PAS MATEMATIKA BLOG

Contoh Soal 4 Transformasi Geometri

$\begin{array}{ll}\\ 16.&\textrm{Bayangan titik A(2,4) dicerminkan }\\ &\textrm{terhadap garis}\: \: y-x=0\: \: \textrm{dilanjutkan}\\ &\textrm{ke garis}\: \: x\sqrt{3}-3y=0\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}A'\left ( 2+\sqrt{3},-1+2\sqrt{3} \right )&&\\ \textrm{b}.\quad A'\left ( 2+\sqrt{3},1-2\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left ( 1-\sqrt{3},-2+\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( -2+\sqrt{3},1+2\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( 2-\sqrt{3},1-2\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ &\begin{cases} x\sqrt{3}-3y=0 & \Leftrightarrow y=\displaystyle \frac{1}{3}\sqrt{3}x\\ &\Leftrightarrow y=\tan 30^{\circ}.x\\\\ x-y=0 & \Leftrightarrow y=x \end{cases}\\\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.30^{\circ} & \sin 2.30^{\circ} \\ \sin 2.30^{\circ} & -\cos 2.30^{\circ} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2} & \displaystyle \frac{1}{2}\sqrt{3}\\ \displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ 4 \end{pmatrix}\\ &=\begin{pmatrix} \sqrt{3}+2\\ -1+2\sqrt{3} \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Jika}\: \: T_{1}=\begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}\: \: \textrm{dan}\: \: T_{2}=\begin{pmatrix} -2 & 5\\ -1 & 3 \end{pmatrix}\\ &\textrm{maka bayangan garis}\: \: x+y+1=0\\ &\textrm{oleh}\: \: T_{2}\circ T_{1}\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}x-2y-1=0&&\\ \textrm{b}.\quad x+2y-1=0&\\ \textrm{c}.\quad x+2y+1=0&\\ \textrm{d}.\quad x-2y+1=0\\ \textrm{e}.\quad x+y-1=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=T_{2}\circ T_{1}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -2 & 5\\ -1 & 3 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 1 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -2+5 & -4+5\\ -1+3 & -2+3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 3 & 1\\ 2 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 3x+y\\ 2x+y \end{pmatrix}\\ \textrm{Dipe}&\textrm{roleh}\\ &\begin{array}{lllllllll}\\ \quad x'&=3x+y\\ \quad y'&=2x+y\qquad\quad-\\\hline x'-y'&=x\\ \Leftrightarrow \quad x&=\color{red}x'-y'\qquad\color{black}....(1)\\ \color{blue}\textrm{maka}\\ \qquad y&=x'-3x\\ &=x'-3(x'-y')\\ &=\color{red}3y'-2x'\quad \color{black}....(2) \end{array}\\ \textrm{Sehin}&\textrm{gga}\\ x+y&+1=0\\ x'-&y'+3y'-2x'+1=0\\ -x'+&2y'+1=0\\ x'-&2y'-1=0\\ \textrm{maka}&\: \textrm{bayangan garisnya}\\ x-2&y-1=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Garis}\: \: 2x+y+4=0\: \: \textrm{ditranslasikan}\\ &\textrm{oleh}\: \: \begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{dilanjutkan transformasi} \\ &\textrm{oleh}\: \: \begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\: \: \textrm{persamaan bayangannya}\\ &\textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2x+y+3=0&&\\ \textrm{b}.\quad \color{red}2x-3y+3=0&\\ \textrm{c}.\quad 2x+3y+3=0&\\ \textrm{d}.\quad 3x+2y+3=0\\ \textrm{e}.\quad 3x-2y+3=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}\: &=\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} -2\\ 5 \end{pmatrix}=\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x-2\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x-2+2y+10\\ y+5 \end{pmatrix}\\ &=\begin{pmatrix} x+2y+8\\ y+5 \end{pmatrix}\\ \textrm{Diper}&\textrm{oleh}\\ &\begin{array}{lllllllll}\\ \quad x''&=x+2y+8\\ \: \: \: 2y''&=2y+10\qquad\quad-\\\hline x''-2y''&=x-2\\ \Leftrightarrow \quad x&=\color{red}x''-2y''+2\: \color{black}....(1)\\ \color{blue}\textrm{maka}\\ \qquad y&=\color{red}y''-5\quad \color{black}\qquad....(2) \end{array}\\ \textrm{sehin}&\textrm{gga}\\ 2x+&y+4=0\\ 2(x''&-2y''+2)+(y''-5)+4=0\\ 2x''-&3y''+3=0\\ \textrm{maka}&\: \textrm{bayangan garisnya}\\ 2x-&3y+3=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Diketahui}\: \: M\: \: \textrm{adalah pencerminan terhadap}\\ &\textrm{garis}\: \: y=-x\: \: \textrm{dan}\: \: T\: \: \textrm{adalah transformasi} \\ &\textrm{yang dinyatakan oleh matriks}\: \: \begin{pmatrix} 2 & 3\\ 0 & -1 \end{pmatrix}\\ &\textrm{Koordinat bayangan titik}\: \: A(2,-8)\: \: \textrm{oleh}\\ &\textrm{transformasi}\: \: M\: \: \textrm{dilanjutkan oleh}\: \: T\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-10,2)&&\\ \textrm{b}.\quad (-2,-10)&\\ \textrm{c}.\quad \color{red}(10,2)&\\ \textrm{d}.\quad (-10,-2)\\ \textrm{e}.\quad (2,10) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix} &=\color{purple}T\circ M\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 2 & 3\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} 0-3 & -2+0\\ 0+1 & 0+0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} -3 & -2\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2\\ -8 \end{pmatrix}\\ &=\begin{pmatrix} -6+16\\ 2+0 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 10\\ 2 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: W\: \: \textrm{adalah transformasi oleh}\\ &\textrm{matriks}\: \: \begin{pmatrix} 1 & 0\\ 3 & 1 \end{pmatrix},\: \: \textrm{maka titik mula}\\ &\textrm{dari}\: \: W'(-2,5)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-11,-2)&&\\ \textrm{b}.\quad (11,-2)&\\ \textrm{c}.\quad \color{red}(-2,11)&\\ \textrm{d}.\quad (2,11)\\ \textrm{e}.\quad (12,11) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Dimi}&\textrm{salkan}:\\ A&=\begin{pmatrix} -2\\ 5 \end{pmatrix},\: \: \textrm{dan}\\ W&=\begin{pmatrix} 1 & 0\\ 3 & 1 \end{pmatrix},\: \: \textrm{serta}\: \: X=\begin{pmatrix} x\\ y \end{pmatrix}\\ \textrm{mak}&\textrm{a}\\ &\begin{array}{|c|}\hline \color{red}\begin{aligned}A&=BX\\ B^{-1}A&=B^{-1}BX\\ B^{-1}A&=I.X\\ B^{-1}A&=X\\ X&=B^{-1}A \end{aligned}\\\hline \end{array}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\displaystyle \frac{1}{\begin{vmatrix} 1 &0 \\ 3 & 1 \end{vmatrix}}\begin{pmatrix} 1 & 0\\ -3 & 1 \end{pmatrix}\begin{pmatrix} -2\\ 5 \end{pmatrix}\\ &=1.\begin{pmatrix} -2+0\\ 6+5 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} -2\\ 11 \end{pmatrix} \end{aligned} \end{array}$

Contoh Soal 3 Transformasi Geometri

$\begin{array}{ll}\\ 11.&\textrm{Titik A(1,-2) dirotasikan sejauh}\: \: 15^{\circ}\\ & \textrm{kemudian dilanjutkan}\: \: 75^{\circ}\: \: \textrm{dengan pusat }\\ &O(0,0)\: \: \textrm{maka bayangan akhir titik A adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-2,1)&&\textrm{d}.\quad \color{red}(2,1)\\ \textrm{b}.\quad (-1,2)&\textrm{c}.\quad (1,2)&\textrm{e}.\quad (-2,-1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos (\theta _{1}+\theta _{2}) & -\sin (\theta _{1}+\theta _{2})\\ \sin (\theta _{1}+\theta _{2}) & \cos (\theta _{1}+\theta _{2}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos (75^{\circ}+15^{\circ})& -\sin (75^{\circ}+15^{\circ})\\ \sin (75^{\circ}+15^{\circ}) & \cos (75^{\circ}+15^{\circ}) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 90^{\circ}&-\sin 90^{\circ}\\ \sin 90^{\circ}&\cos 90^{\circ} \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Jika garis}\: \: 3x-2y+5=0\: \: \textrm{dicerminkan }\\ &\textrm{terhadap garis}\: \: y=-x\: \: \textrm{kemudian}\\ &\textrm{didilatasikan dengan pusat (1,-2) }\\ &\textrm{dengan faktor skala 2, maka persamaan}\\ & \textrm{bayangannya adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x-2y-10=0&&\\ \textrm{b}.\quad x+2y-10=0&\\ \textrm{c}.\quad x-6y+5=0&\\ \textrm{d}.\quad x+2y-12=0\\ \textrm{e}.\quad \color{red}2x-3y+18=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Proses}&\: \textrm{untuk refleksinya}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&-1\\ -1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ -x \end{pmatrix}\\ \textrm{proses}&\: \textrm{dilatasinya}\\ \begin{pmatrix} x''\\ y'' \end{pmatrix}&=\begin{pmatrix} 2&0\\ 0&2 \end{pmatrix}\begin{pmatrix} x'-1\\ y'+2 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-2\\ 2y'+4 \end{pmatrix}+\begin{pmatrix} 1\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2x'-1\\ 2y'+2 \end{pmatrix}\\ &=\begin{pmatrix} 2(-y)-1\\ 2(-x)+2 \end{pmatrix}\\ &\begin{cases} x &=-\displaystyle \frac{1}{2}(y''-2) \\ y &=-\displaystyle \frac{1}{2}(x''+1) \end{cases} \end{aligned}\\ &\begin{aligned}\textrm{Sehingga persam}&\textrm{aan bayangan}\\ \textrm{garisnya adalah}:&\\ 3x&-2y+5=0\\ 3\left ( -\displaystyle \frac{1}{2}(y''-2) \right )&-2\left ( -\displaystyle \frac{1}{2}(x''+1) \right )+5=0\\ -\displaystyle \frac{3}{2}y''+3 &+(x''+1)+5=0\\ 2x&-3y+6+2+10=0\\ 2x&-3y+18=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Titik A(4,-4) dicerminkan terhadap}\\ &\textrm{garis}\: \: y=x\tan 15^{\circ}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: A'(a,b)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}&&\textrm{d}.\quad \color{red}4\sqrt{3}\\ \textrm{b}.\quad 2\sqrt{3}&\textrm{c}.\quad 3\sqrt{3}&\textrm{e}.\quad 6\sqrt{3} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\begin{pmatrix} a\\ b \end{pmatrix}&=\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} \cos 2.15^{\circ}& \sin 2.15^{\circ}\\ \sin 2.15^{\circ} & -\cos 2.15^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \cos 30^{\circ}&\sin 30^{\circ}\\ \sin 30^{\circ}&-\cos 30^{\circ} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} \displaystyle \frac{1}{2}\sqrt{3} & \displaystyle \frac{1}{2}\\ \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} 4\\ -4 \end{pmatrix}\\ &=\begin{pmatrix} 2\sqrt{3}-2\\ 2+2\sqrt{3} \end{pmatrix}\\ &\begin{cases} a &=2\sqrt{3}-2 \\ b &=2+2\sqrt{3} \end{cases}\\ \textrm{mak}&\textrm{a nilai dari}\\ a+b&=\left ( 2\sqrt{3}-2+2+2\sqrt{3} \right )\\ &=4\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Lingkaran}\: \: x^{2}+y^{2}-5x+8y+7=0\\ & \textrm{ditranslasikan oleh}\: \: T=\begin{pmatrix} m\\ n \end{pmatrix}\: \: \textrm{menghasilkan}\\ &\textrm{bayangan}\: \: x^{2}+y^{2}-9x+2y+6=0.\\ & \textrm{Nilai}\: \: m+n=\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2&&\textrm{d}.\quad \color{red}5\\ \textrm{b}.\quad 3&\qquad\textrm{c}.\quad 4\qquad&\textrm{e}.\quad 6 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui sebuah lingkaran dengan persamaan}:\\ & \color{blue}x^{2}+y^{2}-5x+8y+7=0\\ \textrm{kar}&\textrm{ena akibat translasi, maka}\\ &\begin{cases} x & =x'-m \\ y & =y'-n \end{cases}\\ &x^{2}+y^{2}-5x+8y+7=0\\ \textrm{seh}&\textrm{ingga}\\ &\Leftrightarrow \color{purple}(x'-m)^{2}+(y'-n)^{2}-5(x'-m)+8(y'-n)+7=0\\ &\Leftrightarrow \color{purple}x'^{2}+y'^{2}-2mx'-2ny'+m^{2}+n^{2}-5x'+5m+8y'-8n+7=0\\ &\Leftrightarrow \color{purple}x'^{2}+y'^{2}-(2m+5)x'+(8-2n)y'+m^{2}+n^{2}+5m-8n+7=0\\ &\qquad \equiv \: \color{purple}x'^{2}+y'^{2}-9x'+2y'+6=0\qquad (\color{black}\textbf{akhir bayangan})\\ &\begin{cases} 9 &=2m+5 \Rightarrow m=2\\ 2 & =8-2n \: \Rightarrow \, \: n=3 \end{cases}\\ \textrm{Jad}&\textrm{i , nilai}\: \: m+n=2+3=5\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika titik A(-2,1) dicerminkan terhadap garis}\\ & y=-\displaystyle \frac{1}{3}x\sqrt{3}\: ,\: \textrm{maka bayangan dari}\\ &\textrm{titik \textit{A} tersebut adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&&\\ \textrm{b}.\quad \color{red}A'\left ( -1-\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right )&\\ \textrm{c}.\quad A'\left (-1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )&\\ \textrm{d}.\quad A'\left ( 1-\displaystyle \frac{1}{2}\sqrt{3},\displaystyle \frac{1}{2}-\sqrt{3} \right )\\ \textrm{e}.\quad A'\left ( -1+\displaystyle \frac{1}{2}\sqrt{3},-\displaystyle \frac{1}{2}+\sqrt{3} \right ) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ y&=-\displaystyle \frac{1}{3}x\sqrt{3}=\left ( -\displaystyle \frac{1}{3}\sqrt{3} \right )x\\ &=\left (-\tan 30^{\circ} \right )x=\tan \left ( 180^{\circ}-30^{\circ} \right )x\\ &=\tan 150^{\circ}.x\\ \textrm{maka}\: \: \theta &=150^{\circ}\quad \Rightarrow \quad 2\theta =300^{\circ}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\color{purple}\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\color{purple}\begin{pmatrix} \cos 300^{\circ} & \sin 300^{\circ} \\ \sin 300^{\circ} & -\cos 300^{\circ} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\color{purple}\begin{pmatrix} \displaystyle \frac{1}{2} & -\displaystyle \frac{1}{2}\sqrt{3}\\ -\displaystyle \frac{1}{2}\sqrt{3} & -\displaystyle \frac{1}{2} \end{pmatrix}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=\begin{pmatrix} -1-\displaystyle \frac{1}{2}\sqrt{3}\\ \sqrt{3}-\displaystyle \frac{1}{2} \end{pmatrix} \end{aligned} \end{array}$

Contoh Soal 2 Transformasi Geometri

$\begin{array}{ll}\\ 6.&\textrm{Bayangan untuk titik P(2,5) oleh rotasi }\\ &\textrm{dengan pusat}\: \textit{A}(1,3)\: \: \textrm{sejauh}\: \: 180^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (1,0)&&\textrm{d}.\quad (2,0)\\ \textrm{b}.\quad \color{red}(0,1)&\textrm{c}.\quad (0,2)&\textrm{e}.\quad (1,2) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Karena rotasi de}&\textrm{ngan pusat A sebesar}\: \: 180^{\circ},\\ \textrm{maka}\qquad\qquad\: \: \: \: &\\ R\left ( A(1,3),180^{\circ} \right )&=\begin{pmatrix} \cos 180^{\circ} & -\sin 180^{\circ}\\ \sin 180^{\circ} & \cos 180^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\\ \textrm{sehingga bayang}&\textrm{an titik P(2,5)-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\color{blue}\begin{pmatrix} x-a\\ y-b \end{pmatrix}\color{black}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 2-1\\ 5-3 \end{pmatrix}+\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -1\\ -2 \end{pmatrix}+\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 0\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Bayangan kurva}\: \: xy=6\: \: \textrm{oleh rotasi sebesar}\\ & \displaystyle \frac{\pi }{2}\: \: \textrm{dengan pusat}\: \: O(0,0)\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}xy=-6&&\textrm{d}.\quad x(y-x)=6\\ \textrm{b}.\quad xy=6&&\textrm{e}.\quad x(x+y)=-6\\ \textrm{c}.\quad x(x-y)=6&\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\\ \displaystyle \frac{\pi }{2}=90^{\circ},\: \: \textrm{maka}&\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga bayan}&\textrm{gan semua titik yang }\\ \textrm{terletak pada k}& \textrm{urva adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =-x' \end{cases}\\ \textrm{Selanjunya}\: \: \: \textrm{unt}&\textrm{uk bayangan kurvanya }\\ \textrm{adalah}:\qquad\quad&\\ xy&=6\\ y'.(-x')&=6\\ x'y'&=-6\\ \textrm{Jadi , persamaa}&\textrm{n kurva bayangannya}\\ \textrm{adalah}\: &\: \color{red}xy=-6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Sebuah lingkaran yang berpusat di (3,4) }\\ &\textrm{dan menyinggung sumbu-X dicerminkan}\\ &\textrm{terhadap garis}\: \: y=x\: \textrm{, maka persamaan }\\ &\textrm{akhir lingkaran yang terjadi adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}x^{2}+y^{2}-8x-6y+9=0&&\\ \textrm{b}.\quad x^{2}+y^{2}+8x+6y+9=0&\\ \textrm{c}.\quad x^{2}+y^{2}+6x+8y+9=0&\\ \textrm{d}.\quad x^{2}+y^{2}-8x-6y+16=0\\ \textrm{e}.\quad x^{2}+y^{2}+8x+6y+16=0\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Refleksi l}&\textrm{ingkaran yang berpusat di (3,4) }\\ \textrm{dan men}&\textrm{yinggung sumbu-X, }\\ \textrm{dengan}\: \: r&=(y)=4,\\ \textrm{maka}\: \textrm{per}&\textrm{samaan lingkarannya adalah}:\\ (x-3)^{2}+&(y-4)^{2}=4^{2}.\: \textrm{Karena}\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} y\\ x \end{pmatrix}\\ &\begin{cases} x & =y' \\ y & =x' \end{cases}\\ \textrm{selanjutn}&\textrm{ya untuk persamaan bayangan }\\ \textrm{lingkaran} &\textrm{nya adalah}:\\ &(y'-3)^{2}+(x'-4)^{2}=4^{2},\\ & \textbf{menjadi}\\ &(y-3)^{2}+(x-4)^{2}=4^{2},\quad \textrm{atau}:\\ &\color{red}x^{2}+y^{2}-8x-6y+9=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: M_{x}\: \: \textrm{adalah pencerminan terhadap sumbu-X }\\ &\textrm{dan}\: \: M_{y=x}\: \: \textrm{adalah pencerminan terhadap garis}\\ & y=x\: ,\: \textrm{maka matriks transformasi tunggal }\\ &\textrm{yang mewakili}\: \: M_{x}\circ M_{y=x}=\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}\\ \textrm{b}.\quad \color{red}\begin{pmatrix} 0 &1 \\ -1 & 0 \end{pmatrix}&&\textrm{e}.\quad \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\\ \textrm{c}.\quad \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}\: &\textrm{bahwa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ M_{y=x} & =\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \end{cases}\\ M_{x}\circ M_{y=x}&=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ &=\begin{pmatrix} 0+0 & 1+0\\ 0-1 & 0+0 \end{pmatrix}\\ &=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Diketahui vektor}\: \: \vec{x}\: \: \textrm{dirotasikan terhadap titik asal}\\ & O\: \: \textrm{sebesar}\: \: \theta >0\: \: \textrm{searah jarum jam}.\\ &\textrm{Kemudian hasilnya dicerminkan terhadap garis}\: \: y=0\\ & \textrm{menghasilkan vektor}\: \: \vec{y}.\\ &\textrm{Jika}\: \: \vec{y}=A.\vec{x}\: ,\: \textrm{maka matriks}\: \: A-\textrm{nya adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}&&\\ \textrm{b}.\quad \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}&&\\ \textrm{c}.\quad \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}&&\\ \textrm{d}.\quad \color{red}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\\ \textrm{e}.\quad \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui bah}\textrm{wa}:\\ &\begin{cases} M_{x} & = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ R_{-\theta } & =\begin{pmatrix} \cos (-\theta ) & -\sin (-\theta )\\ \sin (-\theta ) & \cos (-\theta ) \end{pmatrix}=\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{cases}\\ &A=M_{x}\circ R_{-\theta }=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \end{aligned} \end{array}$.

Contoh Soal 1 Transformasi Geometri

$\begin{array}{ll}\\ 1.&\textrm{Suatu translasi yang memetakan titik P(9,8) }\\ &\textrm{ke titik}\: \: \textrm{P}'(14,-2)\: \: \textrm{adalah}\: ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\begin{pmatrix} 5\\ -10 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 6\\ 6 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 5\\ 6 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} 23\\ -10 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} 5\\ 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\textrm{T}+\begin{pmatrix} x\\ y \end{pmatrix}\\ \textrm{T}&=\begin{pmatrix} x'-x\\ y'-y \end{pmatrix}\\ &=\begin{pmatrix} 14-9\\ -2-8 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ -10 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ &\textrm{Maka bayangan titik M}(2,-1)\: \: \textrm{adalah}\, ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (7,10)&&\textrm{d}.\quad (1,10)\\ \textrm{b}.\quad (10,7)&\textrm{c}.\quad \color{red}(1,4)&\textrm{e}.\quad (4,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} x' & =2x+3y \\ y' & =3x+2y \end{cases}\\ \left.\begin{matrix} x=2\\ y=-1 \end{matrix}\right\}&\Rightarrow \begin{cases} x' & =2(2)+3(-1)=4-3=1 \\ y' & =3(2)+2(-1)=6-2=4 \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Bayangan untuk titik A(1,3) oleh rotasi }\\ &\textrm{dengan pusat}\: \: \textit{O}(0,0)\textrm{sejauh}\: \: 90^{\circ}\: \: \textrm{adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-1,3)&&\textrm{d}.\quad (1,-3)\\ \textrm{b}.\quad (-1,-3)&\textrm{c}.\quad \color{red}(-3,1)&\textrm{e}.\quad (3,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Karena rotasi d}&\textrm{engan pusat O sebesar}\: \: 90^{\circ},\\ \textrm{maka}\qquad\qquad\: \: &\\ R\left ( O(0,0),90^{\circ} \right )&=\begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ}\\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\\ \textrm{sehingga}\quad\qquad&\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -3\\ 1 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Suatu lingkaran dengan jari-jari 4 }\\ &\textrm{dengan pusat di O(0,0) dtranslasikan}\\ &\textrm{oleh}\: \: \textrm{T}=\begin{pmatrix} 2\\ -3 \end{pmatrix},\: \textrm{maka luas }\\ &\textrm{bayangan lingkaran tersebut adalah}\\ & ....\: \textrm{satuan luas}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \pi &&\textrm{d}.\quad 8\pi \\ \textrm{b}.\quad 2\pi &\textrm{c}.\quad 4\pi &\textrm{e}.\quad \color{red}16\pi \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran berpusat}\\ &\textrm{di O dengan}\: \: r=4.\: \textrm{Karena translasi adalah}\\ &\textrm{termasuk transformasi isometri(kongruen)}\\ &\textrm{maka jari-jari lingkaran bayangannya }\\ &\textrm{akan sama dengan bendanya. Sehingga}\\ &\textrm{ luas bayangan lingkarannya}\\ &=\pi r^{2}=\pi \times 4^{2}=\color{red}16\pi \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Sebuah transformasi yang didefiniskan oleh}\\ & \begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\textrm{Yang merupakan titik invarian (tidak berubah) }\\ &\textrm{adalah}\: ...\\ &\begin{array}{lll}\\ \textrm{a}.\quad (0,0)&&\textrm{d}.\quad (0,-1)\\ \textrm{b}.\quad \color{red}(1,-1)&\textrm{c}.\quad (1,0)&\textrm{e}.\quad (1,1) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{D}&\textrm{iketahui bahwa}:\\ &\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases}\\ &\begin{array}{|c|c|c|c|}\hline \textrm{NO}&\textrm{Titik}&\begin{aligned}&\textrm{Disubstitusikan ke}\\ & \color{blue}\begin{cases} x' & =4-3x \\ y' & =2x-y-4 \end{cases} \end{aligned}&\begin{aligned}&\textrm{Keterangan}\\ &\quad\textrm{Titik} \end{aligned}\\\hline \textrm{a}.&(0,0)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(0)-4=-4 \end{cases}&\textrm{Varian}\\\hline \textrm{b}&(1,-1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(-1)-4=-1 \end{cases}&\color{red}\textbf{Invarian}\\\hline \textrm{c}&(1,0)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(0)-4=-2 \end{cases}&\textrm{Varian}\\\hline \textrm{d}&(0,-1)&\begin{cases} x' & =4-3(0)=4 \\ y' & =2(0)-(-1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \textrm{e}&(1,1)&\begin{cases} x' & =4-3(1)=1 \\ y' & =2(1)-(1)-4=-3 \end{cases}&\textrm{Varian}\\\hline \end{array} \end{aligned} \end{array}$

Transformasi Geometri (XI Matematika Wajib)

A. Pengertian

Transformasi Geometri adalah suatu perubahan objek geometri atau suatu pemetaan dari suatu titik-titik ke himpunan titik-titik yang lain pada bidang kartesius.

Dari pengertian di atas jelas bahwa aturan transformasi sebagaimana fungsi atau pemetaan dan transformasi ini selanjutnya dapat disimbolkan dengan sebuah huruf kapital, misal M, T, R, dan lain sebagainya. Sebagai misal titik P(x,y) oleh transformasi T menghasilkan titik baru yaitu P'(x',y') dan operasi ini dapat dituliskan dengan:

$\begin{aligned}&P(x,y)\overset{T}{\rightarrow}P'(x',y') \end{aligned}$.

B. Matriks Transformasi

Misalkan suatu transfomasi T memetakan sebuah titik A(x,y) ke A'(x',y')

selanjutnya perhatikan ilustrasi berikut:

$\boxed{\begin{aligned}A(x,y)&\xrightarrow[.]{\color{red}Transformasi\, =\: T}A'(x',y')=A'\left ( ax+by,cx+dy \right )\\\\ \Rightarrow &\begin{pmatrix} x'\\ y' \end{pmatrix}=\underset{\underset{transformasi}{Matriks}}{\underbrace{\color{blue}\begin{pmatrix} a & b\\ c & d \end{pmatrix}}}\begin{pmatrix} x\\ y \end{pmatrix} \end{aligned}}$.

C. Jenis-Jenis Transformasi dengan matriks yang sesuaian

1. Translasi (Geseran)

$\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Translasi}&(x,y)\xrightarrow[.]{\begin{pmatrix} a\\ b \end{pmatrix}}(x+a,y+b)&\begin{pmatrix} a\\ b \end{pmatrix}\\\hline \end{array}$.

2. Rotasi (Perputaran)

$\begin{aligned}&\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Rotasi}&&\\\hline \begin{aligned}&\textrm{Pusat rotasi}\\ & \left [ O,\alpha \right ] \end{aligned}&\begin{aligned}&\begin{cases} x' =... \\ y' = ... \end{cases}\\ &\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{biru} \end{aligned} \end{aligned}&\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\\\hline \begin{aligned}&\textrm{Pusat}\\ & (a,b)\: \textrm{sudut}\: \alpha \end{aligned}&\begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=&\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{merah} \end{aligned}\\\hline \end{array}\\ &\color{blue}\begin{cases} x' =x\cos \alpha -y\sin \alpha \\ y' = x\sin \alpha +y\cos \alpha \end{cases}\\ &\color{red}\triangleright \triangleright \triangleright \triangleright \begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}.\begin{pmatrix} x-a\\ y-b \end{pmatrix} \end{aligned}$.

3. Refleksi (Pencerminan)

$\begin{array}{|l|c|c|}\hline \textrm{Refleksi}&&\\\hline \textrm{terhadap sumbu}-\textrm{X}&(x,y)\rightarrow (x,-y)&\begin{pmatrix} 1 &0 \\ 0 & -1 \end{pmatrix}\\\hline \textrm{terhadap sumbu}-\textrm{Y}&(x,y)\rightarrow (-x,y)&\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\\\hline \textrm{terhadap garis y = x}&(x,y)\rightarrow (y,x)&\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = -x}&(x,y)\rightarrow (-y,-x)&\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis x = h}&(x,y)\rightarrow (2h-x,y)&\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} 2h\\ 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = x}&(x,y)\rightarrow (y,x)&\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = -x}&(x,y)\rightarrow (-y,-x)&\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\\\hline \textrm{terhadap garis x = h}&(x,y)\rightarrow (2h-x,y)&\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} 2h\\ 0 \end{pmatrix}\\\hline \textrm{terhadap garis y = k}&(x,y)\rightarrow (x,2k-y)&\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}+\begin{pmatrix} 0\\ 2k \end{pmatrix}\\\hline \textrm{pusat}\: (0,0)\begin{cases} y=mx \\ m=\tan \alpha \end{cases}&&\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}\\\hline \end{array}$.

4. Dilatasi (Perkalian)

$\begin{aligned}&\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Jenis}\\ &\textrm{Transformasi} \end{aligned}&\textrm{Rumus}&\textrm{Matriks}\\\hline \textrm{Dilatasi}&&\\\hline \textrm{Pusat}\: \left [ O,k \right ]&(x,y)\rightarrow (kx,ky)&\begin{pmatrix} k & 0\\ 0 & k \end{pmatrix}\\\hline \begin{aligned}&\textrm{Pusat}\: (a,b)\\ & \textrm{faktor skala}\: k \end{aligned}&\begin{pmatrix} x'-a\\ y'-b \end{pmatrix}&\begin{aligned}&\colorbox{yellow}{Lihat}\\ &\colorbox{yellow}{di bawah}\\ &\colorbox{yellow}{tulisan}\\ &\colorbox{yellow}{warna}\\ &\colorbox{yellow}{merah} \end{aligned}\\\hline \begin{aligned}&\textrm{Luas bangun}\\ &\textrm{ datar} \end{aligned}&\textrm{Misal bangun A}&\textrm{T}=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\\\hline &\textbf{Bangun A}'&= \textrm{det T}\times \textrm{A}\\\hline \end{array}\\ &\color{red}\triangleright \triangleright \triangleright \triangleright \triangleright \triangleright \begin{pmatrix} x'-a\\ y'-b \end{pmatrix}=\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}.\begin{pmatrix} x-a\\ y-b \end{pmatrix} \end{aligned}$.

Catatan:

Translasi, refleksi, dan rotasi suatu objek adalah bagian dari transformasi yang hanya mengubah posisi objek saja, sehingga jenis transformasi-transformasi ini juga disebut dengan transformasi isometri

D. Bayangan Kurva dan Komposisi Transformasi

$\begin{array}{|l|l|}\hline \qquad \color{red}\textrm{Bayangan Kurva}\quad y=f(x)&\qquad\qquad\qquad \color{blue}\textrm{Komposisi Transformasi}\\\hline \begin{aligned}\textrm{Lan}&\textrm{gkah-langkah}:\\ 1.\quad&\textrm{Tentukan bayangan titiknya}\\ &(x,y)\rightarrow \left ( x',y' \right )\\ 2.\quad&\textrm{Salanjutnya tentukan}\: \: x\: \: \textrm{dan}\: \: y\:\\ &\textrm{dalam}\: \: x'\: \: \textrm{dan}\: \: y'\\ 3.\quad&\textrm{Substitusikan}\: \: x\: \: \textrm{dan}\: \: y\\ &\textrm{ke}\: \: \: y=f(x) \end{aligned}&\begin{aligned}\textrm{Lan}&\textrm{gkah-langkah}:\\ 1.\quad&\textrm{Selesaikan sesuai urutan transformasi}\\ &(x,y)\xrightarrow[\qquad.]{T_{1}}(x',y')\xrightarrow[\qquad.]{T_{2}}(x'',y'')\\ 2.\quad&\textrm{Jika dapat disederhanakan kedua transformasi}\\ &\textrm{tersebut di atas, maka cukup dengan}\\ &(x,y)\xrightarrow[\qquad.]{T_{2}\circ T_{1}}(x'',y'') \end{aligned}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah bayangan dari segitiga PQR dengan}\\\ & P(0,4),\: Q(-1,1),\: \textrm{dan}\: \: R(3,6).\\ &\textrm{oleh translasi}\: \: \: T=\begin{pmatrix} 5\\ -2 \end{pmatrix}\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{cases} \begin{pmatrix} x_{P}^{'}\\ y_{P}^{'} \end{pmatrix} &=T+\begin{pmatrix} x_{P}\\ y_{P} \end{pmatrix}=\begin{pmatrix} 5\\ -2 \end{pmatrix}+\begin{pmatrix} 0\\ 4 \end{pmatrix}=\begin{pmatrix} 5+0\\ -2+4 \end{pmatrix}=\begin{pmatrix} 5\\ 2 \end{pmatrix} \\ \begin{pmatrix} x_{Q}^{'}\\ y_{Q}^{'} \end{pmatrix} & =\cdots\qquad \textrm{isilah sendiri} \\ \begin{pmatrix} x_{R}^{'}\\ y_{R}^{'} \end{pmatrix} &= \cdots\qquad \textrm{isilah sendiri} \end{cases} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah bayangan dari garis}\: \: y=2x+4\\ & \textrm{oleh translasi}\: \: T=\begin{pmatrix} -1\\ 2 \end{pmatrix}.\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{array}{|c|c|}\hline \textbf{Bayangan Titik-titik}&\textbf{Bayangan Garis}\\\hline \begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=T+\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -1\\ 2 \end{pmatrix}+\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -1+x\\ 2+y \end{pmatrix}\\ &\begin{cases} x' & =-1+x\Leftrightarrow x=x'+1 \\ y' & =2+y\quad\Leftrightarrow y=y'-2 \end{cases} \end{aligned}&\begin{aligned}y&=2x+4\\ y'-2&=2(x'+1)+4\\ y'&=2x+2+4+2\\ &=2x+8\\ \textrm{Jadi}\, ,&\: \textbf{bayangan garisnya}\\ \textrm{adala}&\textrm{h}:\\ y&=2x+8\\ & \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah bayangan titik A(4,6) oleh rotasi yang berpusat }\\ &\textrm{di titik P(3,-2) dengan sudut putar sebesar}\: \: 90^{\circ} \\\\ &\color{blue}\textbf{Jawab}\\ &\begin{aligned}\textrm{Untuk Ro}&\textrm{tasi yang berpusat di}\: \: (a,b)\: \: \textrm{dengan sudut}\: \: \alpha \: \: \textrm{adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} x-a\\ y-b \end{pmatrix}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} \cos 90^{\circ} & -\sin 90^{\circ}\\ \sin 90^{\circ} & \cos 90^{\circ} \end{pmatrix}\begin{pmatrix} 4-3\\ 6-(-2) \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1\\ 8 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} -8\\ 1 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} -5\\ -1 \end{pmatrix}\\ \textrm{Jadi}\, ,\: &\textrm{bayangan titik A adalah}\: \: \textrm{A}'(-5,-1) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukanlah bayangan titik A(4,6) }\\ &\textrm{oleh dilatasi yang berpusat di titik P(3,-2)}\\ &\textrm{dengan faktor skala}\: \: k=2 \\\\ &\color{blue}\textbf{Jawab}\\ &\begin{aligned}\textrm{Bayangan}&\: \textrm{titik A-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} k & 0\\ 0 & k \end{pmatrix}\begin{pmatrix} x-a\\ y-b \end{pmatrix}+\begin{pmatrix} a\\ b \end{pmatrix}\\ &=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}\begin{pmatrix} 4-3\\ 6-(-2) \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1\\ 8 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 2\\ 16 \end{pmatrix}+\begin{pmatrix} 3\\ -2 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ 14 \end{pmatrix}\\ \textrm{Jadi}\: ,\: &\textrm{bayangan titik A-nya adalah}\: \: \textrm{A}'(5,14) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tentukanlah bayangan titik A(4,6) }\\ &\textrm{oleh translasi}\: \: t\: \: \textrm{dilanjutkan}\: \: s\: \: \textrm{dengan}\\ &\textrm{matriks transformasi berturut-turut }\\ &\textrm{adalah}\: \: T=\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}\: \: \textrm{dan}\: \: S= \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{aligned}\textrm{Bayangan}&\: \textrm{titik A-nya adalah}:\\ \begin{pmatrix} x'\\ y' \end{pmatrix}&=S\times T\times \begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}\begin{pmatrix} 4\\ 6 \end{pmatrix}\\ &=\begin{pmatrix} 2 & 3\\ 1 & 2 \end{pmatrix}\begin{pmatrix} 4\\ 6 \end{pmatrix}\\ &=\begin{pmatrix} 26\\ 16 \end{pmatrix}\\ \textrm{Jadi}\: ,\: &\textrm{bayangan titik A-nya adalah}\: \: \textrm{A}'(26,16) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Suatu kurva}\: \: y=\, ^{3}\log (2x-2)\: \: \textrm{memiliki bayangan}\\ & y=\, ^{3}\log \left ( \displaystyle \frac{2x+3}{3} \right )\: \: \textrm{oleh translasi}\\ & T=\begin{pmatrix} a\\ b \end{pmatrix}.\: \textrm{Tentukanlah nilai}\: \: a+b\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\\ y&=\, ^{3}\log (2x-2)\quad \Leftrightarrow\quad 3^{y}=2x-2\: (\textrm{benda})\\ y&=\, ^{3}\log \left ( \displaystyle \frac{2x+3}{3} \right )\\ & \Leftrightarrow\quad 3^{y}=\left ( \displaystyle \frac{2x+3}{3} \right )\quad (\color{red}\textbf{bayangan})\\ \textrm{sehingga}&\: \textrm{untuk bayangan}\\ 3^{y'-b}&=2(x'-a)-2\quad \Leftrightarrow \quad 3^{y'}.3^{-b}=2(x'-a)-2\\ & \Leftrightarrow\quad 3^{y'}=\displaystyle \frac{2(x'-a)-2}{3^{-b}}=\displaystyle \frac{2x'+3}{3}\\ \textrm{Jadi}\, ,\: &\begin{cases} a &=\displaystyle \frac{5}{2} \\ b &=-1 \end{cases}\\ \textrm{Sehingga}&\: a+b=\displaystyle \frac{5}{2}+(-1)=\displaystyle \frac{3}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tentukanlah bayangan garis}\: \: ax+by+c=0\: \: \textrm{oleh transformasi}\\ &\textrm{yang bersesuaian dengan matriks}\: \: \: \begin{pmatrix} 1&-2\\ 3&-4 \end{pmatrix}\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{array}{|c|c|}\hline \textbf{Proses Awal}&\textbf{Penentuan Bayangan}\\\hline \begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ 3 & -4 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ \begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 1 & -2\\ 3 & -4 \end{pmatrix}^{-1}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\displaystyle \frac{1}{\begin{vmatrix} 1 & -2\\ 3 & -4 \end{vmatrix}}\begin{pmatrix} -4 & 2\\ -3 & 1 \end{pmatrix}\begin{pmatrix} x'\\ y' \end{pmatrix}\\ &=\displaystyle \frac{1}{-4+6}\begin{pmatrix} -4x'+2y'\\ -3x'+y' \end{pmatrix}\\ &=\displaystyle \frac{1}{2}\begin{pmatrix} -4x'+2y'\\ -3x'+y' \end{pmatrix}\\ &\begin{cases} x &=-2x'+y' \\ y &=-\displaystyle \frac{3}{2}x'+\displaystyle \frac{1}{2}y' \end{cases} \end{aligned}&\begin{aligned}ax+by+c&=0\\ a\left ( -2x'+y' \right )+b\left ( -\displaystyle \frac{3}{2}x'+\frac{1}{2}y' \right )+c&=0\\ -2ax'-\displaystyle \frac{3}{2}bx'+ay'+\displaystyle \frac{1}{2}by'+c&=0\\ (-4a-3b)x'+(2a+b)y'+2c&=0\\ &\\ \textbf{Jadi, bayangan garisnya adalah}:&\\ &\\ \color{red}(-4a-3b)x+(2a+b)y+2c&=0\\ &\\ &\\ &\\ & \end{aligned} \\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Diketahui kurva}\: \: y=4x^{2}-9\: \: \textrm{dicerminkan terhadap sumbu-X kemudian}\\ &\textrm{ditranslasikan dengan}\: \: \begin{pmatrix} -1\\ 2 \end{pmatrix}.\: \textrm{Ordinat titik potong terhadap sumbu-Y adalah}....\\\\ &\color{blue}\textbf{Jawab}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\begin{pmatrix} x'\\ y' \end{pmatrix}&=\begin{pmatrix} -1\\ 2 \end{pmatrix}+\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}\\ &=\begin{pmatrix} -1\\ 2 \end{pmatrix}+\begin{pmatrix} x\\ -y \end{pmatrix}\\ &=\begin{pmatrix} -1+x\\ 2-y \end{pmatrix}\\ &\begin{cases} x &= x'-1\\ y &= 2-y' \end{cases} \end{aligned}&\begin{aligned}y&=4x^{2}-9\\ (2-y')&=4(x'-1)^{2}-9\\ -y'&=4(x'^{2}-2x'+1)-9-2\\ -y'&=4x'^{2}-8x'+4-11\\ y'&=-4x'^{2}+8x'+7\\ &\\ \textbf{Maka}\, ,&\, \textbf{persamaan kurva bayangannya}:\\ y&=\color{red}-4x^{2}+8x+7 \end{aligned} \\\hline \end{array}\\ &\begin{aligned}\textrm{Sehingga}&\: \textrm{ordinat dari titik potong terhadap sumbu-Y-nya adalah}:\\ y&=-4x^{2}+8x+7,\qquad \textbf{atau}\\ f(x)&=-4x^{2}+8x+7\\ f(0)&=-4(0)^{2}+8(0)+7\qquad\quad \textrm{saat}\: \: x=0\: (\textrm{karena memotong sumbu-Y})\\ &=7\\ \textrm{Jadi}&\: \textrm{ordinatnya adalah}\: \: y=f(0)=\color{red}7 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Ketujuh

$\begin{array}{ll}\\ 31.&\textrm{Nilai dari}\\ &\quad\quad \cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}-\displaystyle \frac{1}{8} &&&\textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle -\frac{1}{4}&\quad \textrm{c}.&0\quad &\textrm{e}.&\displaystyle \frac{1}{3} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textbf{Alternatif 1}\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\times \displaystyle \frac{2\sin \displaystyle \frac{2\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\left (\sin \displaystyle \frac{4\pi }{7}-\sin 0 \right )\frac{\displaystyle \cos \frac{\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{4\pi }{7}\displaystyle \cos \frac{\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\left ( \sin \displaystyle \frac{5\pi }{7}+\sin \displaystyle \frac{3\pi }{7} \right )\cos \displaystyle \frac{4\pi }{7}}{4\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{5\pi }{7}\cos \displaystyle \frac{4\pi }{7}+\sin \displaystyle \frac{3\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{4\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{9\pi }{7}+\sin \displaystyle \frac{\pi }{7}+\sin \displaystyle \frac{7\pi }{7}+\sin \left (-\displaystyle \frac{\pi }{7} \right )}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{-\sin \displaystyle \frac{2\pi }{7}+\sin \displaystyle \frac{\pi }{7}+0-\sin \displaystyle \frac{\pi }{7}}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{-\sin \displaystyle \frac{2\pi }{7}}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=-\displaystyle \frac{1}{8} \end{aligned}\\ &\textbf{Alternatif 2}\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\\ &=\cos \displaystyle \frac{4\pi }{7}\cos \frac{2\pi }{7}\cos \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( \cos \displaystyle \frac{6\pi }{7}+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( \cos \left ( \pi -\displaystyle \frac{\pi }{7} \right )+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( -\cos \displaystyle \frac{\pi }{7}+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &= \displaystyle \frac{1}{2}\left (-\cos ^{2}\displaystyle \frac{\pi }{7}+\cos \displaystyle \frac{2\pi }{7}\cos \displaystyle \frac{\pi }{7} \right )\\ &=\displaystyle \frac{1}{4}\left ( -\cos \displaystyle \frac{2\pi }{7}-\cos 0+\cos \displaystyle \frac{3\pi }{7}+\cos \displaystyle \frac{\pi }{7} \right )\\ &=\displaystyle \frac{1}{4}\left ( -\cos 0+\color{red}\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7} \color{black}\right )\\ &=\displaystyle \frac{1}{4}\left ( -1+\color{red}\displaystyle \frac{1}{2}\color{black} \right )\\ &=\displaystyle \frac{1}{4}\times \left (-\frac{1}{2} \right )\\ &=-\displaystyle \frac{1}{8} \end{aligned} \end{array}$.
Berikut penjelasan untuk $\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}=\color{red}\displaystyle \frac{1}{2}$.
$\begin{aligned}&\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}\\ &=\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}\times \displaystyle \frac{\left (2\sin\displaystyle \frac{2\pi }{7} \right ) }{\left (2\sin\displaystyle \frac{2\pi }{7} \right )}\\ &=\displaystyle \frac{2\cos\displaystyle \frac{\pi }{7}\sin\displaystyle \frac{2\pi }{7}-2\cos\displaystyle \frac{2\pi }{7}\sin\displaystyle \frac{2\pi }{7}+2\cos\displaystyle \frac{3\pi }{7}\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}-\sin\left (-\displaystyle \frac{\pi }{7} \right )-\left ( \sin\displaystyle \frac{4\pi }{7}-\sin\displaystyle \frac{0\pi }{7} \right )+\sin\displaystyle \frac{5\pi }{7}-\sin\displaystyle \frac{\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}+\sin\displaystyle \frac{\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{5\pi }{7}-\sin\displaystyle \frac{\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{5\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\left (\pi -\displaystyle \frac{4\pi }{7} \right )-\sin\displaystyle \frac{4\pi }{7}+\sin\left (\pi -\displaystyle \frac{2\pi }{7} \right )}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{4\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{1}{2}\qquad \blacksquare \end{aligned}$.

$\begin{array}{ll}\\ 32.&\textrm{Nilai dari}\\ &\quad\quad \sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{16} &&&\textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle \color{red}\frac{1}{8}&\quad \textrm{c}.&\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle 1 \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikan bahwa}\\ \sin \displaystyle \displaystyle \frac{\pi }{14}&=\sin \left (\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14} \right )=\sin \left ( \displaystyle \frac{1}{2}\pi -\frac{6\pi }{14} \right )\\ &=\cos \displaystyle \frac{6\pi }{14} \\ \sin \displaystyle \frac{3\pi }{14}&=...=\cos \displaystyle \frac{4\pi }{14}\\ \sin \displaystyle \frac{9\pi }{14}&=...=\sin \displaystyle \frac{5\pi }{14}=\cos \displaystyle \frac{2\pi }{14} \end{aligned}\\ &...\\ &\sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ &=\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\cos \displaystyle \frac{2\pi }{14}\times \displaystyle \frac{2\sin \displaystyle \frac{2\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\sin \displaystyle \frac{4\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &\textrm{silahkan dilanjutkan}\\ &...\\ &=\displaystyle \frac{1}{8} \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Nilai dari}\\ & \cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{8\pi }{5}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}-\displaystyle \frac{1}{16} &&&\textrm{d}.&\displaystyle \frac{1}{16}\\\\ \textrm{b}.&\displaystyle \frac{1}{8}&\quad \textrm{c}.&\displaystyle 0\quad &\textrm{e}.&\displaystyle \frac{1}{8} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{8\pi }{5}\\ &=\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \left (\pi +\displaystyle \frac{3\pi }{5} \right )\\ &=\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\left (-\cos \displaystyle \frac{3\pi }{5} \right )\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{3\pi }{5}\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos \displaystyle \frac{4\pi }{5}\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos \displaystyle \frac{4\pi }{5}\times \displaystyle \frac{2\sin \displaystyle \frac{\pi }{5}}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\frac{-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\left ( \sin \pi -\sin \displaystyle \frac{3\pi }{5} \right )}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5} \sin \frac{3\pi }{5}}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\left ( \cos \displaystyle \frac{2\pi }{5}\sin \displaystyle \frac{3\pi }{5} \right )}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\left ( \sin \pi -\sin \left ( -\displaystyle \frac{\pi }{5} \right ) \right )}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\sin \displaystyle \frac{\pi }{5}}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\cos \displaystyle \frac{\pi }{5}\sin \displaystyle \frac{\pi }{5}}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\left (\cos \displaystyle \frac{\pi }{5}\sin \displaystyle \frac{\pi }{5} \right )}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\left (\sin \displaystyle \frac{2\pi }{5}-\sin 0 \right )}{8\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\sin \displaystyle \frac{2\pi }{5}}{8\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\sin \pi -\sin \displaystyle \frac{\pi }{5}}{16\sin \displaystyle \frac{\pi }{5}}\\ &=-\displaystyle \frac{\sin \displaystyle \frac{\pi }{5}}{16\sin \displaystyle \frac{\pi }{5}}\\ &=-\displaystyle \frac{1}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 34.&\textrm{Nilai dari}\\ & \qquad\qquad\sin 18^{\circ}\cos 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{6} &&&\textrm{d}.&\displaystyle \frac{1}{3}\\\\ \textrm{b}.&\displaystyle \frac{1}{5}&\quad \textrm{c}.&\displaystyle \color{red}\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sin 18^{\circ}\cos 36^{\circ}\\ &=\sin 18^{\circ}\cos 36^{\circ}\times \displaystyle \frac{2\cos 18^{\circ}}{2\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\left ( \sin 36^{\circ}+\sin 0^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\sin 36^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin 72^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin \left ( 90^{\circ}-18^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 18^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Nilai eksak dari}\: \: \sin 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}&&&\textrm{d}.&\displaystyle \frac{\sqrt{5}-1}{4}\\ \textrm{b}.&\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}&&\quad &\textrm{e}.&\displaystyle \frac{\sqrt{5}-1}{2}\\ \textrm{c}.&\displaystyle \displaystyle \frac{\sqrt{5}+1}{4} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Perhatikan bahwa}\: \: \color{red}\bigtriangleup ABC\: \: \color{black}\textrm{sama kaki}\\ &\textrm{dengan}\: \: AD=DC=CB=1,\: AC=x\\ &\textrm{Diketahui pula}\: \: CD\: \: \textrm{adalah garis bagi}\\ &\textrm{serta}\: \: ABC\: \: \textrm{sebangun}\: \: \bigtriangleup BCD\\ &\textrm{akibatnya}:\\ &\color{red}\textrm{perbandingan sisi yang bersesuaian}\\ &\color{red}\textrm{akan sama},\: \: \color{black}\textrm{maka}\\ &\displaystyle \frac{AB}{BC}=\displaystyle \frac{BC}{AB-AD}\\ &\Leftrightarrow \displaystyle \frac{x}{1}=\frac{1}{x-1}\\ &\Leftrightarrow x(x-1)=1\\ &\Leftrightarrow x^{2}-x-1=0\\ &\Leftrightarrow x=\displaystyle \frac{1\pm \sqrt{5}}{2}\\ &\textrm{akibatnya}\: \: AB=AC=\displaystyle \frac{1+\sqrt{5}}{2}\\ &\textrm{Selanjutnya gunakan}\: \: \color{blue}\textrm{aturan sinus}\\ &\displaystyle \frac{AB}{\sin \angle C}=\frac{BC}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{AB}{BC}=\frac{\sin \angle C}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{ \left (\displaystyle \frac{1+\sqrt{5} }{2} \right )}{1}=\frac{\sin 72^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=\displaystyle \frac{2\sin 36^{\circ}\cos 36^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=2\cos 36^{\circ}\\ &\Leftrightarrow \cos 36^{\circ}=\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{4}\\ &\textrm{Dari fakta di atas kita akan dengan}\\ &\textrm{mudah menentukan nilai sinusnya}\\ &\textrm{yaitu dengan menggunakan}\\ &\textrm{identitas trigonometri berikut}:\\ &\sin ^{2}36^{\circ}+\cos ^{2}36^{\circ}=1\\ &\Leftrightarrow \sin ^{2}36^{\circ}=1-\cos ^{2}36^{\circ}\\ &\Leftrightarrow \sin 36^{\circ}=\sqrt{1-\cos ^{2}36^{\circ}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\left ( \displaystyle \frac{1+\sqrt{5}}{4} \right )^{2}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\displaystyle \frac{6+2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{\displaystyle \frac{10-2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}} \end{aligned}$

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Keenam

$\begin{array}{ll}\\ 26.&\textrm{Bentuk sederhana dari}\\ &\qquad\quad \displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\sin (x-y) &&&\textrm{d}.&\cos (x-y)\\ \textrm{b}.&\displaystyle \color{red}-\tan (x-y)&&&\textrm{e}.&\displaystyle \tan (x-y)\\ \textrm{c}.&\displaystyle \sin (x+y) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}\\ &=\displaystyle \frac{-2\sin (x+y)\sin (x-y)}{2\sin (x+y)\cos (x-y)}\\ &=-\tan (x-y) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 27.&\textrm{Nilai dari}\\ &\quad\quad 8\cos 82,5^{\circ}\sin 37,5^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle 4(\sqrt{3}+\sqrt{2}) &&&\textrm{d}.&\color{red}2(\sqrt{3}-\sqrt{2})\\ \textrm{b}.&\displaystyle 4(\sqrt{3}-\sqrt{2})&&&\textrm{e}.&\displaystyle \sqrt{3}-\sqrt{2}\\ \textrm{c}.&\displaystyle 2(\sqrt{3}+\sqrt{2}) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&8\cos 82,5^{\circ}\sin 37,5^{\circ}\\ &=4\times 2\cos 82,5^{\circ}\sin 37,5^{\circ}\\ &=4\times \left ( \sin (82,5^{\circ}+37,5^{\circ})-\sin (82,5^{\circ}-37,5^{\circ}) \right )\\ &=4\times \left (\sin 120^{\circ}-\sin 45^{\circ} \right )\\ &=4\times \left ( \sin \left ( 180^{\circ}-60^{\circ} \right )-\sin 45^{\circ} \right )\\ &=4\times \left ( \sin 60^{\circ}-\sin 45^{\circ} \right )\\ &=4\times \left ( \displaystyle \frac{1}{2}\sqrt{3}-\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=2\left ( \sqrt{3}-\sqrt{2} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Bentuk lain dari}\\ &\quad\quad -2\cos 5A.\cos 7A\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\cos 6A-\cos A &&&\\ \textrm{b}.&\displaystyle -\cos 6A+\cos A&&&\\ \textrm{c}.&\displaystyle \cos 12A-\cos 2A\\ \textrm{d}.&-\cos 12A+\cos 2A\\ \textrm{e}.&\color{red}\displaystyle -\cos 12A-\cos 2A \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&-2\cos 5A.\cos 7A\\ &=-\left ( 2\cos 5A.\cos 7A \right )\\ &=-\left ( \cos 12A+\cos (-2A) \right )\\ &=-\left ( \cos 12A+\cos 2A \right )\\ &=-\cos 12A-\cos 2A \\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Bentuk sederhana dari}\\ &\quad\quad 4\sin \left ( \displaystyle \frac{1}{4}\pi +x \right )\cos \left ( \displaystyle \frac{1}{4}\pi -x \right )\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}2+2\sin 2x &&&\textrm{d}.&2+2\sin x\\ \textrm{b}.&\displaystyle 2+\sin 2x&&&\textrm{e}.&2+\sin x\\ \textrm{c}.&\displaystyle 2\sin 2x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&4\sin \left ( \displaystyle \frac{1}{4}\pi +x \right )\cos \left ( \displaystyle \frac{1}{4}\pi -x \right )\\ &=2\left ( \sin \left ( \displaystyle \frac{1}{2}\pi \right )+\sin (2x) \right )\\ &=2\left (1+\sin 2x \right )\\ &=2+2\sin 2x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Nilai dari}\\ &\quad\quad \sqrt{3}\sin 80^{\circ}\sin 160^{\circ}\sin 320^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\displaystyle \frac{3}{8} &&&\textrm{d}.&\displaystyle \color{red}\frac{3}{8}\\\\ \textrm{b}.&\displaystyle -\frac{1}{8}&&&\textrm{e}.&\displaystyle \frac{5}{8}\\ \textrm{c}.&\displaystyle \frac{1}{8} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sqrt{3}\sin 80^{\circ}\sin 160^{\circ}\sin 320^{\circ}\\ &=\sqrt{3}\sin 80^{\circ}\sin 20^{\circ}\left (-\sin 40^{\circ} \right )\\ &=-\sqrt{3}\sin 80^{\circ}\sin 40^{\circ}\sin 20^{\circ}\\ &=-\sqrt{3}\sin 80^{\circ}\left ( -\displaystyle \frac{1}{2}\left ( \cos 60^{\circ}-\cos 20^{\circ} \right ) \right )\\ &=-\sqrt{3}\sin 80^{\circ}\left ( -\displaystyle \frac{1}{4}+\displaystyle \frac{\cos 20^{\circ}}{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{2}\sqrt{3}\sin 80^{\circ}\cos 20^{\circ}\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\left ( \sin 100^{\circ}+\sin 60^{\circ} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\left ( \sin 80^{\circ}+\displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}+\displaystyle \frac{1}{8}\sqrt{9}\\ &=\displaystyle \frac{3}{8} \end{aligned} \end{array}$.

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Kelima

$\begin{array}{ll}\\ 21.&\textrm{Nilai}\: \: \cos \displaystyle \frac{5}{12}\pi -\cos \displaystyle \frac{1}{12}\pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\sqrt{6}&&&\textrm{d}.&\displaystyle \frac{1}{2}\sqrt{2}\\\\ \textrm{b}.&-\displaystyle \frac{1}{2}\sqrt{3}&\textrm{c}.&\color{red}-\displaystyle \frac{1}{2}\sqrt{2}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\cos \displaystyle \frac{5}{12}\pi -\cos \displaystyle \frac{1}{12}\pi \\ &=-2\sin \left ( \displaystyle \frac{\displaystyle \frac{5}{12}\pi +\displaystyle \frac{1}{12}\pi }{2} \right )\sin \left ( \displaystyle \frac{\displaystyle \frac{5}{12}\pi -\displaystyle \frac{1}{12}\pi }{2} \right )\\ &=-2\sin \left (\displaystyle \frac{\displaystyle \frac{6}{12}\pi }{2} \right )\sin \left (\displaystyle \frac{\displaystyle \frac{4}{12}\pi }{2} \right ) \\ &=-2\sin \left (\displaystyle \frac{1 }{4}\pi \right )\sin \left (\displaystyle \frac{1 }{6}\pi \right ) \\ &=-2\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \displaystyle \frac{1}{2} \right )\\ &=\color{red}\displaystyle -\frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Bentuk}\: \: \sin \left ( 2x-\displaystyle \frac{3}{2}\pi \right )-\sin \left ( 4x+\displaystyle \frac{1}{2}\pi \right )\\ & \textrm{senilai dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -2\sin 3x.\sin x &&&\textrm{d}.&\color{red}\displaystyle 2\sin 3x.\sin x\\ \textrm{b}.&\displaystyle -2\cos 3x.\sin x&&&\textrm{e}.&\displaystyle 2\cos 3x.\sin x\\ \textrm{c}.&\displaystyle 2\sin 2\left ( x-\pi \right ) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sin \left ( 2x-\displaystyle \frac{3}{2}\pi \right )-\sin \left ( 4x+\displaystyle \frac{1}{2}\pi \right )\\ &=2\cos \left ( \displaystyle \frac{2x-\displaystyle \frac{3}{2}\pi+4x+\displaystyle \frac{1}{2}\pi}{2} \right )\\ &\qquad \times \sin \left ( \displaystyle \frac{2x-\displaystyle \frac{3}{2}\pi-\left (4x+\displaystyle \frac{1}{2}\pi \right )}{2} \right )\\ &=2\cos (3x-\displaystyle \frac{1}{2}\pi )\sin\left ( -x-\pi \right )\\ &=2\cos \left ( \displaystyle \frac{1}{2}\pi -3x \right )\left ( -\sin (\pi +x) \right )\\ &=2\left ( \sin 3x \right )\left ( -(-\sin x) \right )\\ &=2\sin 3x.\sin x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Bentuk}\: \: \displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ & \textrm{senilai dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\tan 6x &&&\textrm{d}.&6\cot x\\ \textrm{b}.&\displaystyle -\cot 6x&&&\textrm{e}.&\displaystyle \color{red}\tan 6x\\ \textrm{c}.&\displaystyle 6\tan x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ &=\displaystyle \frac{\cos 3x-\cos 9x-\sin 6x}{\sin 9x-\sin 3x-\cos 6x}\\ &=\displaystyle \frac{-2\sin 6x\sin (-3x)-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{2\sin 6x\sin 3x-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{\sin 6x(2\sin 3x-1)}{\cos 6x(2\sin 3x-1)}\\ &=\tan 6x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \tan 2x &&\textrm{d}.&\displaystyle \tan 8x\\ \textrm{b}.&\color{red}\displaystyle \tan 4x&\quad&\textrm{e}.&\displaystyle \tan 16x\\ \textrm{c}.&\displaystyle \displaystyle \tan 6x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ &=\displaystyle \frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}\\ &=\displaystyle \frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}\\ &=\displaystyle \frac{2\sin 4x\left ( \cos 3x+\cos x \right )}{2\cos 4x\left ( \cos 3x+\cos x \right )}\\ &=\tan 4x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Bentuk sederhana dari}\\ &\quad\quad \displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \tan A &&&\textrm{d}.&2\cos 2A\\ \textrm{b}.&\displaystyle 2\tan A&&&\textrm{e}.&\displaystyle \color{red}2\tan 2A\\ \textrm{c}.&\displaystyle 2\sin 2A \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}\\ &=\displaystyle \frac{(\cos A+\sin A)^{2}-(\cos A-\sin A)^{2}}{(\cos A-\sin A)(\cos A+\sin A)}\\ &=\displaystyle \frac{(\cos ^{2}A+2\cos A\sin A+\sin ^{2}A)-(\cos ^{2}A-2\cos A\sin A+\sin ^{2}A)}{\cos ^{2}A-\sin ^{2}A}\\ &=\displaystyle \frac{4\cos A\sin A}{\cos ^{2}A-\sin ^{2}A}\\ &=\displaystyle \frac{2\sin 2A}{\cos 2A}\\ &=2\tan 2A \end{aligned} \end{array}$.

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Keempat

$\begin{array}{ll}\\ 16.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\qquad\quad 3\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\sqrt{3} \\ & \textrm{untuk}\: \: 0\leq x\leq \pi \: \: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \left \{ \displaystyle \frac{1}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{12}\pi ,\displaystyle \frac{9}{12}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ \displaystyle \frac{3}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{3}{12}\pi ,\displaystyle \frac{9}{12}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{5}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & 3\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\sqrt{3}\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\frac{\sqrt{3}}{3}\\ &(\textrm{kuadran IV, karena Y negatif, X positif})\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\tan \displaystyle \frac{1}{6}\pi ,\: \: \textbf{menjadi}\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=\tan \left ( 2\pi -\displaystyle \frac{1}{6}\pi \right )=\tan \displaystyle \frac{11}{6}\pi \\ &\left (2x-\displaystyle \frac{1}{3}\pi \right )=\displaystyle \frac{11}{6}\pi\\ &\Leftrightarrow \: \: 2x=\displaystyle \frac{1}{3}\pi+\displaystyle \frac{11}{6}\pi +k.\pi =\displaystyle \frac{13}{6}\pi +k.\pi \\ &\Leftrightarrow \: \: x=\displaystyle \frac{13}{12}\pi +\displaystyle \frac{k.\pi}{2} \\ &k=0\Rightarrow x=\displaystyle \frac{13}{12}\pi=\displaystyle \frac{1}{12}\pi \: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=\displaystyle \frac{13}{12}\pi +\displaystyle \frac{1}{2}\pi =\displaystyle \frac{19}{12}\pi=\displaystyle \frac{7}{12}\pi \quad (\color{blue}\textrm{mm})\\ &k=2\Rightarrow x=\displaystyle \frac{13}{12}\pi +\pi \quad \color{red}\textrm{tidak memenuhi} \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{1}{12}\pi,\: \displaystyle \frac{7}{12}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Salah satu nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\textrm{persamaan}\: \: \cos x+\sin x=\displaystyle \frac{1}{2}\sqrt{6}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{24}\pi &&&\textrm{d}.&\displaystyle \frac{1}{8}\pi \\\\ \textrm{b}.&\displaystyle \frac{1}{15}\pi&\textrm{c}.&\displaystyle \color{red}\frac{1}{12}\pi&\textrm{e}.&\displaystyle \frac{1}{6}\pi \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\textrm{Diketahui bahwa}\\ &\sin x+\cos x=\displaystyle \frac{1}{2}\sqrt{6}\quad \left (\textbf{ingat}:a=1,\: b=1 \right )\\ &\sin x+\cos x=k\cos \left ( x-\theta \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\begin{cases} k & =\sqrt{1^{2}+1^{2}}=\sqrt{2} \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{1}{1}=1\Rightarrow \theta =45^{\circ}=\displaystyle \frac{1}{4}\pi \end{cases}\\ &\qquad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran I, karena}\: \: a,b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&\sin x+\cos x=k\cos \left ( x-\theta \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\Leftrightarrow \sqrt{2}\cos\left ( x-\displaystyle \frac{1}{4}\pi \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\Leftrightarrow \: \: \, \cos \left ( x-\displaystyle \frac{1}{4}\pi \right )=\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{6}}{\sqrt{2}}=\displaystyle \frac{1}{2}\sqrt{3}\\ &\Leftrightarrow \: \: \,\cos \left ( x-\displaystyle \frac{1}{4}\pi \right )=\cos \displaystyle \frac{1}{6}\pi \\ &\Leftrightarrow \quad x-\displaystyle \frac{1}{4}\pi =\pm \displaystyle \frac{1}{6}\pi +k.2\pi \\ &\Leftrightarrow \quad x=\displaystyle \frac{1}{4}\pi\pm \displaystyle \frac{1}{6}\pi+k.2\pi \\ &\Leftrightarrow \quad x_{1}=\displaystyle \frac{5}{12}\pi+k.2\pi \: \: \textbf{atau}\\ &\: \: \: \quad\quad x_{2}=\displaystyle \frac{1}{12}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{5}{12}\pi\qquad (\color{blue}\textrm{memenuhi})\\ &\: \: \qquad\Rightarrow x_{2}=\displaystyle \color{red}\frac{1}{12}\pi\qquad \color{black}(\color{blue}\textrm{memenuhi})\\ &\textrm{Langkah berikutnya tidak diperlukan}\\ &\textrm{karena jawaban sudah kita dapatkan}\\ &\textrm{yaitu}:\: \: \color{red}\displaystyle \frac{1}{12}\pi \end{aligned} \end{array} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Himpunan penyelesaian persamaan}\\ &\qquad\: \: \cos x^{\circ}-\sqrt{3}\sin x^{\circ}=1\\ &\textrm{untuk}\: \: 0\leq x< 360\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}\left \{ 0,240 \right \} &&&\textrm{d}.&\displaystyle \left \{ 180,240 \right \} \\\\ \textrm{b}.&\displaystyle \left \{ 150,270 \right \}&\textrm{c}.&\displaystyle \left \{ 180,300 \right \}&\textrm{e}.&\displaystyle \left \{ 210,270 \right \} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\textrm{Diketahui dari soal bahwa}\\ &\cos x^{\circ}-\sqrt{3}\sin x^{\circ}=1,\\ &\textrm{lalu kita ubah posisinya menjadi}\\\\ &-\sqrt{3}\sin x+\cos x=1\: \: \left (\textbf{ingat}:a=-\sqrt{3},\: b=1 \right )\\ &-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\begin{cases} k & =\sqrt{\left ( -\sqrt{3} \right )^{2}+\left ( 1 \right )^{2}}=\sqrt{4}=2 \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{-\sqrt{3}}{1}=-\sqrt{3}\Rightarrow \theta =300^{\circ} \end{cases}\\ &\quad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran IV, karena}\: \: a<0,\: b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\Leftrightarrow 2\cos\left ( x-300^{\circ} \right )=1\\ &\Leftrightarrow \: \: \, \cos \left ( x-300^{\circ} \right )=\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \: \,\cos \left ( x-300^{\circ} \right )=\cos 60^{\circ}\\ &\Leftrightarrow \quad x-300^{\circ} =\pm 60^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x=300^{\circ}\pm 60^{\circ}+k.360^{\circ}\\ &k=0\Rightarrow x_{1}=300^{\circ}+60^{\circ}=360^{\circ}=0^{\circ}\: \: (\color{blue}\textrm{mm})\\ & \qquad\qquad\quad\color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=300^{\circ}-60^{\circ}=240^{\circ}\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=300^{\circ}\pm 60^{\circ}+360^{\circ}\quad (\color{red}\textrm{tm})\\ \end{aligned} \\ &\textbf{HP}=\left \{0^{\circ},240^{\circ} \right \} \end{array} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Diketahui}\: \: \alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah sudut} \: \: \textbf{lancip}\\ &\textrm{Nilai dari}\: \: \cos \left ( \alpha +\beta \right )\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&1&&&\textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{b}.&\displaystyle \frac{3}{4}&\textrm{c}.&\displaystyle \frac{1}{2}&\textrm{e}.&\color{red}\displaystyle 0 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bullet \quad\alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\bullet \quad\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{sudut} \: \: \textbf{lancip}\\ &\qquad \textrm{akibatnya semua sudut dikuadran I}\\ &\qquad \textrm{sehingga}\color{purple}\begin{cases} \sin & =+ \\ \cos & =+ \\ \tan & =+ \end{cases}\\ &\textrm{ditanya}\: \: \cos \left ( \alpha +\beta \right ),\: \: \textrm{maka}\\ &\textrm{sebagai langkah awal kita adalah}:\\ &\cos \left ( \alpha -\beta \right )=\cos \left ( \displaystyle \frac{\pi }{3} \right )\\ &\Leftrightarrow \: \cos \alpha \cos \beta +\sin \alpha \sin \beta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta+\displaystyle \frac{1}{4} =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta =\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}=\displaystyle \frac{1}{4}\\ &\textbf{Selanjutnya nilai dari}\\ &\cos \left ( \alpha +\beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &=\displaystyle \frac{1}{4}-\displaystyle \frac{1}{4}=0\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Nilai}\: \: \sin 75^{\circ}-\sin 165^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\sqrt{2}&&&\textrm{d}.&\color{red}\displaystyle \frac{1}{2}\sqrt{2}\\\\ \textrm{b}.&\displaystyle \frac{1}{4}\sqrt{3}&\textrm{c}.&\displaystyle \frac{1}{4}\sqrt{6}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\sin 75^{\circ}-\sin 165^{\circ}\\ &=2\cos \left ( \displaystyle \frac{75^{\circ}+165^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{75^{\circ}-165^{\circ}}{2} \right )\\ &=2\cos \displaystyle \frac{240^{\circ}}{2}\sin \left (-\displaystyle \frac{90^{\circ}}{2} \right ) \\ &=2\cos 120^{\circ}\sin \left ( -45^{\circ} \right )\\ &=2\left (-\cos 60^{\circ} \right )\left (- \sin 45^{\circ} \right )\\ &=2\left ( -\displaystyle \frac{1}{2} \right )\left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Ketiga

$\begin{array}{ll}\\ 11.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\cos x \\\\ &\textrm{b}.\quad \displaystyle y=\cos 2x \\\\ &\textrm{c}.\quad \displaystyle y=\cos \displaystyle \frac{1}{2}x \\\\ &\textrm{d}.\quad \displaystyle \color{red}y=2\cos 2x \\\\ &\textrm{e}.\quad \displaystyle y=2\cos \displaystyle \frac{1}{2}x \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\\ & \textbf{atas dan ke bawah}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{2}=180^{\circ}=\pi ,\\ & \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\cos 2x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\sin \left (2x+\pi \right ) \\\\ &\textrm{b}.\quad \displaystyle y=\sin \left (2x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{c}.\quad \displaystyle y=2\sin \left (2x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{d}.\quad \displaystyle y=\sin \left (2x+\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{e}.\quad \displaystyle \color{red}y=2\sin \displaystyle \left (x+\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\: \: \textbf{kiri}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{1}=360^{\circ}=2\pi ,\\ &\textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\sin \left ( x+\displaystyle kx \right )\\ &\textrm{dengan}\: \: +k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kiri}\: \: \displaystyle \frac{1}{2}\pi \: \: \textrm{atau}\: \: 90^{\circ}\\ &\textrm{Jadi},\: \: y=\color{red}2\sin (x+\displaystyle \frac{1}{2}\pi )^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\sin x=\sin \displaystyle \frac{2}{10}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{12}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \sin x=\sin \displaystyle \frac{2}{10}\pi \\ &\Leftrightarrow \: \: x_{1}=\displaystyle \frac{2}{10}\pi+k.2\pi \: \: \: \: \color{blue}\textrm{atau}\\ &\Leftrightarrow \quad x_{2} =\left (\pi -\displaystyle \frac{2}{10}\pi \right )+k.2\pi=\displaystyle \frac{8}{10}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{2}{10}\pi\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{8}{10}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1,2}=....+2\pi \quad (\color{red}\textrm{tidak memenuhi})\\ \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{2}{10}\pi,\: \displaystyle \frac{8}{10}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{1}{3}\pi ,\pi ,\displaystyle \frac{5}{3}\pi ,\displaystyle \frac{7}{3}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{2}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \\ &\Leftrightarrow \: \: 2x-\displaystyle \frac{1}{4}\pi=\displaystyle \frac{1}{4}\pi+k.\pi\\ &\Leftrightarrow \quad 2x =\displaystyle \frac{2}{4}\pi +k.\pi \\ &\Leftrightarrow \quad x =\displaystyle \frac{1}{4}\pi +k.\frac{\pi}{2} \\ &k=0\Rightarrow x=\displaystyle \frac{1}{4}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{\pi}{2}=\displaystyle \frac{3}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=2\Rightarrow x=\displaystyle \frac{1}{4}\pi+\pi =\displaystyle \frac{5}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=3\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{3\pi}{2}=\displaystyle \frac{7}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=4\Rightarrow x=\displaystyle \frac{1}{4}\pi+2\pi =\displaystyle \frac{9}{4}\pi \: \: (\color{red}\textrm{tidak memenuhi}) \\ \end{aligned} \\ &\textbf{HP}=\color{red}\color{red}\left \{\displaystyle \frac{1}{4}\pi,\: \displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\cos 2x-2\cos x=-1\: \: \textrm{untuk}\: \: 0< x< 2\pi \\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{3}{2}\pi, 2\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi, 2\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi ,\displaystyle \frac{3}{2}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \cos 2x-2\cos x=-1\\ &\Leftrightarrow \cos 2x-2\cos x+1=0\\ &\Leftrightarrow \left ( 2\cos ^{2}x-1 \right )-2\cos x+1=0 \\ &\Leftrightarrow 2\cos x\left ( \cos x-1 \right )=0\\ &\Leftrightarrow \cos x=0\: \: \textrm{atau}\: \: \cos x=1\\ &\Leftrightarrow \cos x=\cos \displaystyle \frac{1}{2}\pi \: \: \textrm{atau}\: \: \cos x=\cos 0\\ &\Leftrightarrow x_{1,2}=\pm \displaystyle \frac{1}{2}\pi +k.2\pi \: \: \textrm{atau}\: \: x_{3}=k.2\pi\\ &\textrm{maka}\\ &k=0\Rightarrow x_{1}=-\displaystyle \frac{1}{2}\pi\: \: (\color{red}\textrm{tm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{1}{2}\pi\: \: (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{3}=\displaystyle 0\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1}=\displaystyle \frac{3}{2}\pi \quad (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{2}=\displaystyle \frac{5}{2}\pi\: \: (\color{red}\textrm{tm})\\ &\qquad\qquad x_{3}=\displaystyle 2\pi\: \: (\color{blue}\textrm{mm})\\ \end{aligned} \end{array} \end{array}$

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Kedua

$\begin{array}{ll}\\ 6.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sin 49^{\circ}}{\cos 41^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\sin 73^{\circ}}\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1&&&\textrm{d}.&0,143\\ \textrm{b}.&\displaystyle -0,321&\textrm{c}.&\color{red}0&\textrm{e}.&\displaystyle 0,321 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin 49^{\circ}}{\cos 41^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\sin 73^{\circ}}\\ &=\displaystyle \frac{\sin 49^{\circ}}{\cos \left (90^{\circ}-49^{\circ} \right )}-\displaystyle \frac{\cos 17^{\circ}}{\sin \left (90^{\circ}-17^{\circ} \right )}\\ &=\displaystyle \frac{\sin 49^{\circ}}{\sin 49^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\cos 17^{\circ}}\\ &=1-1\\ &=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\\ &p=r\sin \alpha \cos \beta \\ &q=r\sin \alpha \sin \beta \\ &s=r\cos \alpha \\ &\textrm{maka pernyataan berikut yang}\\ &\textrm{tepat adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\color{red}p^{2}+t^{2}+s^{2}=r^{2}&&&\\ \textrm{b}.&p^{2}-t^{2}+s^{2}=r^{2} \\ \textrm{c}.&p^{2}+t^{2}-s^{2}=r^{2}&\\ \textrm{d}.&-p^{2}+t^{2}+s^{2}=r^{2}&\\ \textrm{e}.&-p^{2}-t^{2}+s^{2}=r^{2} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Saat}\\ &p^{2}+q^{2}\: \: \textrm{maka hasilnya adalah}\\ &\color{purple}\begin{array}{lll}\\ p^{2}&=r^{2}\sin^{2} \alpha \cos^{2} \beta&\\ q^{2}&=r^{2}\sin^{2} \alpha \sin^{2} \beta&+\\\hline &=r^{2}\sin ^{2}\alpha \left ( \cos ^{2}\beta +\sin ^{2}\beta \right )\\ &=r^{2}\sin ^{2}\alpha (1)\\ &=r^{2}\sin ^{2}\alpha \end{array}\\ &\textrm{Dan saat}\\ &p^{2}+q^{2}+s^{2}\: \: \textrm{akan diperoleh hasil}\\ &\color{purple}\begin{array}{lll}\\ p^{2}+q^{2}&=r^{2}\sin ^{2}\alpha &\\ \qquad s^{2}&=r^{2}\cos ^{2}\alpha &+\\\hline &=r^{2}\sin ^{2}\alpha+r^{2}\cos ^{2}\alpha\\ &=r^{2}\left ( \sin ^{2}\alpha+\cos ^{2}\alpha \right )\\ &=r^{2}(1)\\ &=r^{2} \end{array}\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\\ & \displaystyle \frac{\cos \left ( 90^{\circ}+\theta \right )\sec \left ( 2\pi -\theta \right )\tan \left ( \pi -\theta \right )}{\sec \left ( \theta -2\pi \right )\sin \left ( 540^{\circ}+\theta \right )\cot \left ( \theta -90^{\circ} \right )}\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1&&&\textrm{d}.&-\tan \theta \\ \textrm{b}.&\displaystyle 0&\textrm{c}.&\color{red}1&\textrm{e}.&\displaystyle \tan \theta \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\color{purple}\textrm{Ingat kembali sudut-sudut}\\ &\color{purple}\textrm{yang berelasi dari kudran selain I}\\ &\color{purple}\textrm{ke kuadran I beserta tandanya}\\ &\displaystyle \frac{\cos \left ( 90^{\circ}+\theta \right )\sec \left ( 2\pi -\theta \right )\tan \left ( \pi -\theta \right )}{\sec \left ( \theta -2\pi \right )\sin \left ( 540^{\circ}+\theta \right )\cot \left ( \theta -90^{\circ} \right )}\\ &=\displaystyle \frac{\left (-\sin \theta \right ) .\sec \theta .\left (-\tan \theta \right )}{\sec \theta .\left (-\sin \theta \right ). \left (-\tan \theta \right )}\\ &=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Diketahui bahwa}\\ &\sin \theta +\cos \theta =\displaystyle \frac{1}{2} \\ &\textrm{maka nilai dari}\\ &\sin ^{3}\theta +\cos ^{3}\theta \: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{2}&&&\textrm{d}.&\displaystyle \frac{5}{8} \\\\ \textrm{b}.& \displaystyle \frac{3}{4}&\textrm{c}.&\displaystyle \frac{9}{15}&\textrm{e}.&\color{red}\displaystyle \frac{11}{16} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\sin \theta +\cos \theta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \left (\sin \theta +\cos \theta \right )^{2} =\displaystyle \frac{1}{4}\\ &\Leftrightarrow \: \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cos \theta =\displaystyle \frac{1}{4}\\ &\Leftrightarrow \: 1+2\sin \theta \cos \theta =\displaystyle \frac{1}{4}\\ &\Leftrightarrow \: 2\sin \theta \cos \theta =-\displaystyle \frac{3}{4}\\ &\Leftrightarrow \: \sin \theta \cos \theta =-\displaystyle \frac{3}{8}\\ &\textbf{Selanjutnya}\\ &\color{purple}\sin ^{3}\theta +\cos ^{3}\theta\\ &=\color{purple}\left ( \sin \theta +\cos \theta \right )\left ( \sin ^{2}\theta -\sin \theta \cos \theta +\cos ^{2}\theta \right )\\ &=\color{purple}\left ( \displaystyle \frac{1}{2} \right )\left ( 1-\left ( -\displaystyle \frac{3}{8} \right ) \right ) \\ &=\color{purple}\displaystyle \frac{1}{2}\times \displaystyle \frac{11}{8}\\ &=\color{red}\displaystyle \frac{11}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Jika diketahui}\: \: \: \displaystyle \frac{3}{2}\pi <x<2\pi \\ &\textrm{dan}\: \: \: \tan x=m,\\ &\textrm{maka nilai dari}\: \: \sin x \cos x \: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\frac{1}{m^{2}+1}&&&\textrm{d}.&\displaystyle -\frac{m}{m^{2}-1} \\\\ \textrm{b}.& \color{red}\displaystyle -\frac{m}{m^{2}+1}&\textrm{c}.&\displaystyle \frac{m}{m^{2}+1}&\textrm{e}.&\displaystyle \frac{m}{m^{2}-1} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: \displaystyle \frac{3}{2}\pi <x<2\pi\\ &\textrm{ini daerah Kwadran IV, akibatnya adalah nilai}\\ &\begin{cases} \sin x & = -\\ \cos x & =+ \\ \tan x & =- \end{cases}\\ &\textbf{Selanjutnya ada pernyataan}\: \: \tan x=m\\ &\textrm{ini artinya}\: \: \tan x=\displaystyle \frac{m}{1}\\ &\textbf{Perhatikanlah ilustrasi gambar berikut} \end{aligned} \end{array}$.

$.\qquad\begin{aligned} &\textrm{maka nilai dari}\\ &\sin x\cos x\: \: \left (\textrm{ingat yang diminta di Kwadran IV} \right )\\ &=\left (-\displaystyle \frac{m}{\sqrt{m^{2}+1}} \right )\times \left (+\displaystyle \frac{1}{\sqrt{m^{2}+1}} \right )\\ &=\color{red}-\displaystyle \frac{m}{m^{2}+1} \end{aligned}$

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Pertama

$\begin{array}{ll}\\ 1.&\textrm{Nilai}\: \: 105^{\circ}\: \: \textrm{jika dinyatakan ke radian}\\ &\textrm{adalah}\: \: ....\: \: \textrm{radian}\\\\ &\textrm{a}.\quad \displaystyle \frac{1}{3}\pi \\\\ &\textrm{b}.\quad \displaystyle \frac{5}{6}\pi \\\\ &\textrm{c}.\quad \displaystyle \frac{5}{12}\pi \\\\ &\textrm{d}.\quad \color{red}\displaystyle \frac{7}{12}\pi \\\\ &\textrm{e}.\quad \displaystyle \frac{9}{12}\pi \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketah}&\textrm{ui bahwa}\\ 180^{\circ}&=\pi \: \: \: radian\\ 1^{\circ}&=\displaystyle \frac{\pi }{180}\: \: \: radian\\ 105\times 1^{\circ}&=105\times \displaystyle \frac{\pi }{180}\: \: \: radian\\ 105^{\circ}&=\displaystyle \frac{7}{12}\pi \: \: \: radian \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Nilai}\: \: \tan 240^{\circ} \: \: \: \textrm{adalah}\: \: ....\\\\ &\textrm{a}.\quad \displaystyle \color{red}\sqrt{3} \\\\ &\textrm{b}.\quad \displaystyle \frac{1}{3}\sqrt{3} \\\\ &\textrm{c}.\quad \displaystyle -\frac{1}{3}\sqrt{3} \\\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\sqrt{3} \\\\ &\textrm{e}.\quad \displaystyle -\sqrt{3} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\tan 240^{\circ}&=\tan \left ( 180^{\circ}+60^{\circ} \right )\\ &=\tan 60^{\circ}\\ &=\color{red}\sqrt{3}\\ \textbf{catatan}&: \textrm{ingat sudut berelasi} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Perhatikanlah gambar berikut}\\ \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Pada gambar di atas perbandingan}\\ &\sin \theta \: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \sqrt{\displaystyle \frac{a^{2}-d^{2}}{f^{2}+g^{2}}} \\ &\textrm{b}.\quad \sqrt{\displaystyle \frac{a^{2}+b^{2}}{f^{2}+g^{2}}} \\ &\textrm{c}.\quad \sqrt{\displaystyle \frac{a^{2}-b^{2}}{f^{2}-g^{2}}} \\ &\textrm{d}.\quad \sqrt{\displaystyle \frac{a^{2}+b^{2}}{f^{2}-g^{2}}}\\ &\textrm{e}.\quad \color{red}\sqrt{\displaystyle \frac{a^{2}-b^{2}}{f^{2}+g^{2}}} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Dari so}&\textrm{al diketahui bahwa}\\ \sin \theta &=\displaystyle \frac{c}{e}=\displaystyle \frac{\sqrt{a^{2}-b^{2}}}{\sqrt{f^{2}+g^{2}}}\\ &=\color{red}\sqrt{\displaystyle \frac{a^{2}-b^{2}}{f^{2}+g^{2}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Nilai dari}\: \: \left ( \cos ^{2}17^{\circ}-\sin ^{2}73^{\circ} \right ) \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\color{red}0&&&\textrm{d}.&1\\ \textrm{b}.&\displaystyle \frac{1}{3}&\textrm{c}.&\displaystyle \frac{2}{\sqrt{3}}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\left ( \cos ^{2}17^{\circ}-\sin ^{2}73^{\circ} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\left (\sin 73^{\circ} \right )^{2} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\left (\sin \left ( 90^{\circ}-17^{\circ} \right ) \right )^{2} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\cos ^{2}17^{\circ} \right )\\ &=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Jika diketahui}\\ & \displaystyle \frac{x\csc ^{2}30^{\circ}\sec ^{2}45^{\circ}}{8\cos ^{2}45^{\circ}\sin ^{2}60^{\circ}}=\tan ^{2}60^{\circ}-\tan ^{2}30^{\circ},\\ & \textrm{maka nilai}\: \: x\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-2&&&\textrm{d}.&\color{red}1\\ \textrm{b}.&\displaystyle -1&\textrm{c}.&0&\textrm{e}.&\displaystyle 2 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{x\csc ^{2}30^{\circ}\sec ^{2}45^{\circ}}{8\cos ^{2}45^{\circ}\sin ^{2}60^{\circ}}=\tan ^{2}60^{\circ}-\tan ^{2}30^{\circ}\\ &\displaystyle \frac{x\left ( 4 \right )\left ( \displaystyle \frac{4}{2} \right )}{8\left ( \displaystyle \frac{2}{4} \right )\left ( \displaystyle \frac{3}{4} \right )}=3-\left ( \displaystyle \frac{1}{3} \right )\\ &\displaystyle \frac{8x}{3}=\displaystyle \frac{8}{3}\\ &x=1 \end{aligned} \end{array}$