Rumus-Rumus Trigonometri

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 Materi Lawas

A. Rumus Jumlah Dan selisih Sudut

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

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

Perhatikanlah ilustrai berikut

Bukti:

$\begin{aligned}\displaystyle \frac{AC}{\sin 90^{0}}&=\displaystyle \frac{CA'}{\sin \alpha }=\displaystyle \frac{AA'}{\sin \angle C}\\ AA'&=AC.\sin \angle C\\ &=AC.\sin \left ( 90^{0}-\alpha \right )\\ &=AC.\cos \alpha\\ &\textnormal{dengan cara yang kurang lebih }\\ &\textrm{sama akan diperoleh juga}\\ AA'&=AB.\cos \beta\\ &\textnormal{selanjutnya kita tentukan luasnya, yaitu}\\ \left [ ABC \right ]&=\left [ AA'C \right ]+\left [ AA'B \right ]\\ \displaystyle \frac{1}{2}.AB.AC.\sin \left ( \alpha +\beta \right )&=\displaystyle \frac{1}{2}.AC.AA'.\sin \alpha +\displaystyle \frac{1}{2}.AB.AA'.\sin \beta \\ \sin \left ( \alpha +\beta \right )&=\displaystyle \frac{AC.AA'.\sin \alpha }{AB.AC}+\displaystyle \frac{AB.AA'.\sin \beta }{AB.AC}\\ &=\displaystyle \frac{AA'}{AB}.\sin \alpha +\displaystyle \frac{AA'}{AC}.\sin \beta \\ &=\displaystyle \frac{\left ( AB.\cos \beta \right )}{AB}.\sin \alpha +\displaystyle \frac{\left ( AC.\cos \alpha \right )}{AC}.\sin \beta \\ \sin \left ( \alpha +\beta \right )&=\sin \alpha .\cos \beta +\cos \alpha .\sin \beta \quad \blacksquare \end{aligned}$.

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

$\begin{aligned}&\sin \left ( \alpha +(-\beta ) \right )=\sin \left ( \alpha -\color{red}\beta \right )\\ &=\sin\alpha \cos \left ( -\beta \right )+\cos \alpha \sin \left ( -\beta \right )\\ &=\sin \alpha \cos \beta +\cos \alpha \left ( -\sin \beta \right )\\ &=\sin \alpha \cos \beta -\cos \alpha \sin \beta \qquad \blacksquare \end{aligned}$.

Catatan:

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

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa nilai}\: \: \sin 60^{\circ}=\color{red}\displaystyle \frac{1}{2}\sqrt{3}\\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\sin 60^{\circ}&=\sin \left ( 30^{\circ}+30^{\circ} \right )\\ &=\sin 30^{\circ}\cos 30^{\circ}+\cos 30^{\circ}\sin 30^{\circ}\\ &=\sin 30^{\circ}\cos 30^{\circ}+\sin 30^{\circ}\cos 30^{\circ}\\ &=2\sin 30^{\circ}\cos 30^{\circ}\\ &=2\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \color{red}\frac{1}{2}\sqrt{3}\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa nilai}\: \: \sin 90^{\circ}=\color{red}1\\\\ &\begin{aligned}\color{blue}\textbf{Bukti}&\: \: \textbf{pertama}\\ \sin 90^{\circ}&=\sin \left ( 60^{\circ}+30^{\circ} \right )\\ &=\sin 60^{\circ}\cos 30^{\circ}+\cos 60^{\circ}\sin 30^{\circ}\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )+\left (\displaystyle \frac{1}{2} \right )\left (\displaystyle \frac{1}{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{9}+\displaystyle \frac{1}{4}\\ &=\displaystyle \frac{3}{4}+\displaystyle \frac{1}{4}\\ &=\displaystyle \frac{4}{4}\\ &=\displaystyle \color{red}1\qquad \color{black}\blacksquare\\ \color{blue}\textbf{Bukti}&\: \: \textbf{kedua}\\ \sin 90^{\circ}&=\sin \left ( 30^{\circ}+60^{\circ} \right )\\ &=....+....\\ &=....\\ &=....\\ \color{blue}\textbf{Bukti}&\: \: \textbf{ketiga}\\ \sin 90^{\circ}&=\sin \left ( 45^{\circ}+45^{\circ} \right )\\ &=....+....\\ &=....\\ &=....\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukan nilai dari}\: \: \sin 75^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\sin 75^{\circ}&=\sin \left ( 30^{\circ}+45^{\circ} \right )\\ &=\sin 30^{\circ}\cos 45^{\circ}+\cos 30^{\circ}\sin 45^{\circ}\\ &=\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )+\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{2}+\displaystyle \frac{1}{4}\sqrt{6}\\ &=\displaystyle \color{red}\frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukan nilai dari}\: \: \sin 105^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\sin 105^{\circ}&=\sin \left ( 45^{\circ}+60^{\circ} \right )\\ &=\sin 45^{\circ}\cos 60^{\circ}+\cos 45^{\circ}\sin 60^{\circ}\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \displaystyle \frac{1}{2} \right )+\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{2}+\displaystyle \frac{1}{4}\sqrt{6}\\ &=\displaystyle \color{red}\frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Sederhanakan bentuk dari}\: \: \sin \left ( 270^{\circ}+A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\sin \left ( 270^{\circ}+A \right )&=\sin 270^{\circ}\cos A+\cos 270^{\circ}\sin A\\ &=\left ( -1 \right )\cos A+\left ( 0 \right )\sin A\\ &=-\cos A+0\\ &= \color{red}-\cos A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Sederhanakan bentuk dari}\: \: \sin \left ( 270^{\circ}-A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\sin \left ( 270^{\circ}-A \right )&=\sin 270^{\circ}\cos A-\cos 270^{\circ}\sin A\\ &=\left ( -1 \right )\cos A-\left ( 0 \right )\sin A\\ &=-\cos A-0\\ &= \color{red}-\cos A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tunjukkan bahwa}\\ &\begin{array}{lll}\\ \textrm{a}.&\sin \left ( 2\alpha \right )=2\sin \alpha \cos \alpha\\ \textrm{b}.&\sin \alpha =2\sin \left ( \displaystyle \frac{1}{2}\alpha \right ).\cos \left ( \displaystyle \frac{1}{2}\alpha \right ) \end{array}\\\\ &\color{blue}\textbf{Bukti}\\ &\begin{aligned}\textnormal{a.}\quad \sin \left ( \alpha +\beta \right )&=\sin \alpha \cos \beta + \cos \alpha \sin \beta ,&\textnormal{dan jika}\quad \beta =\alpha ,&&\textnormal{maka}\\ \sin \left ( \alpha +\alpha \right )&=\sin \alpha \cos \alpha +\cos \alpha \sin \alpha \\ \sin 2\alpha &=2\sin \alpha \cos \alpha\qquad \blacksquare \\ \textnormal{b.}\qquad\quad \sin 2\alpha &=2\sin \alpha \cos \alpha ,&\textnormal{dan jika}\quad \alpha =\displaystyle \frac{1}{2}\alpha ,&&\textnormal{maka}\\ \sin 2\left ( \displaystyle \frac{1}{2}\alpha \right )&=2\sin \left ( \displaystyle \frac{1}{2}\alpha \right )\cos \left ( \displaystyle \frac{1}{2}\alpha \right )\\ \sin \alpha &=2\sin \left ( \displaystyle \frac{1}{2}\alpha \right )\cos \left ( \displaystyle \frac{1}{2}\alpha \right )\qquad \blacksquare \end{aligned}\\ \end{array}$.

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.