Soal yang Akan Dibahas
Dektahui $ 0 \leq x < \frac{\pi}{2} $. Jika
$ 5\sin 2x + 10\cos ^2 x = 26 \cos 2x $ , maka $ \cos 2x = .... $
A). $ \frac{215}{233} \, $ B). $ \frac{205}{233} \, $ C). $ \frac{169}{233} \, $ D). $ \frac{115}{233} \, $ E). $ \frac{105}{233} $
A). $ \frac{215}{233} \, $ B). $ \frac{205}{233} \, $ C). $ \frac{169}{233} \, $ D). $ \frac{115}{233} \, $ E). $ \frac{105}{233} $
$\spadesuit $ Konsep Dasar Trigonometri :
*). Rumus-rumus dasar trigonometri :
i). $ \sin 2x = 2\sin x \cos x $
ii). $ \cos 2x = \cos ^2 x - \sin ^2 x = (\cos x + \sin x)(\cos x - \sin x) $
iii). $ \cos 2x = 1 - 2\sin ^2 x $
iv). $ \tan x = \frac{depan}{samping} = \frac{\sin x}{\cos x} $ dan $ \sin x = \frac{depan}{miring} $
*). Rumus-rumus dasar trigonometri :
i). $ \sin 2x = 2\sin x \cos x $
ii). $ \cos 2x = \cos ^2 x - \sin ^2 x = (\cos x + \sin x)(\cos x - \sin x) $
iii). $ \cos 2x = 1 - 2\sin ^2 x $
iv). $ \tan x = \frac{depan}{samping} = \frac{\sin x}{\cos x} $ dan $ \sin x = \frac{depan}{miring} $
$\clubsuit $ Pembahasan
*). Mengubah soalnya :
$\begin{align} 5\sin 2x + 10\cos ^2 x & = 26 \cos 2x \\ 5. 2\sin x \cos x + 10\cos ^2 x & = 26(\cos x + \sin x)(\cos x - \sin x) \\ 10\sin x \cos x + 10\cos x \cos x & = 26(\cos x + \sin x)(\cos x - \sin x) \\ 10\cos x( \sin x + \cos x) & = 26(\cos x + \sin x)(\cos x - \sin x) \\ 10\cos x & = 26 (\cos x - \sin x) \\ 26\sin x & = 16\cos x \\ \frac{\sin x}{\cos x} & = \frac{16}{26} \\ \tan x & = \frac{8}{13} \end{align} $
*). Menentukan nilai $ \sin x $ :
Karena nilai $ \tan x = \frac{8}{13} = \frac{depan}{samping} $,
$ miring = \sqrt{8^2 + 13^2 } = \sqrt{64 + 169} = \sqrt{233} $
Sehingga $ \sin x = \frac{depan}{miring} = \frac{8}{\sqrt{233}} $.
*). Menentukan nilai $ \cos 2x $ :
$\begin{align} \cos 2x & = 1 - 2\sin ^2 x \\ & = 1 - 2. \left( \frac{8}{\sqrt{233}} \right)^2 \\ & = 1 - 2. \frac{64}{233} \\ & = \frac{233}{233} - \frac{128}{233} = \frac{105}{233} \end{align} $
Jadi, nilai $ \cos 2x = \frac{105}{233} . \, \heartsuit $
*). Mengubah soalnya :
$\begin{align} 5\sin 2x + 10\cos ^2 x & = 26 \cos 2x \\ 5. 2\sin x \cos x + 10\cos ^2 x & = 26(\cos x + \sin x)(\cos x - \sin x) \\ 10\sin x \cos x + 10\cos x \cos x & = 26(\cos x + \sin x)(\cos x - \sin x) \\ 10\cos x( \sin x + \cos x) & = 26(\cos x + \sin x)(\cos x - \sin x) \\ 10\cos x & = 26 (\cos x - \sin x) \\ 26\sin x & = 16\cos x \\ \frac{\sin x}{\cos x} & = \frac{16}{26} \\ \tan x & = \frac{8}{13} \end{align} $
*). Menentukan nilai $ \sin x $ :
Karena nilai $ \tan x = \frac{8}{13} = \frac{depan}{samping} $,
$ miring = \sqrt{8^2 + 13^2 } = \sqrt{64 + 169} = \sqrt{233} $
Sehingga $ \sin x = \frac{depan}{miring} = \frac{8}{\sqrt{233}} $.
*). Menentukan nilai $ \cos 2x $ :
$\begin{align} \cos 2x & = 1 - 2\sin ^2 x \\ & = 1 - 2. \left( \frac{8}{\sqrt{233}} \right)^2 \\ & = 1 - 2. \frac{64}{233} \\ & = \frac{233}{233} - \frac{128}{233} = \frac{105}{233} \end{align} $
Jadi, nilai $ \cos 2x = \frac{105}{233} . \, \heartsuit $
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