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Trigonometri

A. Tentang Sudut

Sudut adalah pertemuan atau perpotongan 2 buah garis/sinar atau bangun yang dibentuk oleh dua garis yang yang berpotongan di sekitar titik potongnya.

Untuk ukuran sudut, kita mengenal ada beberapa macam, yaitu: derajat, radian, gone/grade.

$\begin{array}{|c|}\hline 1\: keliling\bigcirc =360^{0}=2\pi \: radian=400^{g}\\\hline atau\\\hline \frac{1}{2}\: keliling\bigcirc =180^{0}=\pi \: radian=200^{9}\\\hline \end{array}$ .

Perhatikan ilustrasi berikut

Selanjutnya yang sering dikenalkan adalah sudut dalam ukuran derajat dan radian.

Sebagai catatan:

Ukuran derajat yang diubah ke menit atau detik yang selanjutnya disebut dengan sistem seksagesimal, yaitu:

1 derajat = 60 menit = 3600 detik, atau

$\begin{array}{|c|}\hline 1^{\circ}={60}'={3600}''\\\hline \end{array}$ .

$\fbox{Contoh Soal}$ .

1. $1^{0}=....rad$ .

2. $1\: radian=....^{0}$

3. Jadikanlah sudut $53,24^{0}$ dalam seksagesimal!

4. Jadikanlah sudut $23^{0}{12}'{24}''$ dalam satuan derajat!

Jawab:

1. Perhatikanlah

$\begin{aligned}360^{0}&=2\pi \: rad\\ 360\times 1^{0}&=2\pi \: rad\\ 1^{0}&=\frac{2\pi }{360}\\ &=\frac{\pi }{180}\: rad \end{aligned}$ .

Kaadang dituliskan untuk $\pi \approx \frac{22}{7}\approx 3,14$ , tinggal kita masukkan saja sebagai ganti pi di atas.

2. Dengan cara mirip dengan no.1, yaitu

$\begin{aligned}2\pi \: rad&=360^{0}\\ 2\pi \times 1\: rad&=360^{0}\\ 1\: rad&=\left ( \frac{360}{2\pi } \right )^{0}\\ &=\left ( \frac{180}{\pi } \right )^{0} \end{aligned}$ .

3. Kita ingin menjadikan sudut dari ukuran derajat yang mengandung desimal ke seksagesimal.

Perhatikan langkahnya

$\begin{aligned}53,24^{0}&=53^{0}+0,24^{0}\\ &=53^{0}+0,24\times 1^{0}\\ &=53^{0}+0,24\times {60}'\\ &=53^{0}+{14,4}'\\ &=53^{0}+{14}'+{0,4}'\\ &=53^{0}+{14}'+0,4\times {1}'\\ &=53^{0}+{14}'+0,4\times {60}''\\ &=53^{0}+{14}'+{24}''\\ 53,24^{0}&=53^{0}{14}'{24}'' \end{aligned}$ .

4. Dengan cara yang kurang lebih sama, yaitu

$\begin{aligned}23^{0}{12}'{24}''&=23^{0}+12\times {1}'+24\times {1}''\\ &=23^{0}+12\times \left ( \frac{1}{60} \right )^{0}+24\times \left ( \frac{1}{3600} \right )^{0}\\ &=23^{0}+0,2^{0}+0,00\overline{666}^{0}\\ &=23,20\overline{666}^{0}\end{aligned}$ .

$\fbox{Latihan Soal}$ .

$\begin{array}{cl}\\ 1.&27^{0}=.....rad\\ 2.&28\: rad=....^{0}\\ 3.&29,35^{0}=...^{0}{...}'{...}''\\ 4.&30^{0}{24}'{12}''=....^{0} \end{array}$ .

B. Perbandingan Sudut dalam Segitiga Siku-Siku

Perhatikanlah ilustrasi berikut ini

$\begin{matrix} \sin \alpha =\frac{BC}{AB}\\\\ \cos \alpha =\frac{AC}{AB}\\\\ \tan \alpha =\frac{BC}{AC}\\\\ \csc \alpha =\frac{AB}{BC}\\\\ \sec \alpha =\frac{AB}{AC}\\\\ \cot \alpha =\frac{AC}{BC} \end{matrix}$ .

Untuk Perbandingan Sudut istimewa amati tabel berikut, khususnya sudut $0^{0},30^{0},45^{0},60^{0},90^{0},180^{0}$ .

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\hline \alpha ^{0}&0^{0}&15^{0}&30^{0}&45^{0}&60^{0}&75^{0}&90^{0}&180^{0}&270^{0}&360^{0}\\\hline \sin \alpha ^{0}&0&\frac{1}{4}\left ( \sqrt{6}-\sqrt{2} \right )&\frac{1}{2}&\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{3}&\frac{1}{4}\left ( \sqrt{6}+\sqrt{2} \right )&1&0&-1&0\\\hline \cos \alpha ^{0}&1&\frac{1}{4}\left ( \sqrt{6}+\sqrt{2} \right )&\frac{1}{2}\sqrt{3}&\frac{1}{2}\sqrt{2}&\frac{1}{2}&\frac{1}{4}\left ( \sqrt{6}-\sqrt{2} \right )&0&-1&0&1\\\hline \tan \alpha ^{0}&0&2-\sqrt{3}&\frac{1}{3}\sqrt{3}&1&\sqrt{3}&2+\sqrt{3}&TD&0&TD&0\\\hline \end{array}$ .

Karena pada segitiga siku-siku berlaku teorema Pythagoras, maka ada baiknya kita ingat-ingat tripel Pythagoras di sini yang sering digunakan/dimunculkan .

$\fbox{Contoh Soal}$ .

1. Tentukanlah nilai perbandingan $\sin \alpha ,\: \cos \alpha , \tan \alpha$ untuk segitiga berikut

Jawab:

(a) Untuk sisi miringnya adalah

$\begin{aligned}Sisi\: miring&=\sqrt{3^{2}+4^{2}}\\ &=\sqrt{9+16}\\ &=\sqrt{25}\\ &=5 \end{aligned}$ ,

$\sin \alpha =\frac{3}{5},\quad \cos \alpha =\frac{4}{5}.\quad dan\: \tan \alpha =\frac{3}{4}$ .

(b) Dengan langkah sebagaimana pada langkah (a), kita mendapatkan

$\begin{aligned}Sisi\: miring&=\sqrt{5^{2}+12^{2}}\\ &=\sqrt{25+144}\\ &=\sqrt{169}\\ &=13 \end{aligned}$ ,

$\sin \alpha =\frac{12}{13},\quad \cos \alpha =\frac{5}{13}.\quad dan\: \tan \alpha =\frac{12}{5}$ .

2. Hitunglah nilai dari

$\begin{array}{llll}\\ &&a.&\left ( \tan 30^{0}+\sin 30^{0} \right )\cos 30^{0}\\ &&b.&\left ( \tan 60^{0} \right )^{2}+4\left ( \sin 60^{0} \right )^{2}\\ &&c.&\tan 60^{0}-\sin 60^{0}-\tan 30^{0}\\ &&d.&\displaystyle \frac{1+\sin 30^{0}}{\sin 30^{0}}+\frac{\cos 30^{0}}{1+\sin 30^{0}}\\ &&e.&\displaystyle \frac{2\tan 30^{0}}{1+\tan ^{2}30^{0}} \end{array}$ .

Jawab:

$\begin{array}{llll}\\ &&a.&\left ( \tan 30^{0}+\sin 30^{0} \right )\cos 30^{0}\\ &&&=\displaystyle \left ( \frac{1}{3}\sqrt{3}+\frac{1}{2} \right ).\frac{1}{2}\sqrt{3}\\ &&&=\displaystyle \frac{1}{2}+\frac{1}{4}\sqrt{3} \end{array}$ .

$\begin{array}{llll}\\ &&b.&\left ( \tan 60^{0} \right )^{2}+4\left ( \sin 60^{0} \right )^{2}\\ &&&\displaystyle =\left ( \sqrt{3} \right )^{2}+4\left ( \frac{1}{2}\sqrt{3} \right )^{2}\\ &&&\displaystyle =3+3\\ &&&=6 \end{array}$ .

$\begin{array}{llll}\\ &&c.&\tan 60^{0}-\sin 60^{0}-\tan 30^{0}\\ &&&=\displaystyle \sqrt{3}-\frac{1}{2}\sqrt{3}-\frac{1}{3}\sqrt{3}\\ &&&=\displaystyle \sqrt{3}-\frac{5}{6}\sqrt{3}\\ &&&=\displaystyle \frac{1}{6}\sqrt{3} \end{array}$ .

$\begin{array}{llll}\\ &&d.&\displaystyle \frac{1+\sin 30^{0}}{\sin 30^{0}}+\frac{\cos 30^{0}}{1+\sin 30^{0}}\\ &&&=\displaystyle \frac{1+\frac{1}{2}}{\frac{1}{2}}+\frac{\frac{1}{2}\sqrt{3}}{1+\frac{1}{2}}\\ &&&=\displaystyle 3+\frac{1}{3}\sqrt{3} \end{array}$ .

$\begin{array}{llll}\\ &&e.&\displaystyle \frac{2\tan 30^{0}}{1+\tan ^{2}30^{0}}\\ &&&=\displaystyle \frac{2\times \frac{1}{3}\sqrt{3}}{1+\left ( \frac{1}{3}\sqrt{3} \right )^{2}}\\ &&&=\displaystyle \frac{\frac{2}{3}\sqrt{3}}{1+\frac{1}{3}}\\ &&&=\displaystyle \frac{\frac{2}{3}\sqrt{3}}{\frac{4}{3}}\\ &&&=\displaystyle \frac{1}{2}\sqrt{3} \end{array}$ .

3. Perhatikan ilustrasi berikut

Jika Jarak antara kucing seorang pencatat dan kucing adalah 100 m, maka jarak Pencatat tersebut dengan seorang tentara sebagaimana gambar tersebut di atas adalah?

Jawab:

Perhatikan gambar di atas dengan diberikan tambahan keterangan sebagai berikut

Ditanya berpakah panjang jarak $\left ( x+100 \right )\: meter$ ?

$\begin{aligned}y&=y\\ x.\tan 60^{0}&=\left ( x+100 \right ).\tan 30^{0}\\ x.\sqrt{3}&=\left ( x+100 \right )\frac{1}{3}\sqrt{3}\\ 3x&=x+100\\ 2x&=100\\ x&=50 \end{aligned}$ .

Jadi $x+100=50+100=150$ meter.

4. Tentukanlah perbandingan trigonometri $\angle XOA$ jika $A(3,5)$ .

Jawab:

Perhatikan ilustrasi berikut

Dengan memandang ilustrasi gambar di atas kita mendapatkan $\triangle OAA'$ , dengan menggunakan teorema pythagoras kita mendapatkan

$\begin{aligned}OA^{2}&=\left ( OA' \right )^{2}+\left ( AA' \right )^{2}\\ &=3^{2}+5^{2}\\ &=9+25\\ &=34\\ OA&=\sqrt{34} \end{aligned}$ .

$\begin{array}{llll}\\ &&a.&\displaystyle \sin A'OA=\frac{5}{\sqrt{34}}=\frac{5}{34}\sqrt{34}\\ \\ &&b.&\displaystyle \cos A'OA=\frac{3}{\sqrt{34}}=\frac{3}{34}\sqrt{34}\\ \\ &&c.&\displaystyle \tan A'OA=\frac{5}{3}\\ \\ &&d.&\csc A'OA,\quad \sec A'OA,\quad \cot A'OA\quad silahkan\: \: coba\: \: sendiri\: \: sebagai\: \: latihan \end{array}$ .

C. Koordinat Kartesius dan Koordinat Kutub/Polar

Perhatikanlah ilustrasi berikut

$\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{Koordinat}\\\hline Kartesius\: \rightarrow \: Kutub&Kutub\: \rightarrow \: Kartesius\\\hline P(x,y)\: \rightarrow \: P\left ( r,\alpha ^{0} \right )&P\left ( r,\alpha ^{0} \right )\: \rightarrow \: P(x,y)\\\hline r=\sqrt{x^{2}+y^{2}},\quad \displaystyle \tan \alpha ^{0}=\frac{y}{x},\quad \displaystyle \alpha ^{0}=\arctan \frac{y}{x}&\left\{\begin{matrix} x=r.\cos \alpha ^{0}\\ \\ y=r.\sin \alpha ^{0} \end{matrix}\right.\\\hline \end{array}$ .

D. Perbandingan Trigonometri di Berbagai Kuadran

$\begin{array}{llll}\\ &&(1).&Perbandingan\: Trigonometri\: untuk\: Sudut\: \left ( 90^{0}-\alpha \right )\\ &&&a.\quad \sin \alpha \rightarrow \sin \left ( 90^{0}-\theta \right )=\cos \theta \\ &&&b.\quad \cos \alpha \rightarrow \cos \left ( 90^{0}-\theta \right )=\sin \theta \\ &&&c.\quad \tan \alpha \rightarrow \tan \left ( 90^{0}-\theta \right )=\cot \theta \\ &&&d.\quad \cot \alpha \rightarrow \cot \left ( 90^{0}-\theta \right )=\tan \theta \end{array}$ .

$\begin{array}{llll}\\ &&(2).&Perbandingan\: Trigonometri\: untuk\: Sudut\: \left ( 180^{0}-\alpha \right )\\ &&&a.\quad \sin \alpha \rightarrow \sin \left ( 180^{0}-\theta \right )=\sin \theta \\ &&&b.\quad \cos \alpha \rightarrow \cos \left ( 180^{0}-\theta \right )=-\cos \theta \\ &&&c.\quad \tan \alpha \rightarrow \tan \left ( 180^{0}-\theta \right )=-\tan \theta \\ &&&d.\quad \cot \alpha \rightarrow \cot \left ( 180^{0}-\theta \right )=-\cot \theta \end{array}$ .

$\begin{array}{llll}\\ &&(3).&Perbandingan\: Trigonometri\: untuk\: Sudut\: \left ( 180^{0}+\alpha \right )\\ &&&a.\quad \sin \alpha \rightarrow \sin \left ( 180^{0}+\theta \right )=-\sin \theta \\ &&&b.\quad \cos \alpha \rightarrow \cos \left ( 180^{0}+\theta \right )=-\cos \theta \\ &&&c.\quad \tan \alpha \rightarrow \tan \left ( 180^{0}+\theta \right )=\tan \theta \\ &&&d.\quad \cot \alpha \rightarrow \cot \left ( 180^{0}+\theta \right )=\cot \theta \end{array}$ .

$\begin{array}{llll}\\ &&(4).&Perbandingan\: Trigonometri\: untuk\: Sudut\: \left ( 360^{0}-\alpha \right )\\ &&&a.\quad \sin \alpha \rightarrow \sin \left ( 360^{0}-\theta \right )=-\sin \theta \\ &&&b.\quad \cos \alpha \rightarrow \cos \left ( 360^{0}-\theta \right )=\cos \theta \\ &&&c.\quad \tan \alpha \rightarrow \tan \left ( 360^{0}-\theta \right )=-\tan \theta \\ &&&d.\quad \cot \alpha \rightarrow \cot \left ( 360^{0}-\theta \right )=-\cot \theta \end{array}$ .

Perhatikan ilustrasi berikut

[Sumber]

Untuk sudut $\alpha > 360^{0}$ .

$\begin{array}{llll}\\ &&(5).&Perbandingan\: Trigonometri\: untuk\: Sudutnya\\ &&&a.\quad \sin \alpha \rightarrow \sin \left ( k.360^{0}+\theta \right )=\sin \theta \\ &&&b.\quad \cos \alpha \rightarrow \cos \left ( k.360^{0}+\theta \right )=\cos \theta \\ &&&c.\quad \tan \alpha \rightarrow \tan \left ( k.360^{0}+\theta \right )=\tan \theta \\ \end{array}$ .

Perbandingan trigonometri untuk sudut negatif

$\begin{array}{llll}\\ &&(6).&Perbandingan\: Trigonometri\: untuk\: Sudutnya\\ &&&a.\quad \sin \left ( -\alpha \right ) = -\sin \alpha \\ &&&b.\quad \cos \left ( -\alpha \right ) =\cos \alpha \\ &&&c.\quad \tan \left ( -\alpha \right ) =-\tan \alpha \\ &&&d.\quad \csc \left ( -\alpha \right ) =-\csc \alpha \\ &&&e.\quad \sec \left ( -\alpha \right ) =\sec \alpha \\ &&&f.\quad \cot \left ( -\alpha \right ) =-\cot \alpha \end{array}$ .

$\fbox{\fbox{Contoh Soal}}$ .

1. Tanpa menggunakan tabel/kalkulator tentukanlah nilai $\sin \alpha ,\: \cos \alpha \: dan\: \tan \alpha$ jika diketahui $\alpha =$ .

$\begin{array}{llll}\\ &&a.&120^{0}\qquad b.\: 240^{0}\qquad c.\: 300^{0}\qquad d.\: 1125^{0} \end{array}$ .

Jawab:

$\begin{array}{llll}\\ &&(a).&Perbandingan\: Trigonometri\: untuk\: Sudutnya\: 120^{0}\\ &&&1.\quad \sin 120^{0} = \sin \left ( 180^{0}-60^{0} \right )=\sin 60^{0}=\displaystyle \frac{1}{2}\sqrt{3} \\ &&&2.\quad \cos 120^{0} =\cos \left ( 180^{0}-60^{0} \right )=-\cos 60^{0}=-\displaystyle \frac{1}{2} \\ &&&3.\quad \tan 120^{0} =\tan \left ( 180^{0}-60^{0} \right )=-\tan 60^{0}=-\displaystyle \sqrt{3} \\ \end{array}$ .

$\begin{array}{llll}\\ &&(b).&Perbandingan\: Trigonometri\: untuk\: Sudutnya\: 240^{0}\\ &&&1.\quad \sin 240^{0} = \sin \left ( 180^{0}+60^{0} \right )=-\sin 60^{0}=-\displaystyle \frac{1}{2}\sqrt{3} \\ &&&2.\quad \cos 240^{0} =\cos \left ( 180^{0}+60^{0} \right )=-\cos 60^{0}=-\displaystyle \frac{1}{2} \\ &&&3.\quad \tan 240^{0} =\tan \left ( 180^{0}+60^{0} \right )=\tan 60^{0}=\displaystyle \sqrt{3} \\ \end{array}$ .

$\begin{array}{llll}\\ &&(c).&Perbandingan\: Trigonometri\: untuk\: Sudutnya\: 300^{0}\\ &&&1.\quad \sin 300^{0} = \sin \left ( 360^{0}-60^{0} \right )=-\sin 60^{0}=-\displaystyle \frac{1}{2}\sqrt{3} \\ &&&2.\quad \cos 300^{0} =\cos \left ( 360^{0}-60^{0} \right )=\cos 60^{0}=\displaystyle \frac{1}{2} \\ &&&3.\quad \tan 300^{0} =\tan \left ( 360^{0}-60^{0} \right )=-\tan 60^{0}=-\displaystyle \sqrt{3} \\ \end{array}$ .

$\begin{array}{llll}\\ &&(d).&Perbandingan\: Trigonometri\: untuk\: Sudutnya\: 1125^{0}\\ &&&1.\quad \sin 1125^{0} = \sin \left ( 3\times 360^{0}+45^{0} \right )=\sin 45^{0}=\displaystyle \frac{1}{2}\sqrt{2} \\ &&&2.\quad \cos 1125^{0} =\cos \left ( 3\times 360^{0}+45^{0} \right )=\cos 45^{0}=\displaystyle \frac{1}{2}\sqrt{2} \\ &&&3.\quad \tan 1125^{0} =\tan \left ( 3\times 360^{0}+45^{0} \right )=\tan 45^{0}=1 \\ \end{array}$ .

2. Tunjukkan bahwa

$\begin{array}{llll}\\ &&a.&\displaystyle \frac{\sec \left ( 180^{0}-\alpha \right )}{\tan \left ( 180^{0}+\alpha \right )}=-\csc \alpha \\\\ &&b.&\displaystyle \frac{\sin \left ( 90^{0}-\alpha \right )}{\sec \left ( 180^{0}-\alpha \right )}=-\cos ^{2}\alpha \\\\ &&c.&\displaystyle \frac{\cot 99^{0}}{\cos 198^{0}}\times \frac{\cos 378^{0}}{\cos 81^{0}}=\sec 9^{0}\\\\ &&d.&\tan 71^{0}+\tan 289^{0}+\tan 161^{0}+\tan 199^{0}=0 \end{array}$ .

Jawab:

$\begin{array}{llll}\\ &&a.&\displaystyle \frac{\sec \left ( 180^{0}-\alpha \right )}{\tan \left ( 180^{0}+\alpha \right )} \\ &&&=\displaystyle \frac{-\sec \alpha }{\tan \alpha }=\frac{-\frac{1}{\cos \alpha }}{\frac{\sin \alpha }{\cos \alpha }}=-\frac{1}{\sin \alpha }=-\csc \alpha \end{array}$ .

$\begin{array}{llll}\\ &&b.&\displaystyle \frac{\sin \left ( 90^{0}-\alpha \right )}{\sec \left ( 180^{0}-\alpha \right )} \\ &&&=\displaystyle \frac{\cos \alpha }{-\sec \alpha }=\frac{\cos \alpha }{-\frac{1}{\cos \alpha }}=-\cos ^{2}\alpha \\ \end{array}$ .

$\begin{array}{llll}\\ &&c.&\displaystyle \frac{\cot 99^{0}}{\cos 198^{0}}\times \frac{\cos 378^{0}}{\cos 81^{0}}\\ &&&=\displaystyle \frac{-\cot 81^{0}}{-\cos 18^{0}}\times \frac{\cos \left ( 360^{0}+18^{0} \right ) }{\cos 81^{0}}= \frac{\frac{\cos 81^{0}}{\sin 81^{0}}}{\cos 81^{0}}=\frac{1}{\sin 81^{0}}\\ &&&=\displaystyle \frac{1}{\sin \left ( 90^{0}-9^{0} \right )}=\frac{1}{\cos 9^{0}}=\sec 9^{0} \end{array}$ .

$\begin{array}{llll}\\ &&d.&\tan 71^{0}+\tan 289^{0}+\tan 161^{0}+\tan 199^{0}\\ &&&=\tan 71^{0}+\tan \left ( 360^{0}-71^{0} \right )+\tan \left ( 180^{0}-19^{0} \right )+\tan \left ( 180^{0}+19^{0} \right )\\ &&&=\tan 71^{0}-\tan 71^{0}-\tan 19^{0}+\tan 19^{0}\\ &&&=0 \end{array}$ .

3. Diketahui koordinat kutub titik M adalah $M\left ( 8,60^{0} \right )$ , maka koordinat kartesiusnya adalah….

Jawab:

$Diketahui\quad M\left ( 8,60^{0} \right )\: \rightarrow \: M\left ( x,y \right ).....?\\\\ M\left ( 8,60^{0} \right )\left\{\begin{matrix} r=8\\ \\ \alpha ^{0}=60^{0} \end{matrix}\right.\\\\ \begin{aligned}M\left ( 8,60^{0} \right )&\Rightarrow \\ x&=r.\cos \alpha ^{0} &=8.\cos 60^{0} &=8.\left ( \frac{1}{2} \right ) &=4\\ y&=r.\sin \alpha ^{0} &=8.\sin 60^{0} &=8.\frac{1}{2}\sqrt{3} &=4\sqrt{3} \end{aligned}\\\\ Jadi\quad M\left ( 8,60^{0} \right )\: \rightarrow \: M\left ( 4,4\sqrt{3} \right )$ .

E. Identitas Trigonometri

$\begin{array}{llll}\\ &&1.&\sin ^{2}\alpha +\cos ^{2}\alpha =1\\\\ &&2.&\sec ^{2}\alpha -\tan ^{2}\alpha =1\\\\ &&3.&\csc ^{2}\alpha -\cot ^{2}\alpha =1\\\\ &&4.&\displaystyle \tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{1}{\cot \alpha }\\\\ &&5.&\displaystyle \csc \alpha =\frac{1}{\sin \alpha }\\\\ &&6.&\displaystyle \sec \alpha =\frac{1}{\cos \alpha } \end{array}$ .

F. Fungsi Trigonometri

Perhatikanlah ilustrasi gambar berikut untuk grafik fungsi sinus dan cosinus

[Sumber]

Untuk fungsi tangen,

[sumber]

Fungsi Sinus , f(x)= sin x

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline x&0&\frac{\pi }{6}&\frac{\pi }{4}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{3\pi }{4}&\frac{5\pi }{6}&\pi &\frac{7\pi }{6}&\frac{5\pi }{4}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{7\pi }{4}&\frac{11\pi }{6}&2\pi \\\hline f(x)&0&\frac{1}{2}&\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{3}&1&\frac{1}{2}\sqrt{3}&\frac{1}{2}\sqrt{2}&\frac{1}{2}&0&-\frac{1}{2}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{3}&-1&-\frac{1}{2}\sqrt{3}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}&0\\\hline \end{array}$ .

Fungsi Cosinus , f(x)=cos x

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline x&0&\frac{\pi }{6}&\frac{\pi }{4}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{3\pi }{4}&\frac{5\pi }{6}&\pi &\frac{7\pi }{6}&\frac{5\pi }{4}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{7\pi }{4}&\frac{11\pi }{6}&2\pi \\\hline f(x)&1&\frac{1}{2}\sqrt{3}&\frac{1}{2}\sqrt{2}&\frac{1}{2}&0&-\frac{1}{2}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{3}&-1&-\frac{1}{2}\sqrt{3}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}&0&\frac{1}{2}&\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{3}&1\\\hline \end{array}$ .

Fungsi Tangen , f(x)=tan x

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline x&0&\frac{\pi }{6}&\frac{\pi }{4}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{3\pi }{4}&\frac{5\pi }{6}&\pi &\frac{7\pi }{6}&\frac{5\pi }{4}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{7\pi }{4}&\frac{11\pi }{6}&2\pi \\\hline f(x)&0&\frac{1}{3}\sqrt{3}&1&\sqrt{3}&\infty &-\sqrt{3}&-1&-\frac{1}{3}\sqrt{3}&0&\frac{1}{3}\sqrt{3}&1&\sqrt{3}&\infty &-\sqrt{3}&-1&-\frac{1}{3}\sqrt{3}&0\\\hline \end{array}$ .

Untuk :

$f(x)=\left\{\begin{matrix} a\: \sin k\left ( x\pm \theta \right )\pm c\\\\ a\: \cos k\left ( x\pm \theta \right )\pm c\\\\ a\: \tan k\left ( x\pm \theta \right )\pm c \end{matrix}\right.$ .

$Periode\quad fungsi=\left\{\begin{matrix} \sin \: atau\: \cos=\displaystyle \frac{360^{0}}{\left | k \right |}\\ \\ \tan \: atau\: \cot =\displaystyle \frac{180^{0}}{\left | k \right |} \end{matrix}\right.$ .

$nilai\quad fungsi\quad \sin\: atau\: \cos=\left\{\begin{matrix} f(x)_{max}=\left | a \right |\pm c\\ \\ f(x)_{min}=-\left | a \right |\pm c \end{matrix}\right.$ .

$Amplitudo=\frac{1}{2}\left ( f(x)_{mak}-f(x)_{min} \right )$ .

G. Persamaan Trigonometri Sederhana

$\begin{array}{lllll}\\ &&1.&&Jika\quad \sin x^{0}=\sin \alpha ,\quad maka\\ &&&&(i)\quad x^{0}=\alpha +k.360^{0}\\ &&&&(ii)\quad x^{0}=\left ( 180^{0}-\alpha \right )+k.360^{0}\\\\ &&2.&&Jika\quad \cos x^{0}=\cos \alpha ,\quad maka\\ &&&&x^{0}=\pm \alpha +k.360^{0}\\\\ &&3.&&Jika\quad \tan x^{0}=\tan \alpha ,\quad maka\\ &&&&x^{0}=\alpha +k.180^{0} \end{array}$ .

$\fbox{\LARGE\fbox{Contoh Soal}}$ .

1. Buktikan bahwa $\displaystyle \frac{\cos \alpha }{1+\sin \alpha }=\frac{1-\sin \alpha }{\cos \alpha }$ .

Bukti:

$\begin{aligned}\frac{\cos \alpha }{1+\sin \alpha }&=\frac{\cos \alpha }{1+\sin \alpha }\times \frac{1-\sin \alpha }{1-\sin \alpha }\\ &=\frac{\cos \alpha \times \left ( 1-\sin \alpha \right )}{1-\sin ^{2}\alpha }\\ &=\frac{\cos \alpha \times \left ( 1-\sin \alpha \right )}{\cos ^{2}\alpha }\\ &=\frac{1-\sin \alpha }{\cos \alpha }\quad (\mathbf{Terbukti}) \end{aligned}$ .

2. Tentukanlah himpunan penyelesaian dari persamaan $\displaystyle \cos \left ( 3x-45^{0} \right )=-\frac{1}{2}\sqrt{2},\quad untuk\quad 0^{0}\leq x\leq 360^{0}$ .

Jawab:

$\begin{aligned}\cos \left ( 3x-45^{0} \right )&=-\frac{1}{2}\sqrt{2}\\ \cos \left ( 3x-45^{0} \right )&=\cos 135^{0}\\ 3x-45^{0}&=\pm 135^{0}+k.360^{0}\\ 3x&=45^{0}\pm 135^{0}+k.360^{0}\\ x&=\displaystyle \frac{45^{0}\pm 135^{0}+k.360^{0}}{3}\\ x&=15^{0}\pm 45^{0}+k.120^{0}\\ x&=60^{0}+k.120^{0}\qquad atau\qquad x=-30^{0}+k.120^{0}\\ k=0,\qquad \rightarrow x&=60^{0}\qquad\qquad\qquad atau\qquad x=90^{0}\\ k=1,\qquad \rightarrow x&=180^{0}\qquad\qquad\qquad atau\qquad x=210^{0}\\ k=2,\qquad \rightarrow x&=300^{0}\qquad\qquad\qquad atau\qquad x=330^{0} \end{aligned}$ .

Jadi Himpunan penyelesaiannya adalah = $\left \{ 60^{0}, 90^{0}, 180^{0}, 210^{0}, 300^{0},330^{0} \right \}$ .

3. Lukislah grafik fungsi $\displaystyle f(x)=2\left | \sin x \right |+1,\qquad untuk\quad 0\leq x\leq 2\pi .$ .

Jawab:

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline x&0&\frac{\pi }{6}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{5\pi }{6}&\pi &\frac{7\pi }{6}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{11\pi }{6}&2\pi \\\hline \sin x&0&\frac{1}{2}&\frac{1}{2}\sqrt{3}&1&\frac{1}{2}\sqrt{3}&\frac{1}{2}&0&-\frac{1}{2}&-\frac{1}{2}\sqrt{3}&-1&-\frac{1}{2}\sqrt{3}&-\frac{1}{2}&0\\\hline \left | \sin x \right |&0&\frac{1}{2}&\frac{1}{2}\sqrt{3}&1&\frac{1}{2}\sqrt{3}&\frac{1}{2}&0&\frac{1}{2}&\frac{1}{2}\sqrt{3}&1&\frac{1}{2}\sqrt{3}&\frac{1}{2}&0\\\hline 2\left | \sin x \right |&0&1&\sqrt{3}&2&\sqrt{3}&1&0&1&\sqrt{3}&2&\sqrt{3}&1&0\\\hline 2\left | \sin x \right |+1&1&2&1+\sqrt{3}&3&1+\sqrt{3}&2&1&2&1+\sqrt{3}&3&1+\sqrt{3}&2&1\\\hline \end{array}$ .

Untuk gambar silahkan pembaca melukiskannya sendiri sebagai latihan.

H. Aturan Sinus, Kosinus, dan Luas Segitiga

1. Aturan Sinus

$\begin{array}{|c|}\hline \displaystyle \frac{a}{\sin A}= \frac{b}{\sin B}=\frac{c}{\sin C}=2R\\\hline \end{array}$ .

2. Aturan Kosinus

$\begin{aligned}\cos A&=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ \cos B&=\frac{a^{2}+c^{2}-b^{2}}{2ac}\\ \cos C&=\frac{a^{2}+b^{2}-c^{2}}{2ab} \end{aligned}$ .

3. Luas Segitiga

$\begin{aligned}Luas\: \triangle \: ABC&=\frac{1}{2}.bc.\sin A\\ &=\frac{1}{2}.ac.\sin B\\ &=\frac{1}{2}.ab.\sin C \end{aligned}$ .

atau

$\begin{aligned}Luas\: \triangle \: ABC&=\frac{a^{2}.\sin B.\sin C}{2\sin A}\\ &=\frac{b^{2}.\sin A.\sin C}{2\sin B}\\ &=\frac{c^{2}.\sin A.\sin B}{2\sin C} \end{aligned}$ .

atau

$\begin{aligned}Luas\: \triangle \: ABC&=\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}\qquad dengan\qquad s=\frac{1}{2}\left ( a+b+c \right ) \end{aligned}$ .

$\begin{tabular}{|p{6.0cm}cp{6.0cm}|}\hline &\textbf{Contoh Soal}&\\\hline \end{tabular}$ .

1. Diketahui $\triangle ABC$ dengan panjang sisi AC=10 cm dan BC=16 cm serta luas $\triangle ABC=40\: cm^{2}$ , maka besar $\angle ACB$ jika sudutnya lancip adalah …

Jawab:

Diketahui $\left\{\begin{matrix} AC=10\: cm\\ BC=16\: cm\\ L_{\triangle }=40\: cm^{2} \end{matrix}\right.$ , maka

$\begin{aligned}L_{\triangle ABC}&=\frac{1}{2}.AC.BC.\sin \angle ACB\\ 40&=\frac{1}{2}.10.16.\sin \angle ACB\\ 40&=80.\sin \angle ACB\\ \frac{40}{80}&=\sin \angle ACB\\ \sin \angle ACB&=\frac{1}{2}\\ \sin \angle ACB&=\sin 30^{0}\\ \angle ACB&=30^{0} \end{aligned}$ .

2. Perhatikanlah gambar berikut

Jika $AB+3=BC+2=CD+1=AD=4\: cm$ , maka $\cos \angle BAD$ adalah ….

Jawab:

Perhatikan kembali ilustrasi berikut

Langkah awal kita gunakan garis bantu BD untuk nantinya kita mendapatkan nilai cos dari sudut A, yaitu

$\begin{aligned}BD^{2}&=BA^{2}+DA^{2}-2.BA.DA.\cos \angle A\\ &=1^{2}+4^{2}-2.1.4.\cos \angle A\\ &=17-8\cos \angle A\\ BD^{2}&=BC^{2}+DC^{2}-2.BC.DC.\cos \angle C\\ &=2^{2}+3^{2}-2.2.3.\cos \angle C\\ &=13-12\cos \angle C \end{aligned}\\\\ Perlu\quad diketahui\quad bahwa \angle A+\angle C=\angle B+\angle C=180^{0},\qquad karena\quad ABCD\quad segiempat\quad talibusur\\\\ Sehingga,\\\\ \angle C=180^{0}-\angle A$ .

Selanjutnya,

$\begin{aligned}BD^{2}&=BD^{2}\\ 17-8\cos \angle A&=13-12\cos \angle C\\ 12\cos \angle C-8\cos \angle A&=13-17\\ 12\left ( \cos \left ( 180^{0}-\angle A \right ) \right )-8\cos \angle A&=-4\\ 12\left ( -\cos \angle A \right )-8\cos \angle A&=-4\\ -12\cos \angle A-8\cos \angle A&=-4\\ -20\cos \angle A&=-4\\ \cos \angle A&=\frac{-4}{-20}\\ \cos \angle A&=\frac{1}{5} \end{aligned}$ .

3. Carilah luas $\triangle ABC$ jika diketahui AB=10 cm, AC=14 cm dan BC=16 cm.

Jawab:

$Diketahui\:\: bahwa\:\: L_{\triangle ABC}=\sqrt{s(s-a)(s-b)(s-c)},\quad dengan s=\frac{1}{2}\left ( a+b+c \right )\quad \left\{\begin{matrix} AB=c=10\: cm\\ AC=b=14\: cm\\ BC=a=16\: cm \end{matrix}\right.\\\\ s=\frac{1}{2}\left ( 10+14+16 \right )=\frac{1}{2}.40=20\\\\ \begin{aligned}L_{\triangle ABC}&=\sqrt{20.\left ( 20-10 \right ).\left ( 20-14 \right ).\left ( 20-10 \right )}.\\ &=\sqrt{20(10)(6)(4)}\\ &=\sqrt{4800}\\ &=40\sqrt{3}\quad cm^{2} \end{aligned}$ .

$\fbox{\LARGE \fbox{Latihan Soal}}$ .

1. Perhatikanlah gambar berikut

Tentukanlah nilai $\cos \angle RSP$ .

2. Perhatikanlah ilustrasi gambar berikut

Carilah besar sudut dan panjang sisi yang belum diketahui dari segitiga di atas, kemudian cari pula luasnya?

3. (EBTANAS 2001) Diketahui luas segitiga ABC adalah $\left ( 3+2\sqrt{3} \right )\: cm^{2}$ . Jika panjang sisi $AB=\left ( 6+4\sqrt{3} \right )\: cm$ dan BC=7 cm, maka nilai $\sin \angle ABC$ = ….

4. Diketahui seorang penerjun hendak mendarat secara vertikal sebagaimana ilustrasi berikut

Carilah harga x ?

5. Jika pada jajargenjang ABCD, dua diagonal panjangnya masing-masing 12 cm dan 16 cm dan sudut apitnya adalah $60^{0}$ , maka luas jajargenjang tersebut adalah ….

6. Diketahui $y=1+\cos 2x,\quad dengan\quad 0\leq x\leq 2\pi$ , carilah

$\begin{tabular}{lcp{8.0cm}}\\ &a.&nilai maksimum dan minimum fungsi,\\ &b.&amplitudo, dan\\ &c.&gambarlah sketsa grafiknya \end{tabular}$ .

7. Buktikanlah bahwa

$\begin{array}{llll}\\ &&a.&\sin ^{2}x-\sin ^{2}x.\cos ^{2}x=\sin ^{4}x\\ &&b.&\tan x.\sin x+\cos x=\sec x\\ &&c.&\sin ^{4}x-\cos ^{4}x=\sin ^{2}x-\cos ^{2}x\\ &&d.&\left ( \cos x+\sin x \right )\left ( \cos x-\sin x \right )=1-2\sin ^{2}x\\ &&e.&\left ( 1+\tan ^{2}x \right )\left ( 1-\cos ^{2}x \right )=\tan ^{2}x\\ &&f.&\sqrt{1+2\sin x.\cos x}=\sin x+\cos x\\ &&g.&\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{1}{\cos x.\sin x}\\ &&h.&\displaystyle \frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x}=\cos x+\sin x\\ &&i.&\displaystyle \left ( \frac{\tan x-1}{\tan x+1} \right )\left ( \frac{\sin x+\cos x}{\sin x-\cos x} \right )=1 \end{array}$ .

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Rabu, 23 Agustus 2017

Trigonometri

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