PAS MATEMATIKA BLOG: Lanjuta Rumus trigonometri

$\textrm{Perhatikanlah gambar di bawah ini. Akan ditunjukkan bahwa}\\ \sin \left ( \alpha +\beta \right )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ \textrm{dengan cara menyatakan luas segitiga sebagai }\\ \textrm{jumlah dari dua segitiga lainnya }$

Bukti

Perhatikanlah ΔAA’C dan ΔAA’B

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

Sedangkan untuk bukti

$\cos \left ( \alpha -\beta \right )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$

Perhatikanlah ilustrasi berikut juga

Jika kita ubah menjadi

maka

$\begin{array}{|c|c|c|c|c|c|}\hline \textrm{No}&\textrm{Segitiga}&\textrm{Aturan}\: \textrm{Sinus}&\textrm{No}&\textrm{Segitiga}&\textrm{Aturan}\: \textrm{Sinus}\\\hline 1.&\triangle ACD&\displaystyle \frac{CD}{\sin \angle A }=\displaystyle \frac{AC}{\sin \angle D}=\displaystyle \frac{AD}{\sin \angle C}&2.&\triangle BCD&\displaystyle \frac{CD}{\sin \angle B}=\displaystyle \frac{BC}{\sin \angle D}=\displaystyle \frac{BD}{\sin \angle C}\\ \cline{3-3}\cline{6-6} &&\begin{aligned}CD&=\displaystyle \frac{AC}{\sin 90^{0}}\times \sin \angle A\\ &=\displaystyle \frac{b}{1}\times \sin \alpha \\ &=b\: \sin \alpha\\ &\textnormal{gunakan dalil Pythagoras untuk mencari AD,}\\ AD&=b\: \cos \alpha \end{aligned}&&&\begin{aligned}BD&=\displaystyle \frac{BC}{\sin 90^{0}}\times \sin \angle C\\ &=\displaystyle \frac{a}{1}\times \sin \beta \\ &=a\: \sin \beta \\ &\textnormal{gunakan juga dalil Pythagoras, maka}\\ CD&=a\: \cos \beta \end{aligned}\\\hline \end{array}$

Sehingga

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

Dan untuk formula tangenm misalkan kita diminta untuk menunjukkan bukti berikut

$\begin{array}{lll}\\ &a.&\tan \left ( \alpha +\beta \right )=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ &b.&\tan \left ( \alpha -\beta \right )=\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta } \end{array}$ .

maka sebagai buktinya adalah:

$\begin{aligned}\textnormal{a.}\quad \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\sin \left ( \alpha +\beta \right )}{\cos \left ( \alpha +\beta \right )}\\ &=\displaystyle \frac{\left (\sin \alpha \cos \beta +\cos \alpha \sin \beta \right )\times \left ( \displaystyle \frac{1}{\cos \alpha \cos \beta } \right ) }{\left (\cos \alpha \cos \beta -\sin \alpha \sin \beta \right ) \times \left ( \displaystyle \frac{1}{\cos \alpha \cos \beta } \right )}\\ &=\displaystyle \frac{\displaystyle \frac{\sin \alpha }{\cos \alpha }+\displaystyle \frac{\sin \beta }{\cos \beta }}{1-\displaystyle \frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}\\ &=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }.\qquad \blacksquare \\ \textnormal{b.}\quad \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }&\textnormal{dengan mengganti}\: \beta =-\beta \: \: \textrm{maka,}\\ &=\displaystyle \frac{\tan \alpha +\tan \left ( -\beta \right )}{1-\tan \alpha \tan \left ( -\beta \right )}\\ \tan \left ( \alpha -\beta \right )&=\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }.\qquad \blacksquare \end{aligned}$

PAS MATEMATIKA BLOG

Lanjuta Rumus trigonometri

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