Soal yang Akan Dibahas
Jika $ \alpha + \beta = \frac{\pi}{3} , \, \alpha , \beta \, $ sudut-sudut lancip dan
$ \tan \alpha = \frac{1}{6}\tan \beta $ , maka $ \sin \alpha + \sin \beta = .... $
A). $ \frac{1}{7}\sqrt{7} + \frac{1}{5}\sqrt{5} \, $
B). $ \frac{1}{10} + \frac{1}{5}\sqrt{5} \, $
C). $ \frac{1}{4} + \frac{1}{5}\sqrt{5} \, $
D). $ \frac{1}{14}\sqrt{5} + \frac{1}{5}\sqrt{3} \, $
E). $ \frac{1}{14}\sqrt{7} + \frac{1}{5}\sqrt{5} \, $
A). $ \frac{1}{7}\sqrt{7} + \frac{1}{5}\sqrt{5} \, $
B). $ \frac{1}{10} + \frac{1}{5}\sqrt{5} \, $
C). $ \frac{1}{4} + \frac{1}{5}\sqrt{5} \, $
D). $ \frac{1}{14}\sqrt{5} + \frac{1}{5}\sqrt{3} \, $
E). $ \frac{1}{14}\sqrt{7} + \frac{1}{5}\sqrt{5} \, $
$\spadesuit $ Konsep Dasar Trigonometri :
*). Rumus jumlah sudut :
$ \tan (\alpha + \beta ) = \frac{\tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta} $
*). Rumus perbandingan trigonometri :
$ \sin x = \frac{depan}{miring} \, $ dan $ \tan x = \frac{depan}{samping} $
*). Rumus jumlah sudut :
$ \tan (\alpha + \beta ) = \frac{\tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta} $
*). Rumus perbandingan trigonometri :
$ \sin x = \frac{depan}{miring} \, $ dan $ \tan x = \frac{depan}{samping} $
$\clubsuit $ Pembahasan
*). Menentukan nilai $ \tan \beta $ :
diketahui $ \alpha + \beta = \frac{\pi}{3} = 60^\circ \, $ dan $ \tan \alpha = \frac{1}{6} \tan \beta $
$ \begin{align} \alpha + \beta & = \frac{\pi}{3} \\ \tan (\alpha + \beta ) & = \tan ( \frac{\pi}{3} ) \\ \frac{\tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta} & = \sqrt{3} \, \, \, \, \, \, \, (\text{Subst } \tan \alpha = \frac{1}{6}\tan \beta ) \\ \frac{\frac{1}{6} \tan \beta + \tan \beta }{1 - \frac{1}{6} \tan \beta \tan \beta} & = \sqrt{3} \\ \frac{\frac{7}{6} \tan \beta }{1 - \frac{1}{6} \tan ^2 \beta } & = \sqrt{3} \\ \frac{7}{6} \tan \beta & = \sqrt{3} (1 - \frac{1}{6} \tan ^2 \beta ) \, \, \, \, \, \, \text{(kali 6)} \\ 7\tan \beta & = \sqrt{3} (6 - \tan ^2 \beta ) \, \, \, \, \, \, \text{(kali } \sqrt{3} ) \\ 7\sqrt{3}\tan \beta & = 3 (6 - \tan ^2 \beta ) \\ 7\sqrt{3}\tan \beta & = 18 - 3 \tan ^2 \beta \\ 0 & = 3\tan ^2 \beta + 7\sqrt{3}\tan \beta -18 \\ 0 & = (3\tan \beta - 2\sqrt{3})(\tan \beta + 3\sqrt{3}) \\ \tan \beta & = \frac{2\sqrt{3}}{3} \vee \tan \beta = - 3\sqrt{3} \end{align} $
Karena $ \beta \, $ lancip, maka $ \tan \beta = \frac{2\sqrt{3}}{3} $ yang memenuhi.
Sehingga nilai $ \tan \alpha = \frac{1}{6} . \tan \beta = \frac{1}{6}. \frac{2\sqrt{3}}{3} = \frac{\sqrt{3}}{9} $.
*). Menentukan nilai $ \sin \alpha $ dan $ \sin \beta $ :
*). Menentukan nilai $ \sin \alpha + \sin \beta $ :
$ \begin{align} \sin \alpha + \sin \beta & = \frac{1}{14}\sqrt{7} + \frac{4}{14}\sqrt{7} = \frac{5}{14}\sqrt{7} \end{align} $
Jadi, nilai $ \sin \alpha + \sin \beta = \frac{5}{14}\sqrt{7} . \, \heartsuit $
(tidak ada jawaban pada option)
*). Menentukan nilai $ \tan \beta $ :
diketahui $ \alpha + \beta = \frac{\pi}{3} = 60^\circ \, $ dan $ \tan \alpha = \frac{1}{6} \tan \beta $
$ \begin{align} \alpha + \beta & = \frac{\pi}{3} \\ \tan (\alpha + \beta ) & = \tan ( \frac{\pi}{3} ) \\ \frac{\tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta} & = \sqrt{3} \, \, \, \, \, \, \, (\text{Subst } \tan \alpha = \frac{1}{6}\tan \beta ) \\ \frac{\frac{1}{6} \tan \beta + \tan \beta }{1 - \frac{1}{6} \tan \beta \tan \beta} & = \sqrt{3} \\ \frac{\frac{7}{6} \tan \beta }{1 - \frac{1}{6} \tan ^2 \beta } & = \sqrt{3} \\ \frac{7}{6} \tan \beta & = \sqrt{3} (1 - \frac{1}{6} \tan ^2 \beta ) \, \, \, \, \, \, \text{(kali 6)} \\ 7\tan \beta & = \sqrt{3} (6 - \tan ^2 \beta ) \, \, \, \, \, \, \text{(kali } \sqrt{3} ) \\ 7\sqrt{3}\tan \beta & = 3 (6 - \tan ^2 \beta ) \\ 7\sqrt{3}\tan \beta & = 18 - 3 \tan ^2 \beta \\ 0 & = 3\tan ^2 \beta + 7\sqrt{3}\tan \beta -18 \\ 0 & = (3\tan \beta - 2\sqrt{3})(\tan \beta + 3\sqrt{3}) \\ \tan \beta & = \frac{2\sqrt{3}}{3} \vee \tan \beta = - 3\sqrt{3} \end{align} $
Karena $ \beta \, $ lancip, maka $ \tan \beta = \frac{2\sqrt{3}}{3} $ yang memenuhi.
Sehingga nilai $ \tan \alpha = \frac{1}{6} . \tan \beta = \frac{1}{6}. \frac{2\sqrt{3}}{3} = \frac{\sqrt{3}}{9} $.
*). Menentukan nilai $ \sin \alpha $ dan $ \sin \beta $ :
*). Menentukan nilai $ \sin \alpha + \sin \beta $ :
$ \begin{align} \sin \alpha + \sin \beta & = \frac{1}{14}\sqrt{7} + \frac{4}{14}\sqrt{7} = \frac{5}{14}\sqrt{7} \end{align} $
Jadi, nilai $ \sin \alpha + \sin \beta = \frac{5}{14}\sqrt{7} . \, \heartsuit $
(tidak ada jawaban pada option)
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