Lanjutan 5 Materi Rumus Perkalian Sinus dan Cosinus

C.   Rumus Perkalian Sinus dan Cosinus

Untuk sudut  $\alpha \text{dan}\beta$ berlaku rumus-rumus

$\begin{array}{rl}& \{\begin{array}{ll}2\sin \alpha \cos \gamma & =\sin (\alpha +\gamma )+\sin (\alpha -\gamma )\\ 2\cos \alpha \sin \gamma & =\sin (\alpha +\gamma )-\sin (\alpha -\gamma )\end{array}\\ & \{\begin{array}{ll}2\cos \alpha \cos \gamma & =\cos (\alpha +\gamma )+\cos (\alpha -\gamma )\\ -2\sin \alpha \sin \gamma & =\cos (\alpha +\gamma )-\cos (\alpha -\gamma )\end{array}\end{array}$.

Sebagai buktinya akan ditunjukkan pada tulisan berikut

Bukti untuk nomor pertama

$\begin{array}{rl}\sin & (\alpha +\beta )+\sin (\alpha -\beta )\\ & =\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ & +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ & =2\sin \alpha \cos \beta \end{array}$.

Selanjutnya untuk bukti baris kedua yaitu:

$\begin{array}{rl}\sin & (\alpha +\beta )-\sin (\alpha -\beta )\\ & =\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ & -(\sin \alpha \cos \beta -\cos \alpha \sin \beta )\\ & =2\cos \alpha \sin \beta \end{array}$.

Dan bukti untuk rumus ketiga

$\begin{array}{rl}\cos & (\alpha +\beta )+\cos (\alpha -\beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & +\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ & =2\cos \alpha \cos \beta \end{array}$.

Adapun bukti untuk rumus yang tertakhir adalah:

$\begin{array}{rl}\cos & (\alpha +\beta )-\cos (\alpha -\beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & -(\cos \alpha \cos \beta +\sin \alpha \sin \beta )\\ & =-2\sin \alpha \sin \beta \end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Nyatakanlah ke dalam bentuk jumlah}\\ & \text{atau selisih dari bentuk berikut}\\ & \text{a}.2\cos {80}^{\circ }\sin {50}^{\circ }\\ & \text{b}.2\sin {80}^{\circ }\cos {50}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& 2\cos {80}^{\circ }\sin {50}^{\circ }\\ & =\sin ({80}^{\circ }+{50}^{\circ })-\sin ({80}^{\circ }-{50}^{\circ })\\ & =\sin {130}^{\circ }-\sin {30}^{\circ }\\ & =\sin {130}^{\circ }-{\displaystyle \frac{1}{2}}\\ \text{b}.& 2\sin {80}^{\circ }\cos {50}^{\circ }\\ & =\sin ({80}^{\circ }+{50}^{\circ })+\sin ({80}^{\circ }-{50}^{\circ })\\ & =\sin {130}^{\circ }+\sin {30}^{\circ }\\ & =\sin {130}^{\circ }+{\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Nyatakanlah ke dalam bentuk jumlah}\\ & \text{atau selisih dari bentuk berikut}\\ & \text{a}.2\cos {75}^{\circ }\cos {15}^{\circ }\\ & \text{b}.\sin \left({\displaystyle \frac{3\pi }{8}}\right)\sin \left({\displaystyle \frac{\pi }{8}}\right)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& 2\cos {75}^{\circ }\cos {15}^{\circ }\\ & =\cos ({75}^{\circ }+{15}^{\circ })+\cos ({75}^{\circ }-{15}^{\circ })\\ & =\cos {90}^{\circ }+\cos {60}^{\circ }\\ & =0+{\displaystyle \frac{1}{2}={\displaystyle \frac{1}{2}}}\\ \text{b}.& \sin \left({\displaystyle \frac{3\pi }{8}}\right)\sin \left({\displaystyle \frac{\pi }{8}}\right)\\ & =-{\displaystyle \frac{1}{2}(2\sin \left({\displaystyle \frac{3\pi }{8}}\right)\sin \left({\displaystyle \frac{\pi }{8}}\right))}\\ & =-{\displaystyle \frac{1}{2}(\cos \left({\displaystyle \frac{3\pi }{8}+\frac{\pi }{8}}\right)-\cos \left({\displaystyle \frac{3\pi }{8}-\frac{\pi }{8}}\right))}\\ & =-{\displaystyle \frac{1}{2}\left(\cos {\displaystyle \frac{4\pi }{8}-\cos {\displaystyle \frac{2\pi }{8}}}\right)}\\ & =-{\displaystyle \frac{1}{2}\left(\cos {\displaystyle \frac{1}{2}\pi -\cos {\displaystyle \frac{1}{4}\pi }}\right)}\\ & =-{\displaystyle \frac{1}{2}(0-{\displaystyle \frac{1}{2}\sqrt{2}})}\\ & =-{\displaystyle \frac{1}{2}(-{\displaystyle \frac{1}{2}\sqrt{2}})}\\ & ={\displaystyle \frac{1}{4}\sqrt{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari bentuk berikut}\\ \\ & \text{a}.2\sin 37{\displaystyle {\frac{1}{2}}^{\circ }\cos 7{\displaystyle {\frac{1}{2}}^{\circ }}}\\ \\ & \text{b}.2\cos 82{\displaystyle {\frac{1}{2}}^{\circ }\sin 37{\displaystyle {\frac{1}{2}}^{\circ }}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& 2\sin 37{\displaystyle {\frac{1}{2}}^{\circ }\cos 7{\displaystyle {\frac{1}{2}}^{\circ }}}\\ & =\sin {\left(37{\displaystyle \frac{1}{2}+7{\displaystyle \frac{1}{2}}}\right)}^{\circ }+{\displaystyle \sin {\left(37{\displaystyle \frac{1}{2}-7{\displaystyle \frac{1}{2}}}\right)}^{\circ }}\\ & =\sin {45}^{\circ }+\sin {30}^{\circ }\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+{\displaystyle \frac{1}{2}}}\\ & ={\displaystyle \frac{1}{2}(\sqrt{2}+1)}\\ \text{b}.& 2\cos 82{\displaystyle {\frac{1}{2}}^{\circ }\sin 37{\displaystyle {\frac{1}{2}}^{\circ }}}\\ & =\sin {\left(82{\displaystyle \frac{1}{2}+37{\displaystyle \frac{1}{2}}}\right)}^{\circ }-{\displaystyle \sin {\left(82{\displaystyle \frac{1}{2}-37{\displaystyle \frac{1}{2}}}\right)}^{\circ }}\\ & =\sin {120}^{\circ }-\sin {45}^{\circ }\\ & =\sin ({180}^{\circ }-{60}^{\circ })-\sin {45}^{\circ }\\ & =\sin {60}^{\circ }-\sin {45}^{\circ }\\ & ={\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}}\\ & ={\displaystyle \frac{1}{2}(\sqrt{3}-\sqrt{2})}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa}\\ & \text{a}.2\cos \left({\displaystyle \frac{1}{4}\pi +\theta }\right)\cos \left({\displaystyle \frac{3}{4}\pi -\theta }\right)=\sin 2\theta -1\\ & \text{b}.2\sin ({315}^{\circ }+B)\sin ({45}^{\circ }-B)=\sin 2B-1\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\text{a}.& 2\cos \left({\displaystyle \frac{1}{4}\pi +\theta }\right)\cos \left({\displaystyle \frac{3}{4}\pi -\theta }\right)\\ & =\cos \left({\displaystyle \frac{1}{4}\pi +\theta +{\displaystyle \frac{3}{4}\pi -\theta }}\right)+\cos \left({\displaystyle \frac{1}{4}\pi +\theta -\left({\displaystyle \frac{3}{4}\pi -\theta }\right)}\right)\\ & =\cos \left({\displaystyle \frac{1}{4}\pi +\theta +{\displaystyle \frac{3}{4}\pi -\theta }}\right)+\cos \left({\displaystyle \frac{1}{4}\pi +\theta -{\displaystyle \frac{3}{4}\pi +\theta }}\right)\\ & =\cos \pi +\cos (2\theta -{\displaystyle \frac{1}{2}\pi })\\ & =-1+\cos \left({\displaystyle \frac{1}{2}\pi -2\theta }\right)\\ & =-1+\sin 2\theta (\mathbf{\text{ingat sudut yang berelasi}})\\ & =\sin 2\theta -1◼\\ \text{b}.& 2\sin ({315}^{\circ }+B)\sin ({45}^{\circ }-B)\\ & =-(\cos ({315}^{\circ }+B+{45}^{\circ }-B)-\cos ({315}^{\circ }+B-({45}^{\circ }-B)))\\ & =-(\cos {360}^{\circ }-\cos ({315}^{\circ }-{45}^{\circ }+B+B))\\ & -=-(1-\cos ({270}^{\circ }+2B))\\ & =-(1-\sin 2B)(\mathbf{\text{ingat sudut yang berelasi}})\\ & =\sin 2B-1◼\end{array}\end{array}$