Bank Soal Matematika SMA Fungsi Trigonometri dan Bilangan Real

Diketahui:

Fungsi $g\left(\beta\right)=\cos\beta-\sin2\beta$.

Ditanya:

Nilai dari $g\left(\frac{17}{6}\pi\right)$?

Jawab:

Perlu diingat identitas trigonometri yang berkaitan dengan periode dari fungsi trigonometri dasar, yaitu untuk sembarang bilangan real $r$ dan sembarang bilangan bulat $k$ berlaku

$\sin\left(2k\pi+r\right)=\sin r$

$\cos\left(2k\pi+r\right)=\cos r$

$\tan\left(2k\pi+r\right)=\tan r$

Pada soal akan dicari nilai dari $g\left(\frac{17}{6}\pi\right)$. Diperoleh

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=\cos\left(\frac{17}{6}\pi\right)-\sin\left(2\times\frac{17}{6}\pi\right)$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=\cos\left(\frac{17}{6}\pi\right)-\sin\left(\frac{17}{3}\pi\right)$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=\cos\left(2\frac{5}{6}\pi\right)-\sin\left(5\frac{2}{3}\pi\right)$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=\cos\left(2\pi+\frac{5}{6}\pi\right)-\sin\left(2.2.\pi+1\frac{2}{3}\pi\right)$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=\cos\left(\frac{5}{6}\pi\right)-\sin\left(1\frac{2}{3}\pi\right)$

Perlu diingat pula untuk sudut $\theta$ dengan $\frac{1}{2}\pi<\theta<\pi$ berlaku

$\cos\theta=\cos\left(\pi-\alpha\right)=-\cos\alpha$

dan untuk sudut $\theta$ dengan $\frac{3}{2}\pi<\theta<2\pi$ berlaku

$\sin\theta=\sin\left(2\pi-\alpha\right)=-\sin\alpha$

Didapat

$g\left(\frac{17}{6}\pi\right)=\cos\left(\frac{5}{6}\pi\right)-\sin\left(1\frac{2}{3}\pi\right)$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=\cos\left(\pi-\frac{1}{6}\pi\right)-\sin\left(2\pi-\frac{1}{3}\pi\right)$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=-\cos\left(\frac{1}{6}\pi\right)-\left(-\sin\frac{1}{3}\pi\right)$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=-\cos\left(\frac{1}{6}\pi\frac{180\degree}{\pi}\right)+\sin\frac{1}{3}\pi$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=-\cos\left(\frac{1}{6}180\degree\right)+\sin(\frac{1}{3}\pi\frac{180\degree}{\pi})$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=-\cos\left(30\degree\right)+\sin(\frac{1}{3}180\degree)$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=-\frac{1}{2}\sqrt{3}+\sin60\degree$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=-\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{3}$

$\Leftrightarrow g\left(\frac{17}{6}\pi\right)=0$