Lanjutan Contoh Soal 3

\fbox{21}. Jika diketahui  \sin x=3\cos x  dan   x  terletak di kuadran III, maka nilai dari  \sin x.\cos x = ….

\begin{matrix} a. & \displaystyle \frac{1}{10} & & & c. & \displaystyle \frac{3}{10}\sqrt{10} & & & d. & \displaystyle \frac{1}{10}\sqrt{10}\\ & & & & & & & & & \\ b. & \displaystyle \frac{3}{10} & & & & & & & e. & \sqrt{10} \end{matrix}  .

Jawab:

\begin{aligned}\sin x&=3\cos x\\ \displaystyle \frac{\sin x}{\cos x}&=3\\ \tan x&=3=\frac{3}{1} \end{aligned}.

Perhatikan ilustrasi berikut

218

Sehingga  nilai

\begin{aligned}\sin x.\cos x&=\displaystyle \left ( \frac{y}{r} \right ).\left ( \frac{x}{r} \right )\\ &=\displaystyle \frac{yx}{r^{2}}\\ &=\displaystyle \frac{xy}{x^{2}+y^{2}}\\ &=\displaystyle \frac{3.1}{3^{2}+1^{2}}\\ &=\frac{3}{10} \end{aligned}       .

\fbox{22}. Jika  diketahui  \tan \left ( 3x+60^{0} \right )=\sqrt{3} , maka nilai  x  yang memenuhi persamaan tersebut adalah ….

\begin{matrix} a. & 90^{0} & & & c. & 120^{0} & & & d. & 130^{0}\\ & & & & & & & & & \\ b. & 110^{0} & & & & & & & e. & 230^{0} \end{matrix}   .

Jawab:

\begin{aligned}\tan \left ( 3x+60^{0} \right )&=\sqrt{3}\\ \tan \left ( 3x+60^{0} \right )&=\tan 60^{0}\\ \left ( 3x+60^{0} \right )&=60^{0}+k.180^{0}\\ 3x&=k.180^{0}\\ x&=k.60^{0}\\ k=0\rightarrow x&=0^{0}\\ k=1\rightarrow x&=60^{0}\\ k=2\rightarrow x&=120^{0}\\ k=3\rightarrow x&=180^{0}\\ k=4\rightarrow x&=240^{0}\\ k=5\rightarrow x&=300^{0}\\ k=\cdots \rightarrow x&=\cdots \\ \cdots \cdots \rightarrow &=....dst \end{aligned} .

Jadi nilai  x  yang memenuhi   dari ke 5 opsi  di atas  adalah  120^{0} .

\fbox{23}. Koordinat cartesius untuk titik  \left ( 4,330^{0} \right )  adalah ….

\begin{matrix} a. & \left ( 2\sqrt{3},-2 \right ) & & & c. & \left ( -1,2\sqrt{3} \right ) & & & d. & \left ( -2,2\sqrt{3} \right )\\ & & & & & & & & & \\ b. & \left ( 2\sqrt{3},2 \right ) & & & & & & & e. & \left ( 2,-2\sqrt{3} \right ) \end{matrix}       .

Jawab:

\begin{aligned}\left ( 4,330^{0} \right )&=\left ( x,y \right ),&\textnormal{ingat bahwa}\left\{\begin{matrix} r=4\\ \\ \alpha =330^{0} \end{matrix}\right.\\ &=\left ( r.\cos \alpha ,\: y\sin \alpha \right )\\ &=\left ( 4.\cos 330^{0},\: 4.\sin 330^{0} \right )\\ &=\left ( 4.\cos \left ( 4.90^{0}-30^{0} \right ),\: 4.-\sin \left ( 4.90^{0}-30^{0} \right ) \right )\\ &=\left ( 4.\cos 30^{0},\: 4.-\sin 30^{0} \right )\\ &=\left ( 4.\frac{1}{2}\sqrt{3},\: 4.\left ( -\frac{1}{2} \right ) \right )\\ &=\left ( 2\sqrt{3},-2 \right )\end{aligned}  .

\fbox{24}. Koordinat kutub dari  titik  \left ( -1,-\sqrt{3} \right )  adalah ….

\begin{matrix} a. & \left ( 4,210^{0} \right ) & & & c. & \left ( 6,225^{0} \right ) & & & d. & \left ( 5,240^{0} \right )\\ & & & & & & & & & \\ b. & \left ( 2,240^{0} \right ) & & & & & & & e. & \left ( 2,210^{0} \right ) \end{matrix}   .

Jawab:

Diketahui  sebuah titik yaitu

\left ( -1,-\sqrt{3} \right )\left\{\begin{matrix} x=-1\\ \\ y=-\sqrt{3} \end{matrix}\right.\qquad \Rightarrow \qquad \left ( r,\alpha \right )\\\\ \textrm{Sehingga}\\\\ \begin{aligned}r&=\sqrt{x^{2}+y^{2}}\\ &=\sqrt{\left ( -1 \right )^{2}+\left ( -\sqrt{3} \right )^{2}}\\ &=\sqrt{1+3}\\ &=2 \end{aligned}\quad\quad \textrm{dan}\quad\quad \begin{aligned}\tan \alpha &=\displaystyle \frac{y}{x}\\ &=\displaystyle \frac{-\sqrt{3}}{-1}\\ &=\sqrt{3}\\ &=\tan 60^{0},&\textnormal{bukan termasuk penyelesaian}\\ &=\tan 240^{0},&\textnormal{adalah penyelesaian karena posisinya titik berada di kuadran III}\\\tan \alpha &=\tan 240^{0}\\\alpha &=240^{0} \end{aligned} .

\fbox{25}. Himpunan penyelesaian dari  \tan 2x^{0}=-\displaystyle \frac{1}{3}\sqrt{3}  untuk  0^{0}\leq x^{0}\leq 150^{0}  adalah ….

\begin{matrix} a. & \left \{ 75^{0},165^{0} \right \} & & & c. & \left \{ 45^{0},135^{0} \right \} & & & d. & \left \{ 30^{0},120^{0} \right \}\\ & & & & & & & & & \\ b. & \left \{ 60^{0},150^{0} \right \} & & & & & & & e. & \left \{ 15^{0},105^{0} \right \} \end{matrix}    .

Jawab:

\begin{aligned}\tan 2x^{0}&=-\displaystyle \frac{1}{3}\sqrt{3}\\ \tan 2x^{0}&=\tan \left ( 180^{0}-30^{0} \right ),&\textnormal{nilai tan negatif paling kecil berada di kuadran II}\\ 2x^{0}&=150^{0}+k.180^{0}\\ x^{0}&=75^{0}+k.90^{0}\\ k=0,\Rightarrow x^{0}&=75^{0}\\ k=1,\Rightarrow x^{0}&=75^{0}+90^{0}=165^{0}\\ k=2,\Rightarrow x^{0}&=255^{0},&\textnormal{tidak memenuhi, karena berada di luar batas interval} \end{aligned}\\\\\\ \therefore HP=\left \{ 75^{0},165^{0} \right \}           .

\fbox{26}. Himpunan penyelesaian dari  \sqrt{3}\tan x=-1,\:\: 0^{0}\leq x\leq 360^{0}  adalah ….

\begin{matrix} a. & \left \{ 30^{0},150^{0} \right \} & & & c. & \left \{ 150^{0},330^{0} \right \} & & & d. & \left \{ 120^{0},300^{0} \right \}\\ & & & & & & & & & \\ b. & \left \{150^{0},210^{0} \right \} & & & & & & & e. & \left \{ 60^{0},240^{0} \right \} \end{matrix}   .

Jawab:

\begin{aligned}\sqrt{3}\tan x&=-1\\ \tan x&=-\displaystyle \frac{1}{\sqrt{3}}=-\frac{1}{3}\sqrt{3},&\textnormal{nilai di kuadran II}\\ \tan x&=\tan 150^{0}\\ x&=150^{0}+k.180^{0}\\ k=0,\Rightarrow x&=150^{0}\\ k=1,\Rightarrow x&=330^{0}\\ k=2,\Rightarrow x&=510^{0},\quad \textrm{tidak}\: \textrm{memenuhi}\: (\textrm{tm})\\ k=3,\quad dst&=(\textrm{tm}) \end{aligned}\\\\\\ \therefore HP=\left \{ 150^{0},330^{0} \right \}  .

\fbox{27}. Jika  \cos \left ( x-30^{0} \right )=-\sin 50^{0}\: \: \textrm{dan}\: \: 0^{0}\leq x\leq 360^{0}  , maka himpunan penyelesaiannya adalah ….

\begin{matrix} a. & \left \{ 190^{0} \right \} & & & c. & \left \{ 170^{0},250^{0} \right \} & & & d. & \left \{ 80^{0},280^{0} \right \}\\ & & & & & & & & & \\ b. & \left \{ 140^{0},150^{0} \right \} & & & & & & & e. & \left \{ 20^{0},340^{0} \right \} \end{matrix}   .

Jawab:

\begin{aligned}\cos \left ( x-30^{0} \right )&=-\sin 50^{0}\\ \cos \left ( x-30^{0} \right )&=\cos \left ( 90^{0}+50^{0} \right )\\ \cos \left ( x-30^{0} \right )&=\cos 140^{0}\\ x-30^{0}&=\pm 140^{0}+k.360^{0}\\ x&=30^{0}\pm 140^{0}+k.360^{0}\\ k=0,\Rightarrow x_{1}&=30^{0}+140^{0}=170^{0}\\ \Rightarrow x_{2}&=30^{0}-140^{0}=-110^{0}\: \: (\textrm{tm})\\ k=1,\Rightarrow x_{1}&=30^{0}+140^{0}+360^{0}=530^{0}\: \: (\textrm{tm})\\ \Rightarrow x_{2}&=30^{0}-140^{0}+360^{0}=250^{0} \end{aligned}\\\\\\ \therefore HP=\left \{ 170^{0},250^{0} \right \}   .

\fbox{28}. Pada gambar berikut koordinat titik A dan B masing-masing adalah ….

219

Jawab:

\begin{aligned}y&=y\\ \tan x&=\cos x\\ \displaystyle \frac{\sin x}{\cos x}&=\cos x\\ \sin x&=\cos ^{2}x\\ \sin x&=1-\sin ^{2}x\\ \sin ^{2}x+\sin x-1&=0,&\textnormal{persamaan kuadrat dalam sin, dengan} \left\{\begin{matrix} a=1\\ b=1\\ c=-1 \end{matrix}\right.\\ \sin _{1,2}x&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ \sin _{1,2}x&=\displaystyle \frac{-1\pm \sqrt{1^{2}-4.1.(-1)}}{2.1}\\ &=\displaystyle \frac{-1\pm \sqrt{5}}{2}\\ \sin x&=\displaystyle \frac{-1+\sqrt{5}}{2}\qquad \textrm{atau}\qquad \sin x=\frac{-1-\sqrt{5}}{2}\quad (\textrm{tm}),&\textnormal{karena nilai minimum sin adalah -1}\\ \sin x&=\displaystyle \frac{-1+2,2361}{2},&\textnormal{dengan pendekatan dan dibulatkan sampai 4 desimal}\\ \sin x&=0,61805\\ \sin x&=\sin 38,174^{0},&\textnormal{dengan bantuan kalkulator}\\ x&=38,174^{0}+k.360^{0}\quad \textrm{atau}\\ x&=\left ( 180^{0}-38,174^{0} \right )+k.360^{0}\\ k=0,\Rightarrow x&=38,174^{0}\quad \textrm{atau}\\ x&=141,826^{0}\\ k=1,\Rightarrow x&=\textrm{tidak}\: \: \textrm{perlu} \end{aligned}          .

Sehingga koordinat titik A dan B adalah masing-masing

\textrm{titik}\: A\left ( x,y \right )\: \begin{cases} & x=38,174^{0} \\\\ & y=\cos 38,174^{0} \end{cases}\qquad \textrm{dan}\\\\\\ \textrm{titik}\: B\left ( x,y \right )\: \begin{cases} & x=141,826^{0} \\\\ & y=\cos 141,826^{0} \end{cases} .

\fbox{29}. Tentukanlah besar  \beta  jika diketahui:

\begin{array}{lll}\\ &a.&\sin 2\beta =\cos 3\beta \\ &b.&\tan \left ( \beta +10^{0} \right )=\cot \left ( 3\beta -15^{0} \right ) \end{array}                      .

Jawab:

Poin a dibahas, sedangkan poin b silahkan pembaca coba sendiri sebagai latihan

\begin{aligned}\sin 2\beta &=\cos 3\beta \\ \sin 2\beta &=\sin \left ( 90^{0}-3\beta \right )\\ 2\beta &=90^{0}-3\beta \\ 2\beta +3\beta &=90^{0}\\ \beta&=\displaystyle \frac{90^{0}}{5}\\ \beta &=18^{0} \end{aligned} .

\fbox{30}. Gambarlah grafik fungsi trigonometri berikut untuk  0^{0}\leq x\leq 360^{0}    .

\begin{array}{lll}\\ &a.&y=\sin x^{0}\\ &b.&y=2\sin x^{0}\\ &c.&y=2\sin 2x^{0}\\ &d.&y=2\sin \left ( 2x^{0}-60^{0} \right )\\ &e.&y=2\sin \left ( 2x^{0}-60^{0} \right )+1 \end{array} .

Jawab:

a.  Untuk  y=\sin x^{0}

\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline x&0^{0}&\frac{\pi }{6}=30^{0}&\frac{\pi }{4}=45^{0}&\frac{\pi }{3}=60^{0}&\frac{\pi }{2}=90^{0}&\frac{2\pi }{3}=120^{0}&\frac{3\pi }{4}=135^{0}&\frac{5\pi }{6}=150^{0}&\pi=180^{0} \\\hline y=\sin x^{0}&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 \left ( x,y \right )=\left ( x^{0},\sin x^{0} \right )&\left ( 0,0 \right )&\left ( \frac{\pi }{6},\frac{1}{2} \right )&\left ( \frac{\pi }{4},\frac{1}{2}\sqrt{2} \right )&\left ( \frac{\pi }{3},\frac{1}{2}\sqrt{3} \right )&\left ( \frac{\pi }{2},1 \right )&\left ( \frac{2\pi }{3},\frac{1}{2}\sqrt{3} \right )&\left ( \frac{3\pi }{4},\frac{1}{2}\sqrt{2} \right )&\left ( \frac{5\pi }{6},\frac{1}{2} \right )&\left ( \pi ,0 \right )\\\hline \end{array}\\\\.\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x&\frac{7\pi }{6}=210^{0}&\frac{5\pi }{4}=225^{0}&\frac{4\pi }{3}=240^{0}&\frac{3\pi }{2}=270^{0}&\frac{5\pi }{3}=300^{0}&\frac{7\pi }{4}=315^{0}&\frac{11\pi }{6}=330^{0}&2\pi=360^{0} \\\hline y=\sin 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\\\hline \left ( x,y \right )&\left ( \frac{7\pi }{6},-\frac{1}{2} \right )&\left ( \frac{5\pi }{4},-\frac{1}{2}\sqrt{2} \right )&\left ( \frac{4\pi }{3},-\frac{1}{2}\sqrt{3} \right )&\left ( \frac{3\pi }{2},-1 \right )&\left (\frac{5\pi }{3},-\frac{1}{2}\sqrt{3} \right )&\left ( \frac{7\pi }{4},-\frac{1}{2}\sqrt{2} \right )&\left ( \frac{11\pi }{6},-\frac{1}{2} \right )&\left ( 2\pi ,0 \right )\\\hline \end{array}.

 Dan grafiknya sebagaimana berikut:

225

b. Untuk  y=2\sin x^{0}

\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline x&0^{0}&\frac{\pi }{6}=30^{0}&\frac{\pi }{4}=45^{0}&\frac{\pi }{3}=60^{0}&\frac{\pi }{2}=90^{0}&\frac{2\pi }{3}=120^{0}&\frac{3\pi }{4}=135^{0}&\frac{5\pi }{6}=150^{0}&\pi=180^{0} \\\hline y=\sin x^{0}&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 y=2\sin x^{0}&0&1&\sqrt{2}&\sqrt{3}&2&\sqrt{3}&\sqrt{2}&1&0\\\hline \left ( x,y \right )=\left ( x^{0},2\sin x^{0} \right )&\left ( 0,0 \right )&\left ( \frac{\pi }{6},1 \right )&\left ( \frac{\pi }{4},\sqrt{2} \right )&\left ( \frac{\pi }{3},\sqrt{3} \right )&\left ( \frac{\pi }{2},2 \right )&\left ( \frac{2\pi }{3},\sqrt{3} \right )&\left ( \frac{3\pi }{4},\sqrt{2} \right )&\left ( \frac{5\pi }{6},1 \right )&\left ( \pi ,0 \right )\\\hline \end{array}\\\\.\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x&\frac{7\pi }{6}=210^{0}&\frac{5\pi }{4}=225^{0}&\frac{4\pi }{3}=240^{0}&\frac{3\pi }{2}=270^{0}&\frac{5\pi }{3}=300^{0}&\frac{7\pi }{4}=315^{0}&\frac{11\pi }{6}=330^{0}&2\pi=360^{0} \\\hline y=\sin 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\\\hline y=2\sin x^{0}&-1&-\sqrt{2}&-\sqrt{3}&-2&-\sqrt{3}&-\sqrt{2}&-1&0\\\hline \left ( x,y \right )&\left ( \frac{7\pi }{6},-1 \right )&\left ( \frac{5\pi }{4},-\sqrt{2} \right )&\left ( \frac{4\pi }{3},-\sqrt{3} \right )&\left ( \frac{3\pi }{2},-2 \right )&\left (\frac{5\pi }{3},-\sqrt{3} \right )&\left ( \frac{7\pi }{4},-\sqrt{2} \right )&\left ( \frac{11\pi }{6},-1 \right )&\left ( 2\pi ,0 \right )\\\hline \end{array}.

Gambar grafiknya sebagaimana berikut:

226

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
Pos ini dipublikasikan di Info, Matematika, Pendidikan. Tandai permalink.

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout / Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout / Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout / Ubah )

Foto Google+

You are commenting using your Google+ account. Logout / Ubah )

Connecting to %s