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SMK Teknologi dan Rekayasa

Soal 94 Rumus Trigonometri Jumlah-Selisih Sudut


Diketahui\ tan\ \alpha =-\frac{2}{3},\ \alpha \ sudut\ tumpul.\ Nilai\ sin\ 2\alpha \ adalah\ ....\\\\a.\ -\frac{12}{13}\ \ \ \ \ \ \ \ \ \ b.\ -\frac{6}{13}\ \ \ \ \ \ \ \ \ \ c.\ \ \frac{6}{13}\ \ \ \ \ \ \ \ \ \ d.\ \ \frac{12}{13}\ \ \ \ \ \ \ \ \ \ e.\ \ \frac{12}{5}
Jawaban: a


\begin{array}{rcl}x^{2}&=&2^{2}+3^{2}\\&=&4+9\\&=&13\\x&=&\sqrt{13}\end{array}
sin\ \alpha =\frac{2}{x}=\frac{2}{\sqrt{13}}
cos\ \alpha =-\frac{3}{x}=-\frac{3}{\sqrt{13}}
\begin{array}{rcl}sin\ 2\alpha &=&2\ sin\ \alpha \ cos\ \alpha \\\\&=&2\ \times \ \frac{2}{\sqrt{13}}\ \times \ \left ( -\frac{3}{\sqrt{13}} \right )\\\\&=&-\frac{12}{13}\end{array}
\mathbf{Keterangan}:\\Karena\ \alpha \ sudut\ tumpul\ (sudut\ di\ kuadran\ II),\ maka\ tan\ \alpha \ dan\ cos\ \alpha \\bertanda\ negatif.

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