Lanjutan 1 Materi Rumus-Rumus Trigonometri

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

Dalam penentuan rumus $\cos (\alpha -\beta )$, 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

$\begin{array}{rl}\left[ABC\right]& =\left[ACD\right]+\left[BCD\right]\\ {\displaystyle \frac{1}{2}ab\sin ({90}^0-\alpha +\beta )}& ={\displaystyle \frac{1}{2}ab\cos \alpha \cos \beta +{\displaystyle \frac{1}{2}ab\sin \alpha \sin \beta }}\\ \sin ({90}^0-(\alpha -\beta ))& =\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \cos (\alpha -\beta )& =\cos \alpha \cos \beta +\sin \alpha \sin \beta .◼\end{array}$.

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

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

Catatan:

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

$\text{CONTOH SOAL}$.

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

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\cos {90}^{\circ }=0\\ \\ & \begin{array}{rl}\mathbf{\text{Bukti}}& \mathbf{\text{pertama}}\\ \cos {90}^{\circ }& =\cos ({60}^{\circ }+{30}^{\circ })\\ & =\cos {60}^{\circ }\cos {30}^{\circ }-\sin {60}^{\circ }\sin {30}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)-\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{3}-{\displaystyle \frac{1}{4}\sqrt{3}}}\\ & ={\displaystyle 0◼}\\ \mathbf{\text{Bukti}}& \mathbf{\text{kedua}}\\ \cos {90}^{\circ }& =\cos ({30}^{\circ }+{60}^{\circ })\\ & =....+....\\ & =....\\ & =....\\ \mathbf{\text{Bukti}}& \mathbf{\text{ketiga}}\\ \cos {90}^{\circ }& =\cos ({45}^{\circ }+{45}^{\circ })\\ & =....+....\\ & =....\\ & =....\end{array}\end{array}$.

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

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

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

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

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

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.