Lanjutan 1 Materi Rumus-Rumus Trigonometri

A. 2  Rumus  $\cos \left ( \alpha +\beta \right )$ dan $\cos \left ( \alpha -\beta \right )$.

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

Perhatikanlah ilustrai berikut

Jika diurai gambar di atas adalah 

Mungkin gambarnya ada tang kurang jelas, mari kita perjelas lagi gambar di atas

Perhatikan bahwa

dan

Sehingga

$\begin{aligned}\left [ ABC \right ]&=\left [ ACD \right ]+\left [ BCD \right ]\\ \displaystyle \frac{1}{2}ab\sin \left ( 90^{0}-\alpha +\beta \right )&=\displaystyle \frac{1}{2}ab\cos \alpha \cos \beta +\displaystyle \frac{1}{2}ab\sin \alpha \sin \beta \\ \sin \left ( 90^{0}-\left ( \alpha -\beta \right ) \right )&=\color{red}\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \cos \left ( \alpha -\beta \right )&=\color{red}\cos \alpha \cos \beta +\sin \alpha \sin \beta.\qquad \color{black}\blacksquare \end{aligned}$.

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

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

Catatan:

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

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

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

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa nilai}\: \: \cos 90^{\circ}=\color{red}0\\\\ &\begin{aligned}\color{blue}\textbf{Bukti}&\: \: \textbf{pertama}\\ \cos 90^{\circ}&=\cos \left ( 60^{\circ}+30^{\circ} \right )\\ &=\cos 60^{\circ}\cos 30^{\circ}-\sin 60^{\circ}\sin 30^{\circ}\\ &=\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )-\left (\displaystyle \frac{1}{2}\sqrt{3} \right )\left (\displaystyle \frac{1}{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}-\displaystyle \frac{1}{4}\sqrt{3}\\ &=\displaystyle \color{red}0\qquad \color{black}\blacksquare\\ \color{blue}\textbf{Bukti}&\: \: \textbf{kedua}\\ \cos 90^{\circ}&=\cos \left ( 30^{\circ}+60^{\circ} \right )\\ &=....+....\\ &=....\\ &=....\\ \color{blue}\textbf{Bukti}&\: \: \textbf{ketiga}\\ \cos 90^{\circ}&=\cos \left ( 45^{\circ}+45^{\circ} \right )\\ &=....+....\\ &=....\\ &=....\\ \end{aligned} \end{array}$.

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

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

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

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

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

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.