Topik Bahasan
trigonometri
Sebelumnya kita telah menemukan nilai sin dan cos 3 derajat pada materi "Menentukan nilai sin 3 dan 9 derajat" yaitu :
$ \sin 3^\circ = \frac{1}{8}\left( (-1 + \sqrt{5}). \sqrt{2 + \sqrt{3}} - \sqrt{10 + 2\sqrt{5}} . \sqrt{2 - \sqrt{3}} \right) $
$ \cos 3^\circ = \frac{1}{8}\left( \sqrt{10 + 2\sqrt{5} }. \sqrt{2 + \sqrt{3}} + (-1 + \sqrt{5}) . \sqrt{2 - \sqrt{3}} \right) $
$ \begin{align} & \sin 6^\circ = \sin 2 \times 3^\circ \\ & = 2 \sin 3^\circ \cos 3^\circ \\ & = 2 \times \frac{1}{8}\left( (-1 + \sqrt{5}). \sqrt{2 + \sqrt{3}} - \sqrt{10 + 2\sqrt{5}} . \sqrt{2 - \sqrt{3}} \right) \\ & \times \frac{1}{8}\left( \sqrt{10 + 2\sqrt{5} }. \sqrt{2 + \sqrt{3}} + (-1 + \sqrt{5}) . \sqrt{2 - \sqrt{3}} \right) \\ & = \frac{1}{32}\left( (-1 + \sqrt{5}). \sqrt{2 + \sqrt{3}} - \sqrt{10 + 2\sqrt{5}} . \sqrt{2 - \sqrt{3}} \right) \\ & \times \left( \sqrt{10 + 2\sqrt{5} }. \sqrt{2 + \sqrt{3}} + (-1 + \sqrt{5}) . \sqrt{2 - \sqrt{3}} \right) \\ & = \frac{1}{16}(-1 + \sqrt{5})[ \sqrt{30 + 6\sqrt{5} } + 2] \end{align} $
Jadi, nilai $ \sin 6^\circ = \frac{1}{16}(-1 + \sqrt{5})[ \sqrt{30 + 6\sqrt{5} } + 2] $
*). Menentukan nilai cos 6 derajat,
$ \begin{align} \sin 6^\circ & = \frac{1}{16}(-1 + \sqrt{5})[ \sqrt{30 + 6\sqrt{5} } + 2] \, \, \, \, \, \text{(kuadratkan)} \\ \sin ^2 6^\circ & = [\frac{1}{16}(-1 + \sqrt{5})[ \sqrt{30 + 6\sqrt{5} } + 2] ]^2 \\ & = \frac{1}{32}( 18 - 4\sqrt{5} + (3-\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) \, \, \, \, \, \text{(identitas)} \\ \cos 6^\circ & = \sqrt{1 - \sin ^2 6^\circ} \\ & = \sqrt{1 - \frac{1}{32}( 18 - 4\sqrt{5} + (3-\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } \\ & = \frac{1}{4} \sqrt{ 7 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } \end{align} $
Jadi, nilai $ \cos 6^\circ = \frac{1}{4} \sqrt{ 7 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } $
*). Menentukan nilai cos 12 derajat,
$ \begin{align} & \cos 2A = 2\cos ^2 A - 1 \\ & \cos 12^\circ = \cos 2 \times 6^\circ = 2\cos ^2 6^\circ - 1 \\ & = 2 [\frac{1}{4} \sqrt{ 7 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } ]^2 - 1 \\ & = 2 [\frac{1}{4} \sqrt{ 7 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } ]^2 - 1 \\ & = \frac{1}{8} ( -1 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) \end{align} $
Jadi, nilai $ \cos 12^\circ = \frac{1}{8} ( -1 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) $
*). Menentukan nilai sin 12 derajat,
$ \begin{align} \cos 12^\circ & = \frac{1}{8} ( -1 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) \, \, \, \, \, \text{(kuadratkan)} \\ \cos ^2 12^\circ & = [ \frac{1}{8} ( -1 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) ]^2 \\ & = \frac{1}{64} ( 81 - 28\sqrt{5} + (13 - 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) \, \, \, \, \, \text{(identitas)} \\ \sin A & = \sqrt{1 - \cos ^2 A} \\ \sin 12^\circ & = \sqrt{1 - \cos ^2 12^\circ} \\ & = \sqrt{1 - \frac{1}{64} ( 81 - 28\sqrt{5} + (13 - 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } \\ & = \frac{1}{8} \sqrt{ -17 + 28\sqrt{5} + (-13 + 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } } \end{align} $
Jadi, nilai $ \sin 12^\circ = \frac{1}{8} \sqrt{ -17 + 28\sqrt{5} + (-13 + 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } } $.
Semoga pembahasan soal Menentukan Nilai Sin 6 dan 12 derajat dengan Rumus Trigonometri ini bermanfaat untuk anda. Jika ada pertanyaan atau soal yang ingin di bahas bisa pilih menu tanya soal. Terima kasih dan sampai jumpa di masalah masalah berikutnya guys.
Rumus Dasar Trigonmetri yang digunakan
$ \spadesuit \, $ Rumus trigonometri sudut ganda
$ \sin 2A = 2 \sin A \cos A $
$ \cos 2A = 2\cos ^2 A - 1 $
$ \cos 2A = 1 - 2\sin ^2 A $
$ \clubsuit \, $ Identitas trigonmetri
$ \sin ^2 A + \cos ^2 A = 1 \rightarrow \cos A = \sqrt{1 - \sin ^2 A} $
$ \sin ^2 A + \cos ^2 A = 1 \rightarrow \sin A = \sqrt{1 - \cos ^2 A} $
$ \sin 2A = 2 \sin A \cos A $
$ \cos 2A = 2\cos ^2 A - 1 $
$ \cos 2A = 1 - 2\sin ^2 A $
$ \clubsuit \, $ Identitas trigonmetri
$ \sin ^2 A + \cos ^2 A = 1 \rightarrow \cos A = \sqrt{1 - \sin ^2 A} $
$ \sin ^2 A + \cos ^2 A = 1 \rightarrow \sin A = \sqrt{1 - \cos ^2 A} $
$ \sin 3^\circ = \frac{1}{8}\left( (-1 + \sqrt{5}). \sqrt{2 + \sqrt{3}} - \sqrt{10 + 2\sqrt{5}} . \sqrt{2 - \sqrt{3}} \right) $
$ \cos 3^\circ = \frac{1}{8}\left( \sqrt{10 + 2\sqrt{5} }. \sqrt{2 + \sqrt{3}} + (-1 + \sqrt{5}) . \sqrt{2 - \sqrt{3}} \right) $
Nilai sin 6 derajat dan sin 12 derajat
$ \sin 6^\circ = \frac{1}{16}(-1 + \sqrt{5})[ \sqrt{30 + 6\sqrt{5} } + 2] $
$ \sin 12^\circ = \frac{1}{8} \sqrt{ -17 + 28\sqrt{5} + (-13 + 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } } $
$ \sin 12^\circ = \frac{1}{8} \sqrt{ -17 + 28\sqrt{5} + (-13 + 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } } $
Menentukan Nilai Eksak sin 6 dan 12 derajat :
*). Menentukan nilai sin 6 derajat,$ \begin{align} & \sin 6^\circ = \sin 2 \times 3^\circ \\ & = 2 \sin 3^\circ \cos 3^\circ \\ & = 2 \times \frac{1}{8}\left( (-1 + \sqrt{5}). \sqrt{2 + \sqrt{3}} - \sqrt{10 + 2\sqrt{5}} . \sqrt{2 - \sqrt{3}} \right) \\ & \times \frac{1}{8}\left( \sqrt{10 + 2\sqrt{5} }. \sqrt{2 + \sqrt{3}} + (-1 + \sqrt{5}) . \sqrt{2 - \sqrt{3}} \right) \\ & = \frac{1}{32}\left( (-1 + \sqrt{5}). \sqrt{2 + \sqrt{3}} - \sqrt{10 + 2\sqrt{5}} . \sqrt{2 - \sqrt{3}} \right) \\ & \times \left( \sqrt{10 + 2\sqrt{5} }. \sqrt{2 + \sqrt{3}} + (-1 + \sqrt{5}) . \sqrt{2 - \sqrt{3}} \right) \\ & = \frac{1}{16}(-1 + \sqrt{5})[ \sqrt{30 + 6\sqrt{5} } + 2] \end{align} $
Jadi, nilai $ \sin 6^\circ = \frac{1}{16}(-1 + \sqrt{5})[ \sqrt{30 + 6\sqrt{5} } + 2] $
*). Menentukan nilai cos 6 derajat,
$ \begin{align} \sin 6^\circ & = \frac{1}{16}(-1 + \sqrt{5})[ \sqrt{30 + 6\sqrt{5} } + 2] \, \, \, \, \, \text{(kuadratkan)} \\ \sin ^2 6^\circ & = [\frac{1}{16}(-1 + \sqrt{5})[ \sqrt{30 + 6\sqrt{5} } + 2] ]^2 \\ & = \frac{1}{32}( 18 - 4\sqrt{5} + (3-\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) \, \, \, \, \, \text{(identitas)} \\ \cos 6^\circ & = \sqrt{1 - \sin ^2 6^\circ} \\ & = \sqrt{1 - \frac{1}{32}( 18 - 4\sqrt{5} + (3-\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } \\ & = \frac{1}{4} \sqrt{ 7 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } \end{align} $
Jadi, nilai $ \cos 6^\circ = \frac{1}{4} \sqrt{ 7 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } $
*). Menentukan nilai cos 12 derajat,
$ \begin{align} & \cos 2A = 2\cos ^2 A - 1 \\ & \cos 12^\circ = \cos 2 \times 6^\circ = 2\cos ^2 6^\circ - 1 \\ & = 2 [\frac{1}{4} \sqrt{ 7 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } ]^2 - 1 \\ & = 2 [\frac{1}{4} \sqrt{ 7 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } ]^2 - 1 \\ & = \frac{1}{8} ( -1 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) \end{align} $
Jadi, nilai $ \cos 12^\circ = \frac{1}{8} ( -1 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) $
*). Menentukan nilai sin 12 derajat,
$ \begin{align} \cos 12^\circ & = \frac{1}{8} ( -1 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) \, \, \, \, \, \text{(kuadratkan)} \\ \cos ^2 12^\circ & = [ \frac{1}{8} ( -1 + 2\sqrt{5} + \frac{1}{2} (-3+\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) ]^2 \\ & = \frac{1}{64} ( 81 - 28\sqrt{5} + (13 - 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) \, \, \, \, \, \text{(identitas)} \\ \sin A & = \sqrt{1 - \cos ^2 A} \\ \sin 12^\circ & = \sqrt{1 - \cos ^2 12^\circ} \\ & = \sqrt{1 - \frac{1}{64} ( 81 - 28\sqrt{5} + (13 - 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } ) } \\ & = \frac{1}{8} \sqrt{ -17 + 28\sqrt{5} + (-13 + 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } } \end{align} $
Jadi, nilai $ \sin 12^\circ = \frac{1}{8} \sqrt{ -17 + 28\sqrt{5} + (-13 + 7\sqrt{5}) \sqrt{30 + 6\sqrt{5} } } $.
Semoga pembahasan soal Menentukan Nilai Sin 6 dan 12 derajat dengan Rumus Trigonometri ini bermanfaat untuk anda. Jika ada pertanyaan atau soal yang ingin di bahas bisa pilih menu tanya soal. Terima kasih dan sampai jumpa di masalah masalah berikutnya guys.
Cari Soal dan Pembahasan tentang trigonometri
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