Trigonometri (KTSP MA/SMA Kelas XI)

A. Rumus Trigonometri untuk Jumlah dan Selisih Dua Sudut

  • Rumus  \cos \left ( \alpha \pm \beta \right ) .
    perhatikanlah uraian berikut berkaitan rumus cosinus

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\begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\begin{aligned}\textrm{Diketahui} \: \: &\textrm{bahwa : Pada lingkaran di atas}\\ &\begin{cases} A & =(r,0) \\ B & =\left ( r\cos \alpha ,r\sin \alpha \right ) \\ C & =\left ( r\cos (\alpha +\beta ),r\sin (\alpha +\beta ) \right ) \\ D & =\left ( r\cos \beta ,-r\sin \beta \right ) \end{cases}\\ \end{aligned}}\\\hline \textrm{Jarak}&\multicolumn{2}{|c|}{OA=OB=OC=OD=r}\\\hline &(AC)^{2}&\begin{aligned}(AC)^{2}&=\left ( x_{C}-x_{A} \right )^{2}+\left ( y_{C}-y_{A} \right )^{2}\\ &=\left ( r\cos (\alpha +\beta )-r \right )^{2}+\left ( r\sin (\alpha +\beta )-0 \right )^{2}\\ &=\left ( r^{2}\cos ^{2}(\alpha +\beta )-2r^{2}\cos (\alpha +\beta )+r^{2} \right )+r^{2}\sin ^{2}(\alpha +\beta )\\ &=r^{2}\left ( \cos ^{2}(\alpha +\beta )+\sin ^{2}(\alpha +\beta ) \right )+r^{2}-2r^{2}\cos (\alpha +\beta )\\ &=r^{2}+r^{2}-2r^{2}\cos (\alpha +\beta )\\ &=2r^{2}-2r^{2}\cos (\alpha +\beta ) \end{aligned} \\\cline{2-3} (AC)^{2}=(BD)^{2}&(BD)^{2}&\begin{aligned}(BD)^{2}&=\left ( x_{D}-x_{B} \right )^{2}+\left ( y_{D}-y_{B} \right )^{2}\\ &=\left ( r\cos \alpha - r\cos \beta \right )^{2}+\left ( r\sin \alpha +r\sin \beta \right )^{2}\\ &=...\\ &=...\\ &=...\\ &=2r^{2}-2r^{2}\cos \alpha \cos \beta +2r^{2}\sin \alpha \sin \beta \end{aligned}\\\cline{2-3} &\multicolumn{2}{|c|}{\begin{aligned}(AC)^{2}&=(BD)^{2}\\ 2r^{2}-2r^{2}\cos (\alpha +\beta )&=2r^{2}-2r^{2}\cos \alpha \cos \beta +2r^{2}\sin \alpha \sin \beta\\ \cos (\alpha +\beta )&=\cos \alpha \cos \beta-\sin \alpha \sin \beta \end{aligned}}\\\hline \multicolumn{3}{|c|}{\begin{aligned}\textrm{Jadi},&\: \cos \left ( \alpha +\beta \right )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \end{aligned}}\\\hline \end{array}.

  • Rumus  \sin \left ( \alpha \pm \beta \right ) .
    perhatikanlah uraian berikut berkaitan rumus sinus

356

Perhatikanlah segiempat tali busur di atas.

\begin{aligned}AC\times BD&=BC\times AD+AB\times DC\\ 1\times BD&=\sin x\times \cos y+\cos x\times \sin y,\qquad\textnormal{diameter=1 satuan=AC=BE}\\ BD&=\sin x\times \cos y+\cos x\times \sin y,\qquad \textrm{karena}\: \: \angle BAD=\angle BED,\: \: \textrm{maka}\\ \sin (x+y)&=\sin x\times \cos y+\cos x\times \sin y ,\qquad \angle ABC=\angle ADC=\angle BDE=90^{0}\end{aligned}.

Anda juga dapat membuka kembali lembaran/halaman lama baik sinus , cosinus, dan tangen di sini  bukti dengan bentuk lain.

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{untuk}\: \: \alpha \: \textrm{dan}\: \beta }\\\hline \alpha \neq \beta &\alpha = \beta \\\hline \begin{cases} \textrm{sinus} & (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ \textrm{sinus} & (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ \textrm{cosinus} & (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ \textrm{cosinus} & (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \textrm{tangen} & (\alpha +\beta )=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ \textrm{tangen} & (\alpha -\beta )=\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta } \end{cases}&\begin{aligned}\sin 2\alpha &=2\sin \alpha \cos \alpha \\ \cos 2\alpha &=\cos ^{2}\alpha -\sin ^{2}\alpha \begin{cases} \cos 2\alpha &=2\cos ^{2}\alpha -1 \\ \cos 2\alpha &=1-2\sin ^{2}\alpha \end{cases}\\ \tan 2\alpha &=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \end{aligned}\\\hline \end{array}.

B. Rumus Trigonometri untuk Sudut Ganda

Perhatikan tabel di atas. Saat  \alpha =\beta  dinamakan bersudut ganda

 \LARGE\fbox{\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Gunakan}\: \: \cos (\alpha \pm \beta )\: \: \textrm{untuk menunjukkan bahwa}:\\ &\textrm{a}.\quad \cos (90^{0}-\alpha )=\sin \alpha \\ &\textrm{b}.\quad \cos (90^{0}+\alpha )=-\sin \alpha \\ &\textrm{c}.\quad \cos (180^{0}-\alpha )=-\cos \alpha \\ &\textrm{d}.\quad \cos (180^{0}+\alpha )=-\cos \alpha \\ &\textrm{e}.\quad \cos (270^{0}-\alpha )=-\sin \alpha \\ &\textrm{f}.\quad \cos (270^{0}+\alpha )=\sin \alpha \\ &\textrm{g}.\quad \cos (360^{0}-\alpha )=\cos \alpha \\ &\textrm{h}.\quad \cos (360^{0}+\alpha )=\cos \alpha\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}\textrm{a}.\quad \cos (90^{0}-\alpha )&=\cos 90^{0}.\cos \alpha +\sin 90^{0}.\sin \alpha \\ &=0.\cos \alpha +1.\sin \alpha\\ &=\sin \alpha \qquad\qquad \blacksquare \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \cos (90^{0}+\alpha )&=\cos 90^{0}.\cos \alpha -\sin 90^{0}.\sin \alpha \\ &=0.\cos \alpha -1.\sin \alpha\\ &=-\sin \alpha\qquad\qquad \blacksquare \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad \cos (180^{0}-\alpha )&=\cos 180^{0}.\cos \alpha +\sin 180^{0}.\sin \alpha \\ &=(-1).\cos \alpha -0.\sin \alpha\\ &=-\cos \alpha\qquad\qquad \blacksquare \end{aligned} \\ &\vdots \\ &\\ &\textrm{sebagai pengingat kita}\\ &\qquad \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \cdots \alpha &0^{0}&30^{0}&45^{0}&60^{0}&90^{0}&180^{0}&270^{0}&360^{0}\\\hline \sin \alpha &0&\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0&-1&0\\\hline \cos \alpha &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1&0&1\\\hline \tan \alpha &0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&TD&0&TD&0\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Tanpa menggunakan tabel, tentukanlah nilai eksak untuk}:\\ &\textrm{a}.\quad \cos 80^{0}\cos 10^{0}-\sin 80^{0}\sin 10^{0} \\ &\textrm{b}.\quad \cos 130^{0}\cos 40^{0}+\sin 130^{0}\sin 40^{0} \\ &\textrm{c}.\quad \cos 38^{0}\cos 22^{0}-\sin 38^{0}\sin 22^{0} \\ &\textrm{d}.\quad \cos 70^{0}\cos 25^{0}+\sin 70^{0}\sin 25^{0} \\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}\textrm{a}.\quad \cos 80^{0}\cos 10^{0}-\sin 80^{0}\sin 10^{0}&=\cos \left ( 80^{0}+10^{0} \right )=\cos 90^{0}=0 \end{aligned} \\ &\begin{aligned}\textrm{b}.\quad \cos 130^{0}\cos 40^{0}+\sin 130^{0}\sin 40^{0}&=\cos \left ( 130^{0}-40^{0} \right )=\cos 90^{0}=0 \end{aligned} \\ &\begin{aligned}\textrm{c}.\quad \cos 38^{0}\cos 22^{0}-\sin 38^{0}\sin 22^{0}&=\cos \left ( 38^{0}+22^{0} \right )=\cos 60^{0}=\displaystyle \frac{1}{2} \end{aligned} \\ &\begin{aligned}\textrm{d}.\quad \cos 70^{0}\cos 25^{0}+\sin 70^{0}\sin 25^{0}&=\cos \left ( 70^{0}-25^{0} \right )=\cos 45^{0}=\displaystyle \frac{1}{2}\sqrt{2} \end{aligned} \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Tunjukkan nilai eksak dari}\\ &\textrm{a}.\quad \cos 105^{0}=\displaystyle \frac{1}{4}\sqrt{2}\left ( 1-\sqrt{3} \right )\\ &\textrm{b}.\quad \cos (-15)^{0}=\displaystyle \frac{1}{4}\sqrt{2}\left ( 1+\sqrt{3} \right )\\ &\textrm{c}.\quad \cos 195^{0}=-\displaystyle \frac{1}{4}\sqrt{2}\left ( 1+\sqrt{3} \right )\\ &\textrm{d}.\quad \cos 225^{0}=\displaystyle \frac{1}{4}\sqrt{2}\left ( 1-\sqrt{3} \right )\\\\ &\textrm{Jawab}\\\\ &\begin{aligned}\textrm{a}.\quad \cos 105^{0}&=\displaystyle \cos (45^{0}+60^{0})\\ &=\cos 45^{0}.\cos 60^{0}-\sin 45^{0}.\sin 60^{0}\\ &=\displaystyle \left (\frac{1}{2}\sqrt{2} \right ).\left (\frac{1}{2} \right )-\left (\frac{1}{2}\sqrt{2} \right ).\left (\frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{2}\left ( 1-\sqrt{3} \right )\qquad\qquad \blacksquare \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \cos (-15)^{0}&=\displaystyle \cos (45^{0}-60^{0})\\ &=\cos 45^{0}.\cos 60^{0}+\sin 45^{0}.\sin 60^{0}\\ &=\displaystyle \left (\frac{1}{2}\sqrt{2} \right ).\left (\frac{1}{2} \right )+\left (\frac{1}{2}\sqrt{2} \right ).\left (\frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{2}\left ( 1+\sqrt{3} \right )\qquad\qquad \blacksquare \end{aligned} \\ &\begin{aligned}\textrm{c}.\quad \cos 195^{0}&=\displaystyle \cos (180^{0}+15^{0})\\ &=-\cos 15^{0}\\ &=-\left ( \cos \left ( 60^{0}-45^{0} \right ) \right )\\ &=-\left ( \cos 60^{0}.\cos 45^{0}+\sin 60^{0}.\sin 45^{0} \right )\\ &=-\left ( \displaystyle \frac{1}{2}.\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3} .\frac{1}{2}\sqrt{2}\right )\\ &=-\displaystyle \frac{1}{4}\sqrt{2}\left ( 1+\sqrt{3} \right )\qquad\qquad \blacksquare \end{aligned} \\ &\textrm{d}.\quad \textrm{Silahkan dicoba sendiri sebagai latihan} \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Diketahui}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{sudut-sudut lancip, dengan}\\ &\cos \alpha =\displaystyle \frac{4}{5}\: \: \textrm{dan}\: \: \cos \beta =\frac{12}{13},\: \: \textrm{tentukanlah}:\\ &\textrm{a}.\quad \tan (\alpha +\beta )\\ &\textrm{b}.\quad \tan (\alpha -\beta )\\\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Dalam}\: \, &\textrm{segitiga siku-siku dengan sisi, x, y, dan r}\\ &\textrm{berlaku}:\\ &\qquad\qquad \: \: x^{2}+y^{2}=r^{2}\\ &\textrm{berada di kuadran 1 (sudut lancip)}\\ &\textrm{semua harga bernilai positif}\\ & \end{aligned}}\\\hline 3^{2}+4^{2}=5^{2}&5^{2}+12^{2}=13^{2}\\\hline \cos \alpha =\displaystyle \frac{4}{5}\begin{cases} \sin \alpha =\displaystyle \frac{3}{5} \\ \\ \tan \alpha =\displaystyle \frac{3}{4} \end{cases}&\cos \beta =\displaystyle \frac{12}{13}\begin{cases} \sin \beta =\displaystyle \frac{5}{13} \\ \\ \tan \beta =\displaystyle \frac{5}{12} \end{cases}\\\hline \textbf{a}&\textbf{b}\\\hline \begin{aligned}\tan (\alpha +\beta )&=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }\\ &=\displaystyle \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}}\\ &=\displaystyle \frac{\frac{9}{12}+\frac{5}{12}}{1-\frac{5}{16}}\\ &=\displaystyle \frac{\frac{14}{12}}{\frac{11}{16}}\\ &=\displaystyle \frac{56}{33} \end{aligned}&\begin{aligned}\tan (\alpha -\beta )&=\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha .\tan \beta }\\ &=\displaystyle \frac{\frac{3}{4}-\frac{5}{12}}{1+\frac{3}{4}.\frac{5}{12}}\\ &=\displaystyle \frac{\frac{9}{12}-\frac{5}{12}}{1+\frac{5}{16}}\\ &=\displaystyle \frac{\frac{4}{12}}{\frac{21}{16}}\\ &=\displaystyle \frac{16}{63} \end{aligned}\\\hline \end{array} \end{array}.

C. Rumus Trigonometri untuk Setengah Sudut

\begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\textrm{Berawal dari}}\\\hline \sin ^{2}\alpha =\displaystyle \frac{1-\cos 2\alpha }{2}&\cos ^{2}\alpha =\displaystyle \frac{1+\cos 2\alpha }{2}&\tan \alpha =\displaystyle \frac{\sin \alpha }{\cos \alpha }\\\hline \multicolumn{3}{|c|}{\textrm{Menjadi}}\\\hline \sin \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{2}}&\cos \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1+\cos \alpha }{2}}&\tan \frac{1}{2}\alpha =\begin{cases} \pm \sqrt{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }} \\\\ \pm \displaystyle \frac{\sin \alpha }{1+\cos \alpha } \\\\ \pm \displaystyle \frac{1-\cos \alpha }{\sin \alpha }\\ \end{cases}\\\hline \end{array}.

D. Rumus Trigonometri untuk Perkalian, Penjumlahan, dan Pengurangan Sinus dan Kosinus

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Rumus}}\\\hline \textrm{Perkalian}&\textrm{Penjumlahan dan Pengurangan}\\\hline 2\sin \alpha \cos \beta =\sin (\alpha +\beta )+\sin (\alpha -\beta )&\sin \alpha +\sin \beta =2\sin \frac{1}{2}(\alpha +\beta ).\cos \frac{1}{2}(\alpha -\beta )\\ 2\cos \alpha \sin \beta =\sin (\alpha +\beta )-\sin (\alpha -\beta )&\sin \alpha -\sin \beta =2\cos \frac{1}{2}(\alpha +\beta ).\sin \frac{1}{2}(\alpha -\beta )\\ 2\cos \alpha \cos \beta =\cos (\alpha +\beta )+\cos (\alpha -\beta )&\cos \alpha +\cos \beta =2\cos \frac{1}{2}(\alpha +\beta ).\cos \frac{1}{2}(\alpha -\beta )\\ -2\sin \alpha \sin \beta =\cos (\alpha +\beta )-\cos (\alpha -\beta )&\cos \alpha -\cos \beta =-2\sin \frac{1}{2}(\alpha +\beta ).\sin \frac{1}{2}(\alpha -\beta )\\\hline \end{array}.

E. Identitas Trigonometri 

Selain  sudut-sudut yang berelasi berikut adalah beberapa identitas trigonometri

\begin{array}{|c|c|c|c|}\hline \multicolumn{4}{|c|}{\textrm{Rumus Sudut}}\\\hline \textrm{Setengah}&\textrm{Satu}&\textrm{Dua}&\textrm{Tiga}\\\hline \begin{cases} \sin \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{2}} \\ \cos \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1+\cos \alpha }{2}} \\ \tan \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }} \end{cases}&\begin{cases} \sin \alpha =\displaystyle \frac{1}{\csc \alpha } \\ \cos \alpha =\displaystyle \frac{1}{\sec \alpha } \\ \tan \alpha =\displaystyle \frac{1}{\cot \alpha } \\ \tan \alpha =\displaystyle \frac{\sin \alpha }{\cos \alpha } \\ \cot \alpha =\displaystyle \frac{\cos \alpha }{\sin \alpha } \end{cases}&\begin{cases} \sin 2\alpha =2\sin \alpha \cos \alpha \\ \cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha \\ \tan 2\alpha =\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \end{cases}&\begin{cases} \sin 3\alpha =3\sin \alpha -4\sin ^{3}\alpha \\ \cos 3\alpha =4\cos ^{3}\alpha -3\cos \alpha \\ \tan 3\alpha =\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \end{cases}\\\hline \multicolumn{4}{|c|}{\textrm{Pythagoras}\: \: \begin{cases} \sin ^{2}\alpha +\cos ^{2}\alpha =1 \\ \sec ^{2}\alpha =\tan ^{2}\alpha +1 \\ \csc ^{2}\alpha =\cot ^{2}\alpha +1 \end{cases}}\\\hline \end{array}.

 \LARGE\fbox{\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Tanpa menggunakan kalkulator, hitunglah nilai eksak dari}\\ &\textrm{a}.\quad 2\sin 37\frac{1}{2}^{\circ}\cos 7\frac{1}{2}^{\circ}\\ &\textrm{b}.\quad 2\cos 37\frac{1}{2}^{\circ}\sin 7\frac{1}{2}^{\circ}\\ &\textrm{c}.\quad 2\cos 37\frac{1}{2}^{\circ}\cos 7\frac{1}{2}^{\circ}\\ &\textrm{d}.\quad 2\sin 82\frac{1}{2}^{\circ}\sin 37\frac{1}{2}^{\circ}\\ &\textrm{e}.\quad 2\sin 105^{\circ}\cos 75^{\circ}\\ &\textrm{f}.\quad 2\cos 105^{\circ}\sin 75^{\circ}\\ &\textrm{g}.\quad 2\sin 105^{\circ}\sin 75^{\circ}\\ &\textrm{h}.\quad 2\cos 105^{\circ}\cos 75^{\circ}\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}\textrm{a}.\quad 2\sin 37\frac{1}{2}^{\circ}\cos 7\frac{1}{2}^{\circ}&=\sin \left ( 37\frac{1}{2}^{\circ}+7\frac{1}{2}^{\circ} \right )+\sin \left ( 37\frac{1}{2}^{\circ}-7\frac{1}{2}^{\circ} \right )\\ &=\sin \left ( 45^{\circ} \right )+\sin \left ( 30^{\circ} \right )\\ &=\displaystyle \frac{1}{2}\sqrt{2}+\displaystyle \frac{1}{2}\\ &=\displaystyle \frac{1}{2}\left ( \sqrt{2}+1 \right )\end{aligned}\\ &\begin{aligned}\textrm{b}.\quad 2\cos 37\frac{1}{2}^{\circ}\sin 7\frac{1}{2}^{\circ}&=\sin \left ( 37\frac{1}{2}^{\circ}+7\frac{1}{2}^{\circ} \right )-\sin \left ( 37\frac{1}{2}^{\circ}-7\frac{1}{2}^{\circ} \right )\\ &=\sin \left ( 45^{\circ} \right )-\sin \left ( 30^{\circ} \right )\\ &=\displaystyle \frac{1}{2}\sqrt{2}-\displaystyle \frac{1}{2}\\ &=\displaystyle \frac{1}{2}\left ( \sqrt{2}-1 \right )\end{aligned}\\ &\vdots \\ &\vdots \\ &\vdots \\\\ &\textrm{Untuk soal yang belum dibahas silahkan diselesaikan sendiri sebagai latihan} \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Ubahlah bentuk-bentuk berikut ke dalam bentuk perkalian}\\ &\textrm{a}.\quad \sin 3x+\sin x\\ &\textrm{b}.\quad \sin 5x+\sin x\\ &\textrm{c}.\quad \sin 6y-\sin 2y\\ &\textrm{d}.\quad \cos 6p+\cos 2p\\ &\textrm{e}.\quad \cos 7\alpha +\cos 5\alpha \\ &\textrm{f}.\quad \cos 8\alpha -\cos 2\alpha\\ &\textrm{g}.\quad \sin 48^{\circ}+\sin 22^{\circ}\\ &\textrm{h}.\quad \sin 95^{\circ}-\sin 35^{\circ}\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}\textrm{a}.\quad \sin 3x+\sin x&=2\sin \displaystyle \frac{1}{2}(3x+x)\cos \displaystyle \frac{1}{2}(3x-x)\\ &=2\sin 2x\cos x\end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \sin 5x+\sin x&=2\sin \displaystyle \frac{1}{2}(5x+x)\cos \displaystyle \frac{1}{2}(5x-x)\\ &=2\sin 3x\cos 2x \end{aligned} \\ &\vdots \\ &\vdots \\ &\vdots \\\\ &\textrm{Soal yang belum diselesaikan silahkan dicoba sendiri sebagai latihan} \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Buktikanlah bahwa}\\ &\textrm{a}.\quad \displaystyle \frac{2\tan \alpha }{1+\tan ^{2}\alpha }=\sin 2\alpha \\ &\textrm{b}.\quad \displaystyle \frac{\sin ^{3}\beta \cos ^{3}\beta }{\sin \beta +\cos \beta }=1-\displaystyle \frac{1}{2}\sin 2\beta \\ &\textrm{c}.\quad \displaystyle \frac{\cos 3\alpha -\sin 6\alpha -\cos 9\alpha }{\sin 9\alpha -\cos 6\alpha -\sin 3\alpha }=\tan 6\alpha \\ &\textrm{d}.\quad \displaystyle \frac{\sin 3\beta +\sin 5\beta +\sin 7\beta +\sin 9\beta }{\cos 3\beta +\cos 5\beta +\cos 7\beta +\cos 9\beta }=\tan 6\beta \\ &\textrm{e}.\quad \displaystyle \frac{\sin 2\delta -\sin 4\delta +\sin 6\delta }{\cos 2\delta -\cos 4\delta +\cos 6\delta }=\tan 4\delta \\ &\textrm{f}.\quad 2\cos \left ( \displaystyle \frac{1}{4}\pi +\alpha \right )\cos \left ( \displaystyle \frac{1}{4}\pi -\alpha \right ).\displaystyle \frac{\cos 3\alpha -\cos 5\alpha }{\sin 3\alpha -\sin \alpha }=\sin 4\alpha \\ &\textrm{g}.\quad \sin 2\gamma +\sin 4\gamma +\sin 6\gamma =4\cos \gamma \cos 2\gamma \sin 3\gamma \\\\ &\textrm{Bukti}:\\\\ &\textrm{a}\quad \cdots \\ &\textrm{b}\quad \cdots \\ &\begin{aligned}\textrm{c}.\quad \displaystyle \frac{\cos 3\alpha -\sin 6\alpha -\cos 9\alpha }{\sin 9\alpha -\cos 6\alpha -\sin 3\alpha }&=\displaystyle \frac{(\cos 3\alpha -\cos 9\alpha )-\sin 6\alpha }{(\sin 9\alpha -\sin 3\alpha )-\cos 6\alpha }\\ &=\displaystyle \frac{-2\sin 6\alpha \sin (-3\alpha )-\sin 6\alpha }{2\cos 6\alpha \sin 3\alpha -\cos 6\alpha }\\ &=\displaystyle \frac{2\sin 6\alpha \sin (3\alpha )-\sin 6\alpha }{2\cos 6\alpha \sin 3\alpha -\cos 6\alpha }\qquad \textrm{ingat bahwa}\: \: \sin (-3\alpha )=-\sin 3\alpha \\ &=\displaystyle \frac{\sin 6\alpha (\sin 3\alpha -1)}{\cos 6\alpha (\sin 3\alpha -1)}\\ &=\tan 6\alpha \qquad\qquad \blacksquare \end{aligned} \\ &\vdots \\ &\vdots \\\\ &\textrm{Soal yang belum dibuktikan, silahkan dibuktikan sendiri sebagai latihan} \end{array}.

Sumber Referensi

  1. Setiawan, Teddy. 2009. Trigonometri  123^{+45}  Soal – Jawab dan Pembahasan. Bandung: Yrama Widya.
  2. Wirodikromo, Sartono. 2007. Matematika Jilid 2 IPA untuk Kelas XI. Jakarta: Erlangga.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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