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.
