PAS MATEMATIKA BLOG: Rumus Trigonometri Jumlah dan Selisih Dua Sudut

Perhatikanlah gambar berikut
Untuk mendapatkan rumus $\cos \left ( \alpha \pm \beta \right )$ , perhatikan ilustrasi berikut ini

$\begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\begin{aligned}\textrm{Diketahui} \: \: &\textrm{bahwa : Pada lingkaran di atas}\\ &\begin{cases} A & =(r,0) \\ B & =\left ( r\cos \alpha ,r\sin \alpha \right ) \\ C & =\left ( r\cos (\alpha +\beta ),r\sin (\alpha +\beta ) \right ) \\ D & =\left ( r\cos \beta ,-r\sin \beta \right ) \end{cases}\\ \end{aligned}}\\\hline \textrm{Jarak}&\multicolumn{2}{|c|}{OA=OB=OC=OD=r}\\\hline &(AC)^{2}&\begin{aligned}&(AC)^{2}\\ &=\left ( x_{C}-x_{A} \right )^{2}+\left ( y_{C}-y_{A} \right )^{2}\\ &=\left ( r\cos (\alpha +\beta )-r \right )^{2}+\left ( r\sin (\alpha +\beta )-0 \right )^{2}\\ &=\left ( r^{2}\cos ^{2}(\alpha +\beta )-2r^{2}\cos (\alpha +\beta )+r^{2} \right )+r^{2}\sin ^{2}(\alpha +\beta )\\ &=r^{2}\left ( \cos ^{2}(\alpha +\beta )+\sin ^{2}(\alpha +\beta ) \right )+r^{2}-2r^{2}\cos (\alpha +\beta )\\ &=r^{2}+r^{2}-2r^{2}\cos (\alpha +\beta )\\ &=2r^{2}-2r^{2}\cos (\alpha +\beta ) \end{aligned} \\\cline{2-3} (AC)^{2}=(BD)^{2}&(BD)^{2}&\begin{aligned}&(BD)^{2}\\ &=\left ( x_{D}-x_{B} \right )^{2}+\left ( y_{D}-y_{B} \right )^{2}\\ &=\left ( r\cos \alpha - r\cos \beta \right )^{2}+\left ( r\sin \alpha +r\sin \beta \right )^{2}\\ &=...\\ &=...\\ &=...\\ &=2r^{2}-2r^{2}\cos \alpha \cos \beta +2r^{2}\sin \alpha \sin \beta \end{aligned}\\\cline{2-3} &\multicolumn{2}{|c|}{\begin{aligned}(AC)^{2}&=(BD)^{2}\\ 2r^{2}-2r^{2}\cos (\alpha +\beta )&=2r^{2}-2r^{2}\cos \alpha \cos \beta +2r^{2}\sin \alpha \sin \beta\\ \cos (\alpha +\beta )&=\cos \alpha \cos \beta-\sin \alpha \sin \beta \end{aligned}}\\\hline \multicolumn{3}{|c|}{\begin{aligned}\textrm{Jadi},&\: \cos \left ( \alpha +\beta \right )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \end{aligned}}\\\hline \end{array}$

Sedangkan untuk rumus $\sin \left ( \alpha \pm \beta \right )$

kita uraikan dengan ilustrasi gambar berikut ini

Perhatikanlah untuk segi empat tali busurnya

$\begin{aligned}AC\times BD&=BC\times AD+AB\times DC\\ 1\times BD&=\sin x\times \cos y+\cos x\times \sin y,\qquad\textnormal{diameter=1 satuan=AC=BE}\\ BD&=\sin x\times \cos y+\cos x\times \sin y,\qquad \textrm{karena}\: \: \angle BAD=\angle BED,\: \: \textrm{maka}\\ \sin (x+y)&=\sin x\times \cos y+\cos x\times \sin y ,\qquad \angle ABC=\angle ADC=\angle BDE=90^{0}\end{aligned}$

Selanjutnya

$\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{untuk}\: \: \alpha \: \textrm{dan}\: \beta }\\\hline \alpha \neq \beta &\alpha = \beta \\\hline \begin{cases} \textrm{sinus} & (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ \textrm{sinus} & (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ \textrm{cosinus} & (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ \textrm{cosinus} & (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \textrm{tangen} & (\alpha +\beta )=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ \textrm{tangen} & (\alpha -\beta )=\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta } \end{cases}&\begin{aligned}\sin 2\alpha &=2\sin \alpha \cos \alpha \\ \cos 2\alpha &=\cos ^{2}\alpha -\sin ^{2}\alpha \\ &=\begin{cases} \cos 2\alpha &=2\cos ^{2}\alpha -1 \\ \cos 2\alpha &=1-2\sin ^{2}\alpha \end{cases}\\ \tan 2\alpha &=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \end{aligned}\\\hline \end{array}$

Selanjutnya saat $\alpha =\beta$ disebut sudut ganda

PAS MATEMATIKA BLOG

Rumus Trigonometri Jumlah dan Selisih Dua Sudut

Tidak ada komentar:

Posting Komentar