa contoh soal berikut ini.

1). Tentukan biji dari buram

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

?

Perampungan :

Ada tiga cara nan akan kita sajikan privat mengamankan soal nomor 1 :

Cara I :

*). Rumus Dasar yang kita gunakan adalah rumus pergandaan faedah 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 $

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

Cara II :

*). Rumus Asal yang kita gunakan yaitu 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)] $

*). Memintasi 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} $

jadi, nilai $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.

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Cara III :

*). Rumus Sumber akar 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)] $

$ \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, skor $ \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{8} \sqrt{3} . \, \heartsuit $.

2). Tentukan angka bersumber rancangan

$ \cos 20^\circ \cos 40^\circ \cos 80^\circ $

?

Penyelesaian :

Ada empat cara yang akan kita sajikan privat menyelesaikan tanya nomor 2 :

Cara I :

*). Rumus Sumber akar yang kita gunakan merupakan 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 $

*). Mengendalikan 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} $

bintang sartan, angka $ \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 $

*). Memintasi tanya :

$ \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} $

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

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Cara III :

*). Rumus Asal yang kita gunakan adalah rumus pergandaan kekuatan trigonometri :

$ \cos A . \cos B = \frac{1}{2} [ \cos (A+B) + \cos (A-B)] $

*). Menyelesaikan cak bertanya :

$ \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, poin $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8} . \, \heartsuit $.

Kaidah IV :

*). Rumus Asal yang kita gunakan merupakan “Rumus Trigonometri cak bagi 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 $

*). Memecahkan soal ,

Kita misalkan akhirnya $ P $ alias $ \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{(bikin } \sin 20^\circ ) \\ P & = \frac{1}{8} \end{align} $

jadi, angka $ \cos 20^\circ \cos 40^\circ \cos 80^\circ = P = \frac{1}{8} . \, \heartsuit $.

3). Tentukan nilai mulai sejak
$ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ $
?

(Soal UN Matematika IPA perian 2007)

Penyelesaian :

Cak bertanya ini bisa diselesaikan dengan berbagai cara, diantaranya :

Prinsip 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 $

Biji $ \cos 160^\circ = \cos (180^\circ – 20^\circ ) = – \cos 20^\circ $

*). Menguasai cak bertanya :

$ \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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Cara II :

*). Rumus dasar nan 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 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 pangkal yang digunakan :

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

*). Memecahkan 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 angka dari $ \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 kacamata : $ \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, poin semenjak $ \csc 10^\circ – \sqrt{3} \sec 10^\circ = 4 . \, \heartsuit $.

       Demikian pembahasan materi
Penerapan Rumus Trigonometri pada Soal-soal Adegan 1
dan eksemplar-contohnya. Silahkan juga baca materi lain nan berkaitan dengan trigonometri.

Sin 20 Sin 40 Sin 80

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

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