Rumus-Rumus Trigonometri

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 Materi Lawas

A. Rumus Jumlah Dan selisih Sudut

A. 1  Rumus  $\sin (\alpha +\beta )$ dan $\sin (\alpha -\beta )$.

Dalam penentuan rumus $\sin (\alpha +\beta )$, pada uraian berikut akan ditunjukkan penentuan rumus yang dimaksud dengan bantuan segitiga ABC

Perhatikanlah ilustrai berikut

Bukti:

$\begin{array}{rl}{\displaystyle \frac{AC}{\sin {90}^0}}& ={\displaystyle \frac{C{A}^{\prime }}{\sin \alpha }={\displaystyle \frac{A{A}^{\prime }}{\sin \mathrm{\angle }C}}}\\ A{A}^{\prime }& =AC.\sin \mathrm{\angle }C\\ & =AC.\sin ({90}^0-\alpha )\\ & =AC.\cos \alpha \\ & \text{dengan cara yang kurang lebih }\\ & \text{sama akan diperoleh juga}\\ A{A}^{\prime }& =AB.\cos \beta \\ & \text{selanjutnya kita tentukan luasnya, yaitu}\\ \left[ABC\right]& =\left[A{A}^{\prime }C\right]+\left[A{A}^{\prime }B\right]\\ {\displaystyle \frac{1}{2}.AB.AC.\sin (\alpha +\beta )}& ={\displaystyle \frac{1}{2}.AC.A{A}^{\prime }.\sin \alpha +{\displaystyle \frac{1}{2}.AB.A{A}^{\prime }.\sin \beta }}\\ \sin (\alpha +\beta )& ={\displaystyle \frac{AC.A{A}^{\prime }.\sin \alpha }{AB.AC}+{\displaystyle \frac{AB.A{A}^{\prime }.\sin \beta }{AB.AC}}}\\ & ={\displaystyle \frac{A{A}^{\prime }}{AB}.\sin \alpha +{\displaystyle \frac{A{A}^{\prime }}{AC}.\sin \beta }}\\ & ={\displaystyle \frac{(AB.\cos \beta )}{AB}.\sin \alpha +{\displaystyle \frac{(AC.\cos \alpha )}{AC}.\sin \beta }}\\ \sin (\alpha +\beta )& =\sin \alpha .\cos \beta +\cos \alpha .\sin \beta ◼\end{array}$.

Selanjutnya dengan untuk mendapatkan rumus  $\sin (\alpha -\beta )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{array}{rl}& \sin (\alpha +(-\beta ))=\sin (\alpha -\beta )\\ & =\sin \alpha \cos (-\beta )+\cos \alpha \sin (-\beta )\\ & =\sin \alpha \cos \beta +\cos \alpha (-\sin \beta )\\ & =\sin \alpha \cos \beta -\cos \alpha \sin \beta ◼\end{array}$.

Catatan:

$\left[ABC\right]=\mathbf{\text{luas segitiga ABC}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa nilai}\sin {60}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\sin {60}^{\circ }& =\sin ({30}^{\circ }+{30}^{\circ })\\ & =\sin {30}^{\circ }\cos {30}^{\circ }+\cos {30}^{\circ }\sin {30}^{\circ }\\ & =\sin {30}^{\circ }\cos {30}^{\circ }+\sin {30}^{\circ }\cos {30}^{\circ }\\ & =2\sin {30}^{\circ }\cos {30}^{\circ }\\ & =2\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{3}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\sin {90}^{\circ }=1\\ \\ & \begin{array}{rl}\mathbf{\text{Bukti}}& \mathbf{\text{pertama}}\\ \sin {90}^{\circ }& =\sin ({60}^{\circ }+{30}^{\circ })\\ & =\sin {60}^{\circ }\cos {30}^{\circ }+\cos {60}^{\circ }\sin {30}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)+\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{9}+{\displaystyle \frac{1}{4}}}\\ & ={\displaystyle \frac{3}{4}+{\displaystyle \frac{1}{4}}}\\ & ={\displaystyle \frac{4}{4}}\\ & ={\displaystyle 1◼}\\ \mathbf{\text{Bukti}}& \mathbf{\text{kedua}}\\ \sin {90}^{\circ }& =\sin ({30}^{\circ }+{60}^{\circ })\\ & =....+....\\ & =....\\ & =....\\ \mathbf{\text{Bukti}}& \mathbf{\text{ketiga}}\\ \sin {90}^{\circ }& =\sin ({45}^{\circ }+{45}^{\circ })\\ & =....+....\\ & =....\\ & =....\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari}\sin {75}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin {75}^{\circ }& =\sin ({30}^{\circ }+{45}^{\circ })\\ & =\sin {30}^{\circ }\cos {45}^{\circ }+\cos {30}^{\circ }\sin {45}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)+\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{2}+{\displaystyle \frac{1}{4}\sqrt{6}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{2}+\sqrt{6})}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukan nilai dari}\sin {105}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin {105}^{\circ }& =\sin ({45}^{\circ }+{60}^{\circ })\\ & =\sin {45}^{\circ }\cos {60}^{\circ }+\cos {45}^{\circ }\sin {60}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}}\right)+\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{2}+{\displaystyle \frac{1}{4}\sqrt{6}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{2}+\sqrt{6})}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Sederhanakan bentuk dari}\sin ({270}^{\circ }+A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin ({270}^{\circ }+A)& =\sin {270}^{\circ }\cos A+\cos {270}^{\circ }\sin A\\ & =(-1)\cos A+\left(0\right)\sin A\\ & =-\cos A+0\\ & =-\cos A\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakan bentuk dari}\sin ({270}^{\circ }-A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin ({270}^{\circ }-A)& =\sin {270}^{\circ }\cos A-\cos {270}^{\circ }\sin A\\ & =(-1)\cos A-\left(0\right)\sin A\\ & =-\cos A-0\\ & =-\cos A\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tunjukkan bahwa}\\ & \begin{array}{l}\\ \text{a}.& \sin \left(2\alpha \right)=2\sin \alpha \cos \alpha \\ \text{b}.& \sin \alpha =2\sin \left({\displaystyle \frac{1}{2}\alpha }\right).\cos \left({\displaystyle \frac{1}{2}\alpha }\right)\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rlrlr}\text{a.}\sin (\alpha +\beta )& =\sin \alpha \cos \beta +\cos \alpha \sin \beta ,& \text{dan jika}\beta =\alpha ,& & \text{maka}\\ \sin (\alpha +\alpha )& =\sin \alpha \cos \alpha +\cos \alpha \sin \alpha \\ \sin 2\alpha & =2\sin \alpha \cos \alpha ◼\\ \text{b.}\sin 2\alpha & =2\sin \alpha \cos \alpha ,& \text{dan jika}\alpha ={\displaystyle \frac{1}{2}\alpha ,}& & \text{maka}\\ \sin 2\left({\displaystyle \frac{1}{2}\alpha }\right)& =2\sin \left({\displaystyle \frac{1}{2}\alpha }\right)\cos \left({\displaystyle \frac{1}{2}\alpha }\right)\\ \sin \alpha & =2\sin \left({\displaystyle \frac{1}{2}\alpha }\right)\cos \left({\displaystyle \frac{1}{2}\alpha }\right)◼\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kanginan, M. 2007. Matematika untuk Kelas X Semester 2 Sekolah Menengah Atas. Bandung: GRAFINDO MEDIA PRATAMA.
2. Noormandiri, B.K. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam Berdasarkan Kurikulum 2013 Edisi Revisi 2016. Jakarta: ERLANGGA.