Materi kelas XI, materi kelas XI semester I, trigonometri sma, Uncategorized

Trigonometri

A. Rumus Trigonometri untuk Jumlah dan Selisih Dua Sudut

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

355

\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}

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