Identitas Trigonometri (Kelas XI Peminatan Mat & IA K13 Revisi)

Identitas Trigonometri adalah persamaan-persamaan yang berlaku untuk semua nilai pengganti variabelnya yang mengandung perbandingan trigonometri.

Sebelumnya silahkan lihat kembali

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Pythagoras}\: \: \begin{cases} \sin ^{2}\alpha +\cos ^{2}\alpha =1 \\ \sec ^{2}\alpha =\tan ^{2}\alpha +1 \\ \csc ^{2}\alpha =\cot ^{2}\alpha +1 \end{cases}}\\\hline \textrm{Setengah}&\textrm{Satu}\\\hline \begin{cases} \sin \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{2}} \\ \cos \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1+\cos \alpha }{2}} \\ \tan \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }} \end{cases}&\begin{cases} \sin \alpha =\displaystyle \frac{1}{\csc \alpha } \\ \cos \alpha =\displaystyle \frac{1}{\sec \alpha } \\ \tan \alpha =\displaystyle \frac{1}{\cot \alpha } \\ \tan \alpha =\displaystyle \frac{\sin \alpha }{\cos \alpha } \\ \cot \alpha =\displaystyle \frac{\cos \alpha }{\sin \alpha } \end{cases}\\\hline \textrm{Dua}&\textrm{Tiga}\\\hline \begin{cases} \sin 2\alpha =2\sin \alpha \cos \alpha \\ \cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha \\ \tan 2\alpha =\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \end{cases}&\begin{cases} \sin 3\alpha =3\sin \alpha -4\sin ^{3}\alpha \\ \cos 3\alpha =4\cos ^{3}\alpha -3\cos \alpha \\ \tan 3\alpha =\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \end{cases}\\\hline \end{array}.

\LARGE{\fbox{\LARGE{\fbox{CONTOH SOAL}}}}.

\begin{array}{ll}\\ 1.&\textrm{Dengan identitas}\: \: \sin ^{2}\gamma +\cos ^{2}\gamma =1,\: \textrm{buktikanlah identitas-identitas berikut}!\\ &\begin{array}{lllll}\\ \textrm{a}.&\left ( \sin \gamma -\cos \gamma \right )^{2}=1-\sin 2\gamma &\textrm{f}.&\displaystyle \frac{1+\sin \gamma }{\cos \gamma }+\displaystyle \frac{\cos \gamma }{1+\sin \gamma }=2\sec \gamma \\ \textrm{b}.&\displaystyle \frac{\csc ^{2}\gamma -1}{\csc ^{2}\gamma }=\cos ^{2}\gamma &\textrm{g}.&\displaystyle \frac{1}{1+\sin \gamma }+\displaystyle \frac{1}{1-\sin \gamma }=2\sec ^{2}\gamma \\ \textrm{c}.&\sqrt{\displaystyle \frac{1-\sin ^{2}\gamma }{1-\cos ^{2}\gamma }}=\cot \gamma &\textrm{h}.&\left ( \sec \gamma -\tan \gamma \right )^{2}=\displaystyle \frac{1-\sin \gamma }{1+\sin \gamma }\\ \textrm{d}.&\displaystyle \frac{1+\cos \gamma }{\sin ^{2}\gamma }=\displaystyle \frac{1}{1-\cos \gamma }&\textrm{i}.&\left ( \cot \gamma -\csc \gamma \right )^{2}=\displaystyle \frac{1-\cos \gamma }{1+\cos \gamma }\\ \textrm{e}.&\displaystyle \frac{1+\cos \gamma }{\sin \gamma }+\displaystyle \frac{\sin \gamma }{1+\cos \gamma }=2\csc \gamma &\textrm{j}.&\displaystyle \frac{\sin \gamma -2\sin ^{3}\gamma }{2\cos ^{3}\gamma -\cos \gamma }=\tan \gamma \end{array} \end{array}.

Bukti:

\begin{aligned}1.\textrm{a}.\quad\left ( \sin \gamma -\cos \gamma \right )^{2}&=\left ( \sin \gamma -\cos \gamma \right )\times \left ( \sin \gamma -\cos \gamma \right )\\ &=\sin ^{2}\gamma -2\sin \gamma \cos \gamma +\cos ^{2}\gamma \\ &=\sin ^{2}\gamma +\cos ^{2}\gamma -2\sin \gamma \cos \gamma \\ &=1-\sin 2\gamma \qquad \blacksquare \end{aligned}.

\begin{array}{ll}\\ 2.&\textrm{Dengan identitas}\: \: 1+\tan ^{2}\beta=\sec ^{2}\beta ,\: \textrm{buktikanlah identitas-identitas berikut}!\\ &\begin{array}{lllll}\\ \textrm{a}.&\displaystyle \frac{1+\tan ^{2}\beta }{\csc ^{2}\beta }=\tan ^{2}\beta &\textrm{f}.&1+\tan ^{2}\beta =\displaystyle \frac{\tan ^{2}\beta }{\sin ^{2}\beta }\\ \textrm{b}.&\displaystyle \frac{1}{\sec \beta +\tan \beta }=\sec \beta -\tan \beta &\textrm{g}.&\displaystyle \frac{\tan ^{2}\beta }{1+\tan ^{2}\beta }+\displaystyle \frac{\cot ^{2}\beta }{1+\cot ^{2}\beta }=1 \\ \textrm{c}.&\displaystyle \frac{1-\tan ^{2}\beta }{1+\tan ^{2}\beta }=2\cos ^{2}\beta -1&\textrm{h}.&\displaystyle \frac{1}{\sec \beta +\tan \beta }+\displaystyle \frac{1}{\sec \beta -\tan \beta }=2\sec \beta \\ \textrm{d}.&\displaystyle \frac{\sec \beta +\tan \beta }{\sec \beta -\tan \beta }=\left ( \sec \beta +\tan \beta \right )^{2}&\textrm{i}.&\left ( 1+\tan ^{2}\beta \right )\left ( 1+\displaystyle \frac{1}{\tan ^{2}\beta } \right )=\sec ^{2}\beta \csc ^{2}\beta\\ \textrm{e}.&\displaystyle \frac{\sec \beta -\tan \beta }{\sec \beta +\tan \beta }=\displaystyle \frac{\cos ^{2}\beta }{\left ( 1+\sin \beta \right )^{2}}&\textrm{j}.&\displaystyle \frac{\sec \beta -\tan \beta }{\sec \beta +\tan \beta }=1-2\sec \beta \tan \beta \end{array} \end{array}.

Bukti:

\begin{aligned}2.\textrm{f}.\quad 1+\tan ^{2}\beta &=\sec ^{2}\beta \\ &=\displaystyle \frac{1}{\cos ^{2}\beta }\\ &=\displaystyle \frac{1}{\cos ^{2}\beta }\times \displaystyle \frac{\sin ^{2}\beta }{\sin ^{2}\beta }\\ &=\displaystyle \frac{\sin ^{2}\beta }{\cos ^{2}\beta }\times \displaystyle \frac{1}{\sin ^{2}\beta }\\ &=\tan ^{2}\beta \times \displaystyle \frac{1}{\sin ^{2}\beta }\\ &=\displaystyle \frac{\tan ^{2}\beta }{\sin ^{2}\beta } \qquad \blacksquare \end{aligned}.

\begin{array}{ll}\\ 3.&\textrm{Dengan identitas}\: \: 1+\cot ^{2}\alpha =\csc ^{2}\alpha ,\: \textrm{buktikanlah identitas-identitas berikut}!\\ &\begin{array}{lllll}\\ \textrm{a}.&\left ( \csc \alpha -\cot \alpha \right )^{2}=\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }&\textrm{f}.&\displaystyle \frac{\tan \alpha }{1-\cot \alpha }+\displaystyle \frac{\cot \alpha }{1-\tan \alpha }=1+\tan \alpha +\cot \alpha \\ \textrm{b}.& \sec ^{2}\alpha -\csc ^{2}\alpha =\tan ^{2}\alpha -\cot ^{2}\alpha &\textrm{g}.&1+\displaystyle \frac{\cot ^{2}\alpha }{1+\csc \alpha }=\csc \alpha \\ \textrm{c}.&\sqrt{\sec ^{2}\alpha +\csc ^{2}\alpha }=\tan \alpha +\cot \alpha &\textrm{h}.&\displaystyle \frac{\tan \alpha }{\left ( 1+\tan ^{2}\alpha \right )^{2}}+\displaystyle \frac{\cot \alpha }{\left ( 1+\cot ^{2}\alpha \right )^{2}}=\sin \alpha \cos \alpha \\ \textrm{d}.&\displaystyle \tan ^{2}\alpha +\cot ^{2}\alpha +2=\sec ^{2}\alpha \csc ^{2}\alpha &\textrm{i}.&\left ( \sec ^{2}\alpha -1 \right )\left ( \csc ^{2}\alpha -1 \right )=1\\ \textrm{e}.&\displaystyle \frac{1}{\csc \alpha -\cot \alpha }+\displaystyle \frac{1}{\csc \alpha +\cot \alpha }=2\csc \alpha &\textrm{j}.&\cot ^{2}\alpha \left ( \displaystyle \frac{\sec \alpha -1}{1+\sin \alpha } \right )+\sec ^{2}\alpha \left ( \displaystyle \frac{\sin \alpha -1}{1+\sec \alpha } \right )=0 \end{array} \end{array}.

Bukti:

\begin{aligned}3&.\textrm{j}.\quad \cot ^{2}\alpha \left ( \displaystyle \frac{\sec \alpha -1}{1+\sin \alpha } \right )+\sec ^{2}\alpha \left ( \displaystyle \frac{\sin \alpha -1}{1+\sec \alpha } \right )\\ &=\left ( \csc ^{2}\alpha -1 \right ) \left ( \displaystyle \frac{\sec \alpha -1}{1+\sin \alpha } \right )+\sec ^{2}\alpha \left ( \displaystyle \frac{\sin \alpha -1}{1+\sec \alpha } \right )\\ &=\left ( \displaystyle \frac{1}{\sin ^{2}\alpha } -1 \right ) \left ( \displaystyle \frac{\displaystyle \frac{1}{\cos \alpha } -1}{1+\sin \alpha } \right )+\displaystyle \frac{1}{\cos ^{2}\alpha } \left ( \displaystyle \frac{\sin \alpha -1}{1+\displaystyle \frac{1}{\cos \alpha } } \right )\\ &=\left ( \displaystyle \frac{1-\sin ^{2}\alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{\displaystyle \frac{1-\cos \alpha }{\cos \alpha }}{1+\sin \alpha } \right )+\displaystyle \frac{1}{\cos ^{2}\alpha } \left ( \displaystyle \frac{\sin \alpha -1}{\displaystyle \frac{\cos \alpha +1}{\cos \alpha } } \right )\\ \end{aligned}.

\begin{aligned} .\: &=\left ( \displaystyle \frac{\cos ^{2}\alpha }{\sin ^{2}\alpha .\cos \alpha } \right ) \left ( \displaystyle \frac{1-\cos \alpha }{1+\sin \alpha } \right )+\displaystyle \frac{\cos \alpha }{\cos ^{2}\alpha } \left ( \displaystyle \frac{\sin \alpha -1}{\cos \alpha +1} \right )\\ .\: &=\left ( \displaystyle \frac{\cos \alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{1-\cos \alpha }{1+\sin \alpha } \right )\times \left ( \displaystyle \frac{1-\sin \alpha }{1-\sin \alpha } \right )+\displaystyle \frac{1 }{\cos \alpha } \left ( \displaystyle \frac{\sin \alpha -1}{\cos \alpha +1} \right )\times \left ( \displaystyle \frac{\cos \alpha -1}{\cos \alpha -1} \right )\\ .\: &=\left ( \displaystyle \frac{\cos \alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{1-\sin \alpha -\cos \alpha +\cos \alpha \sin \alpha }{1-\sin^{2} \alpha } \right )+\displaystyle \frac{1 }{\cos \alpha } \left ( \displaystyle \frac{\sin \alpha \cos \alpha -\sin \alpha -\cos \alpha +1}{\cos ^{2}\alpha -1} \right )\\ \end{aligned}.

\begin{aligned} .\: &=\left ( \displaystyle \frac{\cos \alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{1-\sin \alpha -\cos \alpha +\cos \alpha \sin \alpha }{\cos^{2} \alpha } \right )+\displaystyle \frac{1 }{\cos \alpha } \left ( \displaystyle \frac{\sin \alpha \cos \alpha -\sin \alpha -\cos \alpha +1}{-\sin ^{2}\alpha } \right )\\ &=\left ( \displaystyle \frac{1}{\sin ^{2}\alpha \cos \alpha }-\displaystyle \frac{1}{\sin ^{2}\alpha \cos \alpha } \right ) \left ( 1-\sin \alpha -\cos \alpha +\cos \alpha \sin \alpha \right )\\ .\: &=0\qquad \blacksquare \end{aligned}.

\LARGE{\fbox{\LARGE{\fbox{LATIHAN SOAL}}}}.

Untuk soal-soal yang belum ditunjukkan buktinya, silahkan gunakan sebagai latihan mandiri

Sumber Referensi

  1. Kurnia, N.,  dkk. 2017. Jelajah Matematika 2 SMA Kelas XI Peminatan MIPA(Edisi Revisi 2016). Jakarta: Yudistira.
  2. Tampomas, H. 1999. Seribu Pena Matematika SMU Jilid 1 Kelas 1. Jakarta: ERLANGGA.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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