Pembahasan Trigonometri Simak UI 2018 Matematika IPA kode 421

Soal yang Akan Dibahas
Gunakan petunjuk C.
Jika $ \alpha = -\frac{5\pi}{12} $ , maka ....
(1). $ \sin ^4 \alpha - \cos ^4 \alpha = \frac{1}{2}\sqrt{3} \, $
(2). $ \sin ^6 \alpha - \cos ^6 \alpha = \frac{5}{32}\sqrt{3} \, $
(3). $ \cos ^4 \alpha = \frac{7}{16} -\frac{1}{4}\sqrt{3} \, $
(4). $ \sin ^4 \alpha = \frac{7}{16} - \frac{3}{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 ) $
$ \cos ^2 x - \sin ^2 x = \cos 2x $
$ \sin x \cos x = \frac{1}{2} \sin 2x $
$ \sin (-x) = - \sin x \, $ dan $ \cos (-x) = \cos x $
*). Sifat eksponen :
$ a^4 - b^4 = (a^2 - b^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{5\pi}{12} $ dengan $ \pi = 180^\circ $
*). Menentukan beberapa nilai :
$\begin{align} \cos 2 \alpha & = \cos 2 . (-\frac{5\pi}{12}) = \cos 150^\circ = -\frac{1}{2}\sqrt{3} \\ \sin 2 \alpha & = \sin 2 .(- \frac{5\pi}{12}) = -\sin 150^\circ = -\frac{1}{2} \\ (\sin ^2 \alpha - \cos ^2 \alpha) & = -(\cos ^2 \alpha - \sin ^2 \alpha) \\ & = -\cos 2 \alpha = - (-\frac{1}{2}\sqrt{3} ) = \frac{1}{2}\sqrt{3} \\ \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{1}{2}\sqrt{3} \, $ ?
$ \begin{align} \sin ^4 \alpha - \cos ^4 \alpha & = (\sin ^2 \alpha - \cos ^2 \alpha)(\sin ^2 \alpha + \cos ^2 \alpha) \\ & = (\sin ^2 \alpha - \cos ^2 \alpha)(1) \\ & = (\sin ^2 \alpha - \cos ^2 \alpha) = \frac{1}{2}\sqrt{3} \end{align} $
Pernyataan (1) BENAR.

(2). $ \sin ^6 \alpha - \cos ^6 \alpha = \frac{5}{32}\sqrt{3} \, $ ?
$ \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{1}{2}\sqrt{3} )(1) +(-\frac{1}{4})^2 (\frac{1}{2}\sqrt{3} ) \\ & = \frac{1}{2}\sqrt{3} + \frac{1}{16} . \frac{1}{2}\sqrt{3} \\ & = \frac{1}{2}\sqrt{3} + \frac{1}{32} \sqrt{3} \\ & = \frac{16}{32}\sqrt{3} + \frac{1}{32} \sqrt{3} \\ & = \frac{17}{32}\sqrt{3} \end{align} $
Pernyataan (2) SALAH.

(3). $ \cos ^4 \alpha = \frac{7}{16} -\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) BENAR.

(4). $ \sin ^4 \alpha = \frac{7}{16} - \frac{3}{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 + \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) SALAH.

Sehingga pernyataan (1) dan (3) yang BENAR, jawabannya B.
Jadi, yang BENAR adalah pernyataan (1) dan (3) $ . \, \heartsuit $

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