Pembahasan Trigonometri Simak UI 2018 Matematika IPA kode 416

Soal yang Akan Dibahas
Gunakan petunjuk C.
Jika $ \alpha = -\frac{\pi}{12} $ , maka ....
(1). $ \sin ^4 \alpha + \cos ^4 \alpha = \frac{6}{8} \, $
(2). $ \sin ^6 \alpha + \cos ^6 \alpha = \frac{12}{16} \, $
(3). $ \cos ^4 \alpha = \frac{1}{2} -\frac{1}{4}\sqrt{3} \, $
(4). $ \sin ^4 \alpha = \frac{7}{16} - \frac{1}{4}\sqrt{3} \, $

$\spadesuit $ Konsep Dasar
*). identitas trigonometri :
$ \sin ^2 x + \cos ^2 x = 1 $
*). Rumus sudut rangkap :
$ \cos ^2 x = \frac{1}{2}(1 + \cos 2x ) $
$ \sin ^2 x = \frac{1}{2}(1 - \cos 2x ) $
$ \sin x \cos x = \frac{1}{2} \sin 2x $
*). Sudut negatif :
$ \sin (-x) = -\sin x $ dan $ \cos (-x) = \cos x $
*). Sifat eksponen :
$ a^4 + b^4 = (a^2 + b^2 )^2 - 2 (a.b)^2 $
$ a^6 + b^6 = (a^4 + b^4)(a^2 + b^2) - (ab)^2(a^2 + b^2) $

$\clubsuit $ Pembahasan
*). Diketahui $ \alpha = -\frac{\pi}{12} $ dengan $ \pi = 180^\circ $
*). Menentukan beberapa nilai :
$\begin{align} \cos 2 \alpha & = \cos 2 . (-\frac{\pi}{12}) = \cos 30^\circ = \frac{1}{2}\sqrt{3} \\ \sin 2 \alpha & = \sin 2 . (-\frac{\pi}{12}) = -\sin 30^\circ = -\frac{1}{2} \\ \sin \alpha . \cos \alpha & = \frac{1}{2} \sin 2 \alpha = \frac{1}{2} . (-\frac{1}{2}) = -\frac{1}{4} \end{align} $

*). Kita cek keempat pernyataan :
(1). $ \sin ^4 \alpha + \cos ^4 \alpha = \frac{6}{8} \, $ ?
$ \begin{align} \sin ^4 \alpha + \cos ^4 \alpha & = (\sin ^2 \alpha + \cos ^2 \alpha)^2 - 2(\sin \alpha . \cos \alpha )^2 \\ & = (1)^2 - 2(- \frac{1}{4} )^2 \\ & = 1 - 2( \frac{1}{16} ) = 1 - \frac{1}{8} = \frac{7}{8} \end{align} $
Pernyataan (1) SALAH.

(2). $ \sin ^6 \alpha + \cos ^6 \alpha = \frac{12}{16} \, $ ?
$ \begin{align} & \sin ^6 \alpha + \cos ^6 \alpha \\ & = (\sin ^4 \alpha + \cos ^4 \alpha )(\sin ^2 \alpha + \cos ^2 \alpha ) - (\sin x \cos x)^2 (\sin ^2 \alpha + \cos ^2 \alpha) \\ & = ( \frac{7}{8} )(1) - ( -\frac{1}{4} )^2 (1) \\ & = \frac{7}{8} - \frac{1}{16} = \frac{13}{16} \end{align} $
Pernyataan (2) SALAH.

(3). $ \cos ^4 \alpha = \frac{1}{2} -\frac{1}{4}\sqrt{3} \, $ ?
$ \begin{align} \cos ^4 \alpha & = \cos ^2 \alpha . \cos ^2 \alpha \\ & = \frac{1}{2}(1 + \cos 2 \alpha ) . \frac{1}{2}(1 + \cos 2 \alpha ) \\ & = \frac{1}{4}(1 + \cos 2 \alpha )^2 \\ & = \frac{1}{4}(1 + \frac{1}{2}\sqrt{3} )^2 \\ & = \frac{1}{4}(1 + \sqrt{3} + \frac{3}{4} ) \\ & = \frac{1}{4}(\frac{7}{4} + \sqrt{3} ) \\ & = \frac{7}{16} + \frac{1}{4}\sqrt{3} \end{align} $
Pernyataan (3) SALAH.

(4). $ \sin ^4 \alpha = \frac{7}{16} - \frac{1}{4}\sqrt{3} \, $ ?
$ \begin{align} \sin ^4 \alpha & = \sin ^2 \alpha . \sin ^2 \alpha \\ & = \frac{1}{2}(1 - \cos 2 \alpha ) . \frac{1}{2}(1 - \cos 2 \alpha ) \\ & = \frac{1}{4}(1 - \cos 2 \alpha )^2 \\ & = \frac{1}{4}(1 - \frac{1}{2}\sqrt{3} )^2 \\ & = \frac{1}{4}(1 - \sqrt{3} + \frac{3}{4} ) \\ & = \frac{1}{4}(\frac{7}{4} - \sqrt{3} ) \\ & = \frac{7}{16} - \frac{1}{4}\sqrt{3} \end{align} $
Pernyataan (4) BENAR.

Sehingga pernyataan (4) yang BENAR, jawabannya D.
Jadi, yang BENAR adalah pernyataan (4) $ . \, \heartsuit $

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