Geometri Ruang (Dimensi Tiga)

 $\text{A. Kedudukan Titik, Garis, dan Bidang}$

Perhatikanlah gambar bangunan gedung MA Futuhiyah Jeketro (gambar diambil sekitar awal tahun 2020) berikut

Jika tiang-tiang yang ada pada gambar di atas diasumsikan sebagai garis lurus dan tembok atau lantai atau pun halaman madrasah diasumsikan sebagai luasan maka dari kedua hal tersebut kita dapat melihat kedudukan antara garis dengan garis atau antara garis dengan bidang yang akhirnya mendasari bagaimana hubungan sebuah titik dengan titik yang lain yang tidak berimpit.

Dan perhatikanlah pula gambar kubus ABCD.EFGH berikut ini

Semakin jelas bahwa kedudukan 3 unsur utama yaitu titik, garis an bidang sangat dominan dalam pembahasan selanjutnya

Selanjutnya tempat kedudukan dalam hal ini akan terbagi dalam beberapa bagian dalam bangun ruang dimensi tiga yang lebih detil yaitu:

• hubungan titik dengan titik
• hubungan titik dengan garis 
• hubungan titik dengan bidang 
• hubungan garis dengan garis
• hubungan garis dengan bidang, dan
• hubungan bidang dengan bidang

$\text{B. Menghitung Jarak}$.

$\text{B.1 Kedudukan Suatu Titik Terhadap Titik Lain}$.

Jarak antara suatu titik dengan dengan suatu titik yang lain adalah panjang garis yang menghubungkan kedua titik itu

Selanjutnya untuk menentukan jarak antara titik beberapa pemecahan masalah dibutuihkan dalil Pythagoras.

Perhatikanlah ilustrasi gambar kubus ABCD.EFGH berikut ini yang telah disertakan jaraknya berbantuan dalil Pythagoras

Pada ilustrasi gambar di atas sudah dimunculkan jarak antar titik di antaranya sebagai berikut

$\begin{array}{|ccl|}\hline \text{No}& \text{Kedudukan}& \text{Besar jarak}\\ 1& \text{A ke B}& \text{AB}=\text{AE}=a\\ 2& \text{A ke C}& \text{AC}=\text{AF}=a\sqrt{2}\\ 3& \text{A ke G}& \text{AG}=\text{CE}=a\sqrt{3}\\ 4& \text{A ke Q}& \text{AQ}=\text{BQ}={\displaystyle \frac{1}{2}a\sqrt{6}}\\ 5& \text{A ke P}& \text{AP}=\text{EP}={\displaystyle \frac{3}{2}a}\\ 6& \text{C ke R}& \text{CR}=\text{BR}={\displaystyle \frac{1}{2}a\sqrt{5}}\\ \hline\end{array}$.

$\text{B. 1. 1 Penamaan Bangun Ruang}$.

perhatikanlah gambar kubus di atas. Penamaan bangun ruang model kubus itu dinamakan dengan istilah kubus ABCD.EFGH atau  ${\displaystyle \frac{\text{EFGH}}{\text{ABCD}}}$. Dan penulisan ABCD posisi di depan atau di bawah menunjukkan posisi bidang alas dan EFGH dituliskan setelah ABCD atau EFGH posisi di atas menunjukkan bahwa bidang EFGH adalah bidang atas. Urutan penyebutan titik pada bangun ruang itupun harus mengikuti arah putar jarum jam atau sebaliknya, sebagi misal pada bidang alas kubus tidak boleh disebutkan bidang ABDC tetapi ditulis dengan arah putar berkebalikan arah putar jarum jam yaitu ABCD. Jika Bagian alas dituliskan menggunakan kebalikan arah putar jarum jam, maka bidang alasnyapun mengikuti dengan diikuti peletakaan tanda titik yang menunjukkan perbedaan posisi bidangnya, sehingga bangun ruang kubus biasa disebutkan dengan istilah ABCD.EFGH. Hal penamaan tersebut akan berlaku pula pada bangun ruang seperti: Balok, Prisma dan semacamnya. Khusus untuk Limas penamaan di awali dengan posisi puncaknya terlebih dahulu sebagai contoh limas D.ABC berikut

$\text{B. 1. 2 Proyeksi Titik}$.

Perhatikanlah ilustrasi berikut

Ilustrasi di atas adalah suasana ketika tengah hari dan posisi matahari tepat di atas sehingga posisi matahari tepat memebentuk sudut ${90}^{\circ }$ terhadap permukaan bumi, maka bayangan benda akan tepat tegak lurus sebagaimana ilustrasi di atas. 

Ilustrasi di atas adalah sebagai gambaran proyeksi suatu benda pada benda lain. Proyeksi adalah bayangan yang terbentuk dari suatu bangun pada bidang datar dengan arah bayangan dengan bidang datar tersebut sebagai bidang proyeksi membentuk sudut ${90}^{\circ }$ jika dilukiskan.  

$\text{a. Proyeksi titik pada garis}$.

$\text{b. Proyeksi titik pada bidang}$.

$\text{c. Proyeksi garis pada bidang}$.

$\text{B. 1. 3 Alat Bantu Hitung Bangun Ruang}$.

$\text{a. Luas bangun datar beraturan}$.

$\begin{array}{|cll|}\hline \text{No}& \text{Bangun Datar}& \text{Luas}\\ 1& \text{Segitiga}& \begin{array}{rl}& {\displaystyle \frac{\text{alas}\times \text{tinggi}}{2}}\\ & \text{atau}\\ & \sqrt{s(s-a)(s-b)(s-c)}\\ & \text{dengan}\\ & s\text{adalah keliling segitiga}\\ & \text{atau}\\ & s=a+b+c\\ & \text{atau}\\ & \left[ABC\right]={\displaystyle \frac{1}{2}ab\sin \mathrm{\angle }C}\\ & \left[ABC\right]={\displaystyle \frac{1}{2}ac\sin \mathrm{\angle }B}\\ & \left[ABC\right]={\displaystyle \frac{1}{2}bc\sin \mathrm{\angle }A}\\ & \text{dengan}\\ & \left[ABC\right]=\text{luas segitiga ABC}\end{array}\\ 2& \text{Persegi}& \begin{array}{rl}& \text{sisi}\times \text{sisi}\\ & \text{atau}\\ & {\displaystyle \frac{\text{diagonal 1}\times \text{diagonal 2}}{2}}\end{array}\\ 3& \text{Persegi panjang}& \text{panjang}\times \text{lebar}\\ 4& \text{Belah ketupat}& {\displaystyle \frac{\text{diagonal 1}\times \text{diagonal 2}}{2}}\\ 5& \text{Layang-Layang}& {\displaystyle \frac{\text{diagonal 1}\times \text{diagonal 2}}{2}}\\ 6& \text{Trapesium}& {\displaystyle \frac{\text{jumlah sisi sejajar}\times \text{tinggi}}{2}}\\ 7& \text{Lingkaran}& \begin{array}{rl}& \pi \times {\text{jari-jari}}^2\\ & \text{atau}\\ & {\displaystyle \frac{1}{4}\times \pi \times {\text{diagonal}}^2}\end{array}\\ \hline\end{array}$

$\text{b. Dalil Pythagoras untuk segitiga siku-siku}$.

$\begin{array}{c}\\ \begin{array}{rl}& \text{Dalil/rumus Pythagoras}\\ & a^2+b^2=c^2\\ & \text{atau}\\ & c=\sqrt{a^2+b^2}\end{array}& \begin{array}{rl}& \sin \mathrm{\angle }ACB={\displaystyle \frac{a}{c}}\\ & \cos \mathrm{\angle }ACB={\displaystyle \frac{b}{c}}\\ & \tan \mathrm{\angle }ACB={\displaystyle \frac{a}{b}={\displaystyle \frac{\sin \mathrm{\angle }ACB}{\cos \mathrm{\angle }ACB}}}\\ & \csc \mathrm{\angle }ACB={\displaystyle \frac{c}{a}}\\ & \sec \mathrm{\angle }ACB={\displaystyle \frac{c}{b}}\\ & \cot \mathrm{\angle }ACB={\displaystyle \frac{b}{a}={\displaystyle \frac{\cos \mathrm{\angle }ACB}{\sin \mathrm{\angle }ACB}}}\end{array}\end{array}$

$\text{c. Identitas trigonometri pada segitiga siku-siku}$.

$\begin{array}{rl}& \text{Dalil/rumus Pythagoras}\\ & a^2+b^2=c^2\\ & \text{Perhatikan lagi gambar di poin c di atas}\\ & \begin{array}{|cl|}\hline 1.& \text{Rumus saat dibagi dengan}{c}^2\\ & {\displaystyle \frac{a^2}{c^2}+{\displaystyle \frac{b^2}{c^2}={\displaystyle \frac{c^2}{c^2}\iff {\displaystyle \frac{a^2}{c^2}+{\displaystyle \frac{b^2}{c^2}=1}}}}}\\ & \\ & \text{menjadi}{\left({\displaystyle \frac{a}{c}}\right)}^2+{\left({\displaystyle \frac{b}{c}}\right)}^2=1\\ & {\sin }^2\mathrm{\angle }ACB+{\cos }^2\mathrm{\angle }ACB=1\\ 2& \text{Rumus saat dibagi dengan}{b}^2\\ & {\displaystyle \frac{a^2}{b^2}+{\displaystyle \frac{b^2}{b^2}={\displaystyle \frac{c^2}{b^2}\iff {\displaystyle \frac{a^2}{b^2}+1={\displaystyle \frac{c^2}{b^2}}}}}}\\ & \\ & \text{menjadi}{\left({\displaystyle \frac{a}{b}}\right)}^2+1={\left({\displaystyle \frac{c}{b}}\right)}^2\\ & {\tan }^2\mathrm{\angle }ACB+1={\sec }^2\mathrm{\angle }ACB\\ 3& \text{Rumus saat dibagi dengan}{a}^2\\ & {\displaystyle \frac{a^2}{a^2}+{\displaystyle \frac{b^2}{a^2}={\displaystyle \frac{c^2}{a^2}\iff 1+{\displaystyle \frac{b^2}{a^2}={\displaystyle \frac{c^2}{a^2}}}}}}\\ & \\ & \text{menjadi}1+{\left({\displaystyle \frac{b}{a}}\right)}^2={\left({\displaystyle \frac{c}{a}}\right)}^2\\ & 1+{\cot }^2\mathrm{\angle }ACB={\csc }^2\mathrm{\angle }ACB\\ \hline\end{array}\end{array}$

$\text{d. Tabel trigonometri nilai sudut istimewa}$.

$\begin{array}{|ccccccc|}\hline \alpha & 0^{\circ }& 30& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }& {180}^{\circ }\\ \sin \alpha & 0& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}\sqrt{3}}& 1& 0\\ \cos \alpha & 1& {\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}}& 0& -1\\ \tan \alpha & 0& {\displaystyle \frac{1}{3}\sqrt{3}}& 1& \sqrt{3}& \text{TD}& 0\\ \hline\end{array}$.

$\text{e. Aturan sinus pada segitiga sebarang}$.

${\displaystyle \frac{BC}{\sin \mathrm{\angle }A}={\displaystyle \frac{AC}{\sin \mathrm{\angle }B}={\displaystyle \frac{AB}{\sin \mathrm{\angle }C}}}}$

$\text{f. Aturan cosinus pada segitiga sebarang}$.

Perhatikanlah gmabar pada poin e di atas, aturan cosinusnya adalah:

$\begin{array}{rl}\bullet & \cos \mathrm{\angle }A={\displaystyle \frac{b^2+c^2-a^2}{2bc}}\\ \bullet & \cos \mathrm{\angle }B={\displaystyle \frac{a^2+c^2-a^2}{2ac}}\\ \bullet & \cos \mathrm{\angle }C={\displaystyle \frac{a^2+b^2-a^2}{2ab}}\end{array}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui kubus ABCD.EFGH dengan rusuk 8 cm}.\\ & \text{Jika titik P, Q dan R berturut-turut terletak pada}\\ & \text{pertengahan garis AB, BC, dan bidang ADHE,}\\ & \text{tentukanlah jarak antara}\\ & \text{a}.\text{titik P ke titik R}\\ & \text{b}.\text{titik Q ke titik R}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah gambar kubus berikut}\end{array}$.

$.\begin{array}{rl}\text{a}.\text{PR}& =...(\text{gunakan rumus Pythagoras}\\ & \text{pada segitiga PAR, yaitu}:)\\ {\text{PR}}^2& ={\text{PA}}^2+{\text{AR}}^2\\ & \bullet \text{dengan PA}={\displaystyle \frac{1}{2}sisi=\frac{1}{2}.8=4cm}\\ & \bullet \text{AR}={\displaystyle \frac{1}{2}.\text{diagonal sisi kubus}=\frac{1}{2}\left(8\sqrt{2}\right)=4\sqrt{2}}\\ \text{seh}& \text{ingga panjang PR dapat dihitung, yaitu:}\\ {\text{PR}}^2& ={\text{PA}}^2+{\text{AR}}^2\\ \text{PR}& =\sqrt{{\text{PA}}^2+{\text{AR}}^2}\\ & =\sqrt{4^2+{\left(4\sqrt{2}\right)}^2}=\sqrt{16+16.2}=\sqrt{48}=\sqrt{16.3}\\ & =4\sqrt{3}cm\\ \text{b}\text{QR}& =.....\\ & \text{perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{rl}\text{QR}& =...(\text{gunakan rumus Pythagoras}\\ & \text{pada segitiga qR'R, yaitu}:)\\ {\text{QR}}^2& ={\text{(QR')}}^2+{\text{(R'R)}}^2\\ & \bullet \text{dengan QR'}=sisikubus=8cm\\ & \bullet \text{R'R}={\displaystyle \frac{1}{2}.\text{sisi kubus}=\frac{1}{2}\left(8\right)=4cm}\\ \text{seh}& \text{ingga panjang QR dapat dihitung, yaitu:}\\ {\text{QR}}^2& ={\text{(QR')}}^2+{\text{(R'R)}}^2\\ \text{QR}& =\sqrt{{\text{(QR')}}^2+{\text{(R'R)}}^2}\\ & =\sqrt{8^2+{\left(4\right)}^2}=\sqrt{64+16}=\sqrt{80}=\sqrt{16.5}\\ & =4\sqrt{5}cm\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui limas beraturan T.ABCD dengan }.\\ & \text{panjang semua rusuknya adalah sama yaitu }\\ & \text{8 cm. Jika titik P terletak pada pada rusuk }\\ & \text{tegak TB, tentukanlah jarak titik A ke titik P}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah gambar kubus berikut}\end{array}$.

Kita gambarkan posisi titik P yang berada di rusuk tegak TB, yaitu

Selanjutnya gambar kita partisi lagi supaya kita lebih fokus menjadi

Tampak bahwa dengan memandang sebuah segitiga ABP dalam segitiga besar ABT kita akan dengan mudah menentukan jarak titik A ke titik P dengan aturan cosinus.

Perhatikanlah gambar berikut

$\begin{array}{rl}\text{Aturan}& \text{Cosinus}:\\ \cos \mathrm{\angle }B& ={\displaystyle \frac{A{B}^2+B{P}^2-A{P}^2}{2.AB.AP}}\\ \cos {60}^{\circ }& ={\displaystyle \frac{8^2+4^2-x^2}{2.8.4}}\\ {\displaystyle \frac{1}{2}}& ={\displaystyle \frac{64+16-x^2}{64}}\\ 32& =80-x^2\\ x^2& =80-32\\ x& =\sqrt{80-32}=\sqrt{48}=\sqrt{16.3}\\ & =4\sqrt{3}cm\end{array}$.

Jadi, jarak titik A ke titik P adalah sebesar  $4\sqrt{3}$   cm.

$\begin{array}{l}\\ 3.& \text{Diketahui kubus ABCD.EFGH dengan rusuk}acm.\\ & \text{Jika titik P merupakan proyeksi titik C pada}\\ & \text{bidang AFH, tentukanlah jarak dari titik A }\\ & \text{ke titik S}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah gambar kubus berikut}\end{array}$.

$.\begin{array}{rl}\text{Diket}& \text{hui bahwa}:\\ \text{proye}& \text{ksi titik C pada bidang AFH adalah S}\\ \text{Ketik}& \text{a dilukiskan tampak sebagaimana ilustrasi}\\ \text{gamb}& \text{ar di atas. Dan diketahui pula, masih terkait}\\ \text{pemb}& \text{ahasan materi ini beberapa komponen ruas}\\ \text{garis}& \text{telah diketahui panjangnya, yaitu}:\\ \bullet AP& =PC={\displaystyle \frac{1}{2}a\sqrt{6}cm}\\ \bullet AC& =a\sqrt{2}cm\\ \bullet PQ& =acm\\ \bullet CS& =...cm(\text{belum diketahui, demikian pula})\\ \bullet AS& =...cm(\text{juga belum diketahui})\end{array}$.

$.\begin{array}{rl}\text{Pandang}& \text{lah}△APC\\ \text{dengan}& \text{menghitung luas}△APC,\text{kita akan}\\ \text{menentu}& \text{kan jarak titik C ke S, yaitu}:\\ \left[APC\right]& =\left[APC\right]\\ {\displaystyle \frac{alas\times t}{2}}& ={\displaystyle \frac{alas\times t}{2}\iff alas\times t=alas\times t}\\ \text{misalka}& \text{n}{t}_{CS}\text{adalah jarak titik C ke S}\\ ala{s}_{AP}\times & t_{CS}=ala{s}_{AC}\times t_{PQ}\\ t_{CS}& ={\displaystyle \frac{ala{s}_{AC}\times t_{PQ}}{ala{s}_{AP}}}\\ t_{CS}& ={\displaystyle \frac{AC\times PQ}{AP}}\\ & ={\displaystyle \frac{a\sqrt{2}\times a}{{\displaystyle \frac{1}{2}a\sqrt{6}}}=\frac{2a}{\sqrt{3}}}\\ & ={\displaystyle \frac{2a}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2a}{3}\sqrt{3}cm}\\ \text{Jelas jug}& \text{a bahwa}:\text{AP}\perp \text{CS},\text{maka}△ASC\\ \text{siku-siku}& \text{karena semua ini efek proyeksi, akibatnya}\\ \text{berlaku}& \text{rumus Pythagoras, yaitu}:\\ A{C}^2& =A{S}^2+S{C}^2\\ A{S}^2& =A{C}^2-S{C}^2\\ AS& =\sqrt{A{C}^2-S{C}^2}\\ & =\sqrt{{\left(a\sqrt{2}\right)}^2-{\left({\displaystyle \frac{2a}{\sqrt{3}}}\right)}^2}\\ & =\sqrt{2a^2-{\displaystyle \frac{4a^2}{3}}}\\ & =\sqrt{{\displaystyle \frac{6a^2-4a^2}{3}}}=\sqrt{{\displaystyle \frac{2a^2}{3}}}\\ & =\sqrt{{\displaystyle \frac{2a^2}{3}\times \frac{3}{3}}}=\frac{a}{3}\sqrt{6}cm\end{array}$

DAFTAR PUSTAKA

1. Sharma, S.N., dkk. 2017. Jelajah Matematika SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA.
2. Thohir, Ahmad. 2020. Geometri Ruang. http:https://ahmadthohir1089.wordpress.com/2020/07/25/geometri-ruang-xii-matematika-wajib/

Lanjutan Persamaan Trigonometri

$\begin{array}{rl}& \text{f. Menentukan Nilai Perbandingan Trigonometri}\\ & \text{pada Segitiga Siku-Siku}\end{array}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui}\tan \theta ={\displaystyle \frac{a}{x}}\\ & \text{Tentukanlah nilai}{\displaystyle \frac{x}{\sqrt{a^2+x^2}}}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah gambar segitiga AOX berikut}\end{array}$

$.\begin{array}{rl}& \text{Dengan rumus Pythagoras dapatr ditentukan}\\ & \text{panjang ruas}\text{AX, yaitu}:\\ & A{O}^2+O{X}^2=A{X}^2\\ & \text{atau}\\ & A{X}^2=A{O}^2+O{X}^2\\ & AX=\sqrt{A{O}^2+O{X}^2}\\ & =\sqrt{x^2+a^2},\\ & \text{maka}\\ & \bullet \sin \theta ={\displaystyle \frac{a}{\sqrt{x^2+a^2}}}\\ & \bullet \cos \theta ={\displaystyle \frac{x}{\sqrt{x^2+a^2}}}\\ & \text{Jadi, nilai}{\displaystyle \frac{x}{\sqrt{x^2+a^2}}=\cos \theta }\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}\sin \beta +\cos \beta ={\displaystyle \frac{6}{5},\text{tentukanlah}}\\ & \text{a}.\sin \beta \cos \beta \\ & \text{b}.{\sin }^3\beta +{\cos }^3\beta \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \sin \beta +\cos \beta ={\displaystyle \frac{6}{5}}\\ & \text{saat masing-masing ruas dikuadratkan,}\\ & \text{maka}\\ & {(\sin \beta +\cos \beta )}^2={\left({\displaystyle \frac{6}{5}}\right)}^2\\ & {\sin }^2\beta +2\sin \beta \cos \beta +{\cos }^2\beta ={\displaystyle \frac{36}{25}}\\ & {\sin }^2\beta +{\cos }^2\beta +2\sin \beta \cos \beta ={\displaystyle \frac{36}{25}}\\ & 1+2\sin \beta \cos \beta ={\displaystyle \frac{36}{25}}\\ & 2\sin \beta \cos \beta ={\displaystyle \frac{36}{25}-1}\\ & 2\sin \beta \cos \beta ={\displaystyle \frac{36-25}{25}=\frac{11}{25}}\\ & \sin \beta \cos \beta ={\displaystyle \frac{11}{50}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\sin }^3\beta +{\cos }^3\beta \\ & =(\sin \beta +\cos \beta )({\sin }^2\beta +{\cos }^2\beta -\sin \beta \cos \beta )\\ & =(\sin \beta +\cos \beta )(1-\sin \beta \cos \beta )\\ & =\left({\displaystyle \frac{6}{5}}\right).(1-{\displaystyle \frac{11}{50}})\\ & =\left({\displaystyle \frac{6}{5}}\right).\left({\displaystyle \frac{50-11}{50}}\right)\\ & =\left({\displaystyle \frac{6}{5}}\right).\left({\displaystyle \frac{39}{50}}\right)\\ & ={\displaystyle \frac{3\times 39}{5\times 25}=\frac{117}{125}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}\tan \alpha ={\displaystyle \frac{1}{\sqrt{7}},\text{tentukanlah}}\\ & \left({\displaystyle \frac{{\csc }^2\alpha -{\sec }^2\alpha }{{\csc }^2\alpha +{\sec }^2\alpha }}\right)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}:\\ \tan \alpha & ={\displaystyle \frac{1}{\sqrt{7}},\text{dan ingat juga bahwa}}\\ {\sec }^2\alpha & ={\tan }^2\alpha +1={\left({\displaystyle \frac{1}{\sqrt{7}}}\right)}^2+1=\frac{1}{7}+1=\frac{8}{7}\\ \text{Demik}& \text{ian juga},\cot \alpha ={\displaystyle \frac{1}{\tan \alpha }={\displaystyle \frac{1}{\left(\frac{1}{\sqrt{7}}\right)}=\sqrt{7},}}\\ \text{maka},& {\csc }^2\alpha ={\cot }^2\alpha +1={\left(\sqrt{7}\right)}^2+1=7+1=8\\ \text{Selanj}& \text{utnya}\\ & \left({\displaystyle \frac{{\csc }^2\alpha -{\sec }^2\alpha }{{\csc }^2\alpha +{\sec }^2\alpha }}\right)=\left({\displaystyle \frac{8-{\displaystyle \frac{8}{7}}}{8+{\displaystyle \frac{8}{7}}}}\right)\\ & ={\displaystyle \frac{{\displaystyle \frac{56-8}{7}}}{{\displaystyle \frac{56+8}{7}}}}\\ & ={\displaystyle \frac{48}{64}}\\ & ={\displaystyle \frac{3}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Jika}\beta \text{sudut lancip dan}\cos \beta ={\displaystyle \frac{3}{5},}\\ & \text{tentukan nilai dari}{\displaystyle \frac{\sin \beta \tan \beta -1}{2{\tan }^2\beta }}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \\ \cos \beta & ={\displaystyle \frac{3}{5}\Rightarrow {\sin }^2\beta +{\cos }^2\beta =1}\\ {\sin }^2\beta & +{\cos }^2\beta =1\\ \sin \beta & =\sqrt{1-{\cos }^2\beta }=\sqrt{1-{\left({\displaystyle \frac{3}{5}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{9}{25}}}=\sqrt{{\displaystyle \frac{16}{25}}}={\displaystyle \frac{4}{5}}\\ \text{Sehingga}& \tan \beta ={\displaystyle \frac{\sin \beta }{\cos \beta }=\frac{{\displaystyle \frac{4}{5}}}{{\displaystyle \frac{3}{5}}}=\frac{4}{3}}\\ & {\displaystyle \frac{\sin \beta \tan \beta -1}{2{\tan }^2\beta }={\displaystyle \frac{{\displaystyle \frac{4}{5}\times \frac{4}{3}-1}}{2{\left({\displaystyle \frac{4}{3}}\right)}^2}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{16}{15}-1}}{{\displaystyle \frac{32}{9}}}={\displaystyle \frac{{\displaystyle \frac{1}{15}}}{{\displaystyle \frac{32}{9}}}={\displaystyle \frac{9}{32\times 15}}}}\\ & ={\displaystyle \frac{3}{32\times 5}}\\ & ={\displaystyle \frac{3}{160}}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

Persamaan Trigonometri

 $\text{A. 1 Identitas Trigonometri}$.

A. 1. 1  Nilai Trigonometri Sudut

$\text{a. Perbandingan Trigonometri dalam Segitiga Siku-Siku}$.

Perhatikanlah ilustrasi sebuah segitiga siku-siku sama kaki berikut

Diketahui pula bahwa :

$\begin{array}{c}\bullet \sin {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}}\\ \bullet \cos {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}}\\ \bullet \tan {45}^{\circ }=1\end{array}$.

$\begin{array}{c}\bullet \csc {45}^{\circ }={\displaystyle \sqrt{2}}\\ \bullet \sec {45}^{\circ }={\displaystyle \sqrt{2}}\\ \bullet \cot {45}^{\circ }=1\end{array}$.

Berikut ilustrasi segitiga dengan sudut istimewa yang lain yaitu ${30}^{\circ }$ dan  ${60}^{\circ }$.

$\begin{array}{|cc|}\hline \begin{array}{c}\bullet \sin {30}^{\circ }={\displaystyle \frac{1}{2}}\\ \bullet \cos {30}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \bullet \tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}}\\ \bullet \sin {60}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \bullet \cos {60}^{\circ }={\displaystyle \frac{1}{2}}\\ \bullet \tan {30}^{\circ }={\displaystyle \sqrt{3}}\end{array}& \begin{array}{c}\bullet \csc {30}^{\circ }={\displaystyle 2}\\ \bullet \sec {30}^{\circ }={\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}\\ \bullet \cot {30}^{\circ }={\displaystyle \sqrt{3}}\\ \bullet \csc {60}^{\circ }={\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}\\ \bullet \sec {60}^{\circ }={\displaystyle 2}\\ \bullet \cot {30}^{\circ }={\displaystyle \frac{1}{3}\sqrt{3}}\end{array}\\ \hline\end{array}$

Perhatikan segitiga ABC siku-siku di C berikut

Perhatikanlah segitiga OAB berikut

$\begin{array}{rl}\text{a}.& \sin \alpha ={\displaystyle \frac{y}{r}}\\ \text{b}.& \cos \alpha ={\displaystyle \frac{x}{r}}\\ \text{c}.& \tan \alpha ={\displaystyle \frac{y}{x}}\\ \text{d}.& \csc \alpha ={\displaystyle \frac{r}{y}}\\ \text{e}.& \sec \alpha ={\displaystyle \frac{r}{x}}\\ \text{f}.& \cot \alpha ={\displaystyle \frac{x}{y}}\end{array}$.

A. 1. 2  Identitas Trigonometri Dasar

$\text{a. Dalil Pythagoras Segitiga Siku-Siku}$.

$\begin{array}{c}\\ \begin{array}{rl}& \text{Dalil/rumus Pythagoras}\\ & a^2+b^2=c^2\\ & \text{atau}\\ & c=\sqrt{a^2+b^2}\end{array}& \begin{array}{rl}& \sin \mathrm{\angle }ACB={\displaystyle \frac{a}{c}}\\ & \cos \mathrm{\angle }ACB={\displaystyle \frac{b}{c}}\\ & \tan \mathrm{\angle }ACB={\displaystyle \frac{a}{b}={\displaystyle \frac{\sin \mathrm{\angle }ACB}{\cos \mathrm{\angle }ACB}}}\\ & \csc \mathrm{\angle }ACB={\displaystyle \frac{c}{a}}\\ & \sec \mathrm{\angle }ACB={\displaystyle \frac{c}{b}}\\ & \cot \mathrm{\angle }ACB={\displaystyle \frac{b}{a}={\displaystyle \frac{\cos \mathrm{\angle }ACB}{\sin \mathrm{\angle }ACB}}}\end{array}\end{array}$

$\text{b. Identitas trigonometri pada segitiga siku-siku}$.

$\begin{array}{rl}& \text{Dalil/rumus Pythagoras}\\ & a^2+b^2=c^2\\ & \text{Perhatikan lagi gambar di poin c di atas}\\ & \begin{array}{|cl|}\hline 1.& \text{Rumus saat dibagi dengan}{c}^2\\ & {\displaystyle \frac{a^2}{c^2}+{\displaystyle \frac{b^2}{c^2}={\displaystyle \frac{c^2}{c^2}\iff {\displaystyle \frac{a^2}{c^2}+{\displaystyle \frac{b^2}{c^2}=1}}}}}\\ & \\ & \text{menjadi}{\left({\displaystyle \frac{a}{c}}\right)}^2+{\left({\displaystyle \frac{b}{c}}\right)}^2=1\\ & {\sin }^2\mathrm{\angle }ACB+{\cos }^2\mathrm{\angle }ACB=1\\ 2& \text{Rumus saat dibagi dengan}{b}^2\\ & {\displaystyle \frac{a^2}{b^2}+{\displaystyle \frac{b^2}{b^2}={\displaystyle \frac{c^2}{b^2}\iff {\displaystyle \frac{a^2}{b^2}+1={\displaystyle \frac{c^2}{b^2}}}}}}\\ & \\ & \text{menjadi}{\left({\displaystyle \frac{a}{b}}\right)}^2+1={\left({\displaystyle \frac{c}{b}}\right)}^2\\ & {\tan }^2\mathrm{\angle }ACB+1={\sec }^2\mathrm{\angle }ACB\\ 3& \text{Rumus saat dibagi dengan}{a}^2\\ & {\displaystyle \frac{a^2}{a^2}+{\displaystyle \frac{b^2}{a^2}={\displaystyle \frac{c^2}{a^2}\iff 1+{\displaystyle \frac{b^2}{a^2}={\displaystyle \frac{c^2}{a^2}}}}}}\\ & \\ & \text{menjadi}1+{\left({\displaystyle \frac{b}{a}}\right)}^2={\left({\displaystyle \frac{c}{a}}\right)}^2\\ & 1+{\cot }^2\mathrm{\angle }ACB={\csc }^2\mathrm{\angle }ACB\\ \hline\end{array}\end{array}$

$\text{c. Tabel trigonometri nilai sudut istimewa}$.

$\begin{array}{|ccccccc|}\hline \alpha & 0^{\circ }& 30& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }& {180}^{\circ }\\ \sin \alpha & 0& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}\sqrt{3}}& 1& 0\\ \cos \alpha & 1& {\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}}& 0& -1\\ \tan \alpha & 0& {\displaystyle \frac{1}{3}\sqrt{3}}& 1& \sqrt{3}& \text{TD}& 0\\ \hline\end{array}$.

$\text{d. Aturan sinus pada segitiga sebarang}$.

${\displaystyle \frac{BC}{\sin \mathrm{\angle }A}={\displaystyle \frac{AC}{\sin \mathrm{\angle }B}={\displaystyle \frac{AB}{\sin \mathrm{\angle }C}}}}$

$\text{e. Aturan cosinus pada segitiga sebarang}$.

Perhatikanlah gmabar pada poin e di atas, aturan cosinusnya adalah:

$\begin{array}{rl}\bullet & \cos \mathrm{\angle }A={\displaystyle \frac{b^2+c^2-a^2}{2bc}}\\ \bullet & \cos \mathrm{\angle }B={\displaystyle \frac{a^2+c^2-a^2}{2ac}}\\ \bullet & \cos \mathrm{\angle }C={\displaystyle \frac{a^2+b^2-a^2}{2ab}}\end{array}$.

$\begin{array}{rl}& \text{Macam-Macam Identitas Trigonometri Dasar}\\ & 1.\csc \alpha ={\displaystyle \frac{1}{\sin \alpha }5.\tan \alpha ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}}\\ & 2.\sec \alpha ={\displaystyle \frac{1}{\cos \alpha }6.{\tan }^2\alpha +1={\sec }^2\alpha }\\ & 3.\cot \alpha ={\displaystyle \frac{1}{\tan \alpha }7.{\cot }^2\alpha +1={\csc }^2\alpha }\\ & 4.\cot \alpha ={\displaystyle \frac{\cos \alpha }{\sin \alpha }8.{\sin }^2\alpha +{\cos }^2=1}\end{array}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa}\\ & \tan \alpha ={\displaystyle \frac{\sin \alpha \cos \alpha }{1-{\sin }^2\alpha }}\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}\tan \alpha & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}\\ & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }\times \frac{\cos \alpha }{\cos \alpha }}\\ & ={\displaystyle \frac{\sin \alpha \cos \alpha }{{\cos }^2\alpha }}\\ & ={\displaystyle \frac{\sin \alpha \cos \alpha }{1-{\sin }^2\alpha }◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1}{\sqrt{{\tan }^2\beta }}\times \sin \beta =\cos \beta }\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{\sqrt{{\tan }^2\beta }}\times \sin \beta }& ={\displaystyle \frac{1}{\tan \beta }\times \sin \beta }\\ & ={\displaystyle \frac{\cos \beta }{\sin \beta }\times \sin \beta }\\ & =\cos \beta ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{{\cos }^2\gamma }{1-\sin \gamma }=1+\sin \gamma }\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{{\cos }^2\gamma }{1-\sin \gamma }}& ={\displaystyle \frac{1-{\sin }^2\gamma }{1-\sin \gamma }}\\ & ={\displaystyle \frac{(1-\sin \gamma )(1+\sin \gamma )}{1-\sin \gamma }}\\ & =1+\sin \gamma ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }={\cos }^2\theta -{\sin }^2\theta }\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }}& ={\displaystyle \frac{1-{\tan }^2\theta }{{\sec }^2\theta }={\displaystyle \frac{1-{\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }}}{{\displaystyle \frac{1}{{\cos }^2\theta }}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta }}}{{\displaystyle \frac{1}{{\cos }^2\theta }}}}\\ & ={\cos }^2\theta -{\sin }^2\theta ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tunjukkan bahwa}\\ & {\cos }^4\alpha -{\sin }^4\alpha =1-2{\sin }^2\alpha \\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\cos }^4\alpha -{\sin }^4\alpha & ={({\cos }^2\alpha )}^2-{({\sin }^2\alpha )}^2\\ & =({\cos }^2\alpha -{\sin }^2\alpha )({\cos }^2\alpha +{\sin }^2\alpha )\\ & =({\cos }^2\alpha -{\sin }^2\alpha )\times 1\\ & ={\cos }^2\alpha -{\sin }^2\alpha \\ & =(1-{\sin }^2\alpha )-{\sin }^2\alpha \\ & =1-2{\sin }^2\alpha ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{\sin \beta \sec \beta }{{\sin }^2\beta -{\tan }^2\beta }=-{\displaystyle \frac{\cos \beta }{{\sin }^3\beta }}}\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{\sin \beta \sec \beta }{{\sin }^2\beta -{\tan }^2\beta }}& ={\displaystyle \frac{\sin \beta \left({\displaystyle \frac{1}{\cos \beta }}\right)}{{\sin }^2\beta -{\displaystyle \frac{{\sin }^2\beta }{{\cos }^2\beta }}}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{\sin \beta }{\cos \beta }}\right)}{{\sin }^2\beta (1-{\displaystyle \frac{1}{{\cos }^2\beta }})}\times \frac{{\cos }^2\beta }{{\cos }^2\beta }}\\ & ={\displaystyle \frac{\sin \beta \cos \beta }{{\sin }^2\beta ({\cos }^2\beta -1)}}\\ & ={\displaystyle \frac{\cos \beta }{\sin \beta (-{\sin }^2\beta )}}\\ & =-{\displaystyle \frac{\cos \beta }{{\sin }^3\beta }◼}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Noormandiri, B. K. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
2. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

Lanjutan Limit Fungsi Trigonometri

 $\text{C. Menentukan Nilai Limit Fungsi Trigonometri}$

Dalam bahasan ini yang akan dibahas adalah nilai limit mendekati $a$ atau nilai $x$ di sekitar $a$. Ada 3 cara yang populer digunakan untuk menentukan nilai limit fungsi trigonometri ini dengan salah satunya yang paling sering digunakan adalah substitusi langsung di antara cara-cara penyelesaian lainnya. Jika dengan cara substitusi langsung nantinya mendfapatkan nilai bentuk tak tentu yaitu ${\displaystyle \frac{0}{0}}$, maka cara Anda harus menggunakan cara yang lainnya sampai Anda temukan nilai limitnya. Selanjutnya 3 cara yang dimaksud di atas adalah sebagai berikut:

$\text{C. 1 dengan substitusi langsung}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai limit dari fungsi-sungsi}\\ & \text{berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}(\sin x+\tan x)\\ \text{b}& \underset{x\to \pi }{\text{lim}}\left({\displaystyle \sin x+\cos x}\right)\\ \text{c}& \underset{x\to {\displaystyle \frac{\pi }{4}}}{\text{lim}}\left({\displaystyle \frac{\sin x+\cos x}{\tan x}}\right)\\ \text{d}& \underset{x\to {\displaystyle 0}}{\text{lim}}\left({\displaystyle \frac{1+\cos 2x}{1+2\cos x}}\right)\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{De}& \text{ngan substitusi langsung didapatkan}\\ \text{a}.& \underset{x\to 0}{\text{lim}}(\sin x+\tan x)\\ & =\sin 0+\tan 0=0+0=0\\ \text{b}.& \underset{x\to \pi }{\text{lim}}\left({\displaystyle \sin x+\cos x}\right)\\ & =\sin \pi +\cos \pi =0+(-1)=-1\\ \text{c}.& \underset{x\to {\displaystyle \frac{\pi }{4}}}{\text{lim}}\left({\displaystyle \frac{\sin x+\cos x}{\tan x}}\right)\\ & =\left({\displaystyle \frac{\sin {\displaystyle \frac{\pi }{4}+\cos {\displaystyle \frac{\pi }{4}}}}{\tan {\displaystyle \frac{\pi }{4}}}}\right)={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{2}+{\displaystyle \frac{1}{2}\sqrt{2}}}}{1}}\\ & ={\displaystyle \frac{\sqrt{2}}{1}=\sqrt{2}}\\ \text{d}.& \underset{x\to 0}{\text{lim}}\left({\displaystyle \frac{1+\cos 2x}{1+2\cos x}}\right)\\ & =\left({\displaystyle \frac{1+\cos 2(0)}{1+2\cos (0)}}\right)={\displaystyle \frac{1+1}{1+2.1}={\displaystyle \frac{2}{3}}}\end{array}\end{array}$

$\text{C. 2 dengan menyederhanakan}$

Langkah ini ditempuh setelah langkah substitusi langsung tidak memungkinkan atau ketemu bentuk tak tentu  ${\displaystyle \frac{0}{0}}$.

Gunakanlah identitas-identitas trigonometri yang Anda dapatkan di kelas XI  dan akan sering digunakan nantinya di antaranya, yaitu:

$\begin{array}{rl}\bullet \sin 2x& =2\sin x\cos x\\ \bullet \cos 2x& ={\cos }^{2}x-{\sin }^{2}x\\ & =1-2{\sin }^{2}x\\ & =2{\cos }^{2}x-1\\ \bullet \tan 2x& ={\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}\end{array}$

Demikian juga

$\begin{array}{rl}\bullet \sin A+\sin B& =2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\\ \bullet \sin A-\sin B& =2\cos \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)\\ \bullet \cos A+\cos B& =2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\\ \bullet \cos A-\cos B& =-2\sin \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)\end{array}$.

Masih banyak bentuk identitas trigonometri selain di atas, karenanya sekiranya perlu maka hafalkanlah

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai limit dari fungsi-sungsi}\\ & \text{berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2x}{\sin x}}\\ \text{b}& \underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{\sin 4x}{\sin x\cos x}}\\ \text{c}& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{{\tan }^{2}x}}\\ \text{d}& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{1-\cos x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{De}& \text{ngan menyederhankan bentuk trigonometri}\\ \text{ak}& \text{an didapatkan bentuk yang lebih sederhana}\\ \text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2x}{\sin x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{\sin 0}{\sin 0}={\displaystyle \frac{0}{0},\text{maka perlu disederhanakan}}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2x}{\sin x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2\sin x\cos x}{\sin x}=\underset{x\to 0}{\text{lim}}{\displaystyle 2\cos x}}}\\ & \text{baru digunakan substitusinya}\\ & =2\cos 0=2.1=2\\ \text{b}.& \underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{\sin 4x}{\sin x\cos x}}\\ & \text{Sama seperti langkah di atas, yaitu}:\\ & =\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{\sin 4x}{\sin x\cos x}=\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{2\sin 2x\cos 2x}{\sin x\cos x}}}\\ & =\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{4\sin x\cos x\cos 2x}{\sin x\cos x}=\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle 4\cos 2x}}\\ & =4\cos 2\left({\displaystyle \frac{\pi }{2}}\right)=4\cos \pi =4.(-1)=-4\\ \text{c}.& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{{\tan }^{2}x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{1-{\cos }^{2}0}{{\tan }^{2}0}={\displaystyle \frac{1-1}{0}=\frac{0}{0},}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{{\tan }^{2}x}=\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{\left({\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}\right)}}}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle {\cos }^{2}x={\cos }^{2}0=1^2=1}\\ \text{d}.& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{1-\cos x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{{\sin }^{2}0}{1-\cos 0}={\displaystyle \frac{0}{1-1}=\frac{0}{0},}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{1-\cos x}=\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{1-\cos x}}}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{(1+\cos x)(1-\cos x)}{1-\cos x}}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle (1+\cos x)}\\ & =1+\cos 0=1+1=2\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai limit dari fungsi-sungsi}\\ & \text{berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos x-\cos 3x}{1-\cos 2x}}\\ \text{b}& \underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{De}& \text{ngan menyederhankan bentuk trigonometri}\\ \text{ak}& \text{an didapatkan bentuk yang lebih sederhana}\\ \text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos x-\cos 3x}{1-\cos 2x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{\cos 0-\cos 0}{1-\cos 0}={\displaystyle \frac{1-1}{1-1}={\displaystyle \frac{0}{0},}}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos x-\cos 3x}{1-\cos 2x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{-2\sin \left({\displaystyle \frac{x+3x}{2}}\right)\sin \left({\displaystyle \frac{x-3x}{2}}\right)}{1-(1-2{\sin }^{2}x)}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{-2\sin 2x\sin (-x)}{2{\sin }^{2}x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2\sin 2x\sin x}{2\sin x.\sin x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2(2\sin x\cos x)\sin x}{2\sin x.\sin x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle 2\cos x}\\ & \text{baru digunakan substitusinya}\\ & ={\displaystyle 2\cos 0=2.1=2}\\ \text{b}.& \underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{1-\tan x}{\sin x-\cos x}=\frac{1-1}{\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}}={\displaystyle \frac{0}{0},}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}=\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\left({\displaystyle \frac{\sin x}{\cos x}}\right)}{\sin x-\cos x}}}\\ & =\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{-(\sin x-\cos x)}{\cos x(\sin x-\cos x)}}\\ & =\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{-1}{\cos x}={\displaystyle \frac{-1}{\cos \left({\displaystyle \frac{\pi }{4}}\right)}={\displaystyle \frac{-1}{\frac{1}{\sqrt{2}}}=-\sqrt{2}}}}\end{array}\end{array}$

$\text{C. 3 dengan rumus limit fungsi trigonometri}$

Berikut rumus limit fungsi trigonometri yang akan kita gunakan

$\begin{array}{rl}\bullet \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin ax}{ax}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{ax}{\sin ax}=\frac{a}{a}=1}\\ \bullet \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan ax}{ax}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{ax}{\tan ax}=\frac{a}{a}=1}\end{array}$.

BUKTINYA ada di sini

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 4x}{7x}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{2x}{\tan 9x}}\\ \text{c}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 8x}{\sin 3x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 4x}{7x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{4}{7}\times {\displaystyle \frac{\sin 4x}{4x}={\displaystyle \frac{4}{7}}}}}\\ \text{b}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{2x}{\tan 9x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2}{9}\times {\displaystyle \frac{9x}{\tan 9x}={\displaystyle \frac{2}{9}}}}}\\ \text{c}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 8x}{\sin 3x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 8x}{8x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{3x}{\sin 3x}\times \frac{8}{3}={\displaystyle \frac{8}{3}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{24x^2}{8{\sin }^{2}x}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{5x^2}{15\tan (-9x)\sin 3x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{24x^2}{8{\sin }^{2}x}={\displaystyle \frac{24}{8}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}}}}}\\ & =3\times 1\times 1\\ & =3\\ \text{b}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{5x^2}{15\tan (-9x)\sin 3x}}\\ & ={\displaystyle \frac{5}{15}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan (-9x)}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan 3x}}}}\\ & ={\displaystyle \frac{1}{3}\times (-\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{9x}{\tan 9x}\times \frac{1}{9}})\times \times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{3x}{\tan 3x}\times \frac{1}{3}}}\\ & =-{\displaystyle \frac{1}{3}\times \frac{1}{9}\times \frac{1}{3}=-\frac{1}{81}}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-\cos 6x}{x^2}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos 4x-\cos 2x}{x^2}}\\ \text{c}& \underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x^2-5x+6)\sin (x-3)}{{(x^2-7x+12)}^2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-\cos 6x}{x^2}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-(1-2{\sin }^{2}3x)}{x^2}}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2{\sin }^{2}3x}{x^2}=2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{x}}}}\\ & =2\times 3\times 3=18\\ \text{b}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos 4x-\cos 2x}{x^2}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{-2\sin \left({\displaystyle \frac{4x+2x}{2}}\right)\sin \left({\displaystyle \frac{4x-2x}{2}}\right)}{x^2}}\\ & =-2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x.\sin x}{x^2}}\\ & =-2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{x}=-2\times 3\times 1=-6}}\\ \text{c}.& \underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x^2-5x+6)\sin (x-3)}{{(x^2-7x+12)}^2}}\\ & =\underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x-2)(x-3)\sin (x-3)}{{((x-3)(x-4))}^2}}\\ & =\underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x-2)(x-3)\sin (x-3)}{(x-3)(x-3)(x-4{)}^2}}\\ & =\underset{x\to 3}{\text{lim}}{\displaystyle {\displaystyle \frac{x-2}{(x-4{)}^2}\times \underset{x\to 3}{\text{lim}}{\displaystyle \frac{\sin (x-3)}{(x-3)}}}}\\ & ={\displaystyle \frac{(3-2)}{(3-4{)}^2}\times 1={\displaystyle \frac{1}{(-1{)}^2}=\frac{1}{1}=1}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tentukanlah nilai limit dari}\underset{x\to \frac{1}{2}}{\text{lim}}{\displaystyle \frac{\sin (4x-2)}{\tan 2x-1}}\\ \\ & \text{Jawab}:\\ & \text{Kita misalkan}a=2x-1,\\ & \text{ketika}x\to {\displaystyle \frac{1}{2},\text{maka akan didapatkan}a\to 0}\\ & (\text{textbf{dibaca}: saat nilai}x\text{mendekati}{\displaystyle \frac{1}{2},\text{maka nilai}}\\ & a\text{akan mendekati nilai}0).\\ & \text{Selanjutnya kita buatkan penyesuaian, yaitu}:\\ & \begin{array}{rl}\underset{x\to \frac{1}{2}}{\text{lim}}{\displaystyle \frac{\sin (4x-2)}{\tan 2x-1}}& =\underset{x\to \frac{1}{2}}{\text{lim}}{\displaystyle \frac{\sin 2(2x-1)}{\tan 2x-1}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2a}{\tan a}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2a}{2a}\times 2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{a}{\tan a}}}\\ & =2\times 1=2\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selidikilah limit fungsi berikut, apakah}\\ & \text{memiliki nilai limit atau tidak}\\ & \text{a}.\underset{x\to 0}{\text{lim}}2x-1\\ & \text{b}.\underset{x\to 0}{\text{lim}}{x}^2-x-2\\ & \text{c}.\underset{x\to 1}{\text{lim}}f(x),\text{dengan}f(x)=\{\begin{array}{ll}2x& ;x<1\\ 4x-1& ;x\ge 1\end{array}\\ & \text{ d}.\underset{x\to 2}{\text{lim}}\sqrt{f(x)},\text{dengan}f(x)=\{\begin{array}{ll}4x-1& ;x<2\\ 2x+5& ;x\ge 2\end{array}\\ & \text{e}.\underset{x\to 0}{\text{lim}}(x+\cos x)\\ & \text{f}.\underset{x\to 0}{\text{lim}}x\tan x\\ & \text{g}.\underset{x\to {\displaystyle \frac{\pi }{2}}}{\text{lim}}(\sin x+2\cos x)\\ & \text{h}.\underset{x\to {\displaystyle \frac{\pi }{2}}}{\text{lim}}(2\tan x-\sin 2x)\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 5x}{x}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{4x}{\sin x}}\\ \text{c}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 6x}{8x}}\\ \text{d}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{2x}{\tan 7x}}\\ \text{e}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{\sin 2x}}\\ \text{f}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 4x}{\tan 8x}}\\ \text{9}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{{\sin }^{2}5x}{2x^2}}\\ \text{h}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 5x\tan 6x}{x\tan 7x}}\\ \text{i}& \underset{x\to y}{\text{lim}}{\displaystyle \frac{\sin x-\sin y}{x-y}}\\ \text{j}& \underset{x\to 2}{\text{lim}}{\displaystyle \frac{\sin (x^2-4)}{x-2}}\\ \text{k}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin mx-\sin nx}{\cos mx-\cos nx}}\\ \text{l}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-\cos 3x\cos x}{x^2}}\end{array}\\ \end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
3. Yuana, A.R., Indriyastuti. 2017. Perspektif Matematika untuk Kelas XII SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI PUSTAKA MANDIRI.

Limit Fungsi Trigonometri

 $\text{A. Pendahuluan}$

Mengingat kembali definisi limit yang telah dipelajari sebelumnya di kelas XI, yaitu limit fungsi aljabar $f(x)$ yang didefinisikan dengan:

$\begin{array}{rl}\underset{x\to a}{\text{lim}}f(x)=L& \mathbf{\text{adalah}}\text{Jika}x\text{mendekati}a\\ & \text{dengan tidak sama dengan}a,\\ & \text{maka nilai}f(x)\text{mendekati}L\end{array}$.

Perhatikan definisi di atas istilah  $x\text{mendekati}a$ dituliskan dengan simbol  $(x\to a)$. Suatu nilai limit dianggap ada jika nilai $f(x)$ mendekati  $a$ dari arah kiri sama dengan nilai $f(x)$ mendekati  $a$ dari arah kanan dengan nilai yang sama misalnya $L$. Jika disimbolkan pernyataan ini menjadi berikut

$\begin{array}{rl}\underset{x\to a^{-}}{\text{lim}}f(x)=\underset{x\to a^{+}}{\text{lim}}f(x)=\underset{x\to a}{\text{lim}}f(x)=L& \end{array}$.

$\begin{array}{rl}\text{Perlu di}& \text{perhatikan bahwa didekati dari}\\ \bullet \mathbf{\text{kiri}}& \text{disimbolkan dengan}\underset{x\to a^{-}}{\text{lim}}f(x),\text{dan}\\ \bullet \mathbf{\text{kan}}& \mathbf{\text{an}}\text{disimbolkan dengan}\underset{x\to a^{+}}{\text{lim}}f(x)\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai limit dari}f(x)={\displaystyle \frac{x^2-4}{x-2}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika fungsi}{\displaystyle \frac{x^2-4}{x-2}}\\ & \text{di sekitar}x=2\text{sebagaimana dalam tabel}\\ & \text{berikut}\end{array}\end{array}$.

$\begin{array}{rl}.& \text{Jadi, nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}=2\text{atau dapat dikatakan}}\\ & \text{nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}\mathbf{\text{ada}}}\\ & \text{meskipun nilai substitusi langsung}x=2\text{yaitu}\\ & f(0)={\displaystyle \frac{0^2-0}{0-0}=\frac{0}{0}\text{berupa bentuk tak}}\\ & \text{tentu. Berikut ilustrasinya}\end{array}$

$\begin{array}{l}\\ 2.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 5}{\text{lim}}f(x),\text{untuk}f(x)=\{\begin{array}{ll}x& \text{saat}x<5\\ 5-x& \text{saat}x\ge 5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 5^{-},\text{maka}\underset{x\to 5^{-}}{\text{lim}}f(x)=x\\ & \text{atau}\underset{x\to 5^{-}}{\text{lim}}f(x)=5\\ & \text{boleh juga dituliskan dengan}\\ & \underset{x\to 5^{-}}{\text{lim}}f(x)=x=5.\text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 5^{+},\\ & \text{maka}\underset{x\to 5^{+}}{\text{lim}}f(x)=\underset{x\to 5^{+}}{\text{lim}}(5-x)=5-5=0.\\ & \text{Karena nilai}\underset{x\to 5^{-}}{\text{lim}}f(x)\ne \underset{x\to 5^{+}}{\text{lim}}f(x),\text{maka}\\ & \text{nilai atau harga}\underset{x\to 5}{\text{lim}}f(x)\mathbf{\text{tidak ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 0}{\text{lim}}f(x),\text{untuk}f(x)=\cos x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 0^{-},\text{maka}\underset{x\to 0^{-}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& -0,5& -0,4& -0,3& -0,2& -0,1& 0\\ \cos x& ...& ...& 0,999986& 0,999994& 0,9999985& 1\\ \hline\end{array}\\ & \text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 0^{+},\\ & \text{maka}\underset{x\to 0^{+}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& 0& 0,1& 0,2& 0,3& 0,4& 0,5\\ \cos x& 1& 0,9999985& 0,999994& 0,999986& ...& ...\\ \hline\end{array}\\ & \text{Karena nilai}\underset{x\to 0^{-}}{\text{lim}}f(x)=\underset{x\to 0^{+}}{\text{lim}}f(x)=1,\text{maka}\\ & \text{nilai}\underset{x\to 0}{\text{lim}}\cos x\mathbf{\text{ ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\text{B. Sifat-Sifat Limit Fungsi}$

$\begin{array}{rl}& \text{Misalkan}f\text{dan}g\text{adalah fungsi-fungsi yang}\\ & \text{mempunyai nilai limit di titik sekitar}x=a\\ & \text{atau}(x\to a)\text{dan}c\text{adalah suatu konstanta}\\ & \text{serta}n\text{adalah suatu bilangan bulat positif},\\ & \text{maka berlaku sifat-sifat berikut}:\\ & \begin{array}{l}\\ 1.& \underset{x\to a}{\text{lim}}{\displaystyle c=c}\\ 2.& \underset{x\to a}{\text{lim}}{\displaystyle x^n=a^n}\\ 3.& \underset{x\to a}{\text{lim}}c.f(x)=c.\underset{x\to a}{\text{lim}}f(x)\\ 4.& \underset{x\to a}{\text{lim}}(f(x)\pm g(x))=\underset{x\to a}{\text{lim}}f(x)\pm \underset{x\to a}{\text{lim}}g(x)\\ 5.& \underset{x\to a}{\text{lim}}(f(x)\times g(x))=\underset{x\to a}{\text{lim}}f(x)\times \underset{x\to a}{\text{lim}}g(x)\\ 6.& \underset{x\to a}{\text{lim}}{\displaystyle \frac{f(x)}{g(x)}={\displaystyle \frac{\underset{x\to a}{\text{lim}}f(x)}{\underset{x\to a}{\text{lim}}g(x)}}}\\ 7.& \underset{x\to a}{\text{lim}}{(f(x))}^n={[\underset{x\to a}{\text{lim}}f(x))]}^n\\ 8.& \underset{x\to a}{\text{lim}}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)},\text{dengan}\underset{x\to a}{\text{lim}}f(x)\ge 0\\ & \text{dan}n\text{genap}\end{array}\end{array}$

Fungsi Eksponen

  $\text{A. Bilangan Pangkat Positif}$

Misalkan diketahui bahwa $a$ adalah suatu bilangan tidak nol dan $m$ adalah bilangan asli, maka bilangan ekponen atau bilangan berpangkat dedefinisikan dengan:

$a^m=\underset{m}{\underbrace{a\times a\times \times a\times ...\times a}}$

$\begin{array}{rl}\text{Bilangan}& :\\ a& \text{disebut basis atau bilangan pokok}\\ n& \text{disebut sebagai bilangan pangkat/eksponen}\end{array}$.

$\text{ CONTOH SOAL}$.

$(1).3^4=3\times 3\times 3\times 3=81$

$(2).5^4=5\times 5\times 5\times 5=625$

$(3).2^6=2\times 2\times 2\times 2\times 2\times 2=64$

$(4).6^7=6\times 6\times 6\times 6\times 6\times 6\times 6=279936$

$(5).(-3{)}^3=(-3)\times (-3)\times (-3)=-27$

$(6).(-2{)}^4=(-2)\times (-2)\times (-2)\times (-2)=16$

$(7).{\left({\displaystyle \frac{1}{5}}\right)}^3=\left({\displaystyle \frac{1}{5}}\right)\times \left({\displaystyle \frac{1}{5}}\right)\times \left({\displaystyle \frac{1}{5}}\right)={\displaystyle \frac{1}{125}}$

$(8).{(-{\displaystyle \frac{1}{2}})}^3=(-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})=-{\displaystyle \frac{1}{8}}$

$\text{B. Sifat-Sifat Bilangan Pangkat Positif}$

$\begin{array}{r}\\ 1.& a^m.a^n=a^{m+n}\\ 2.& a^m:a^n=a^{m-n}\\ 3.& {\left(a^m\right)}^n=a^{m.n},\text{syarat}a\ne 0\\ 4.& {\left(ab\right)}^n=a^n.b^n\\ 5.& {\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n},\text{syarat}b\ne 0\end{array}$

Beberpa hal yang perlu diketahui juga, yaitu

$\begin{array}{r}\\ 1.& (a+b{)}^2=a^2+2ab+b^2\\ 2.& (a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3\\ 3.& {(a+\frac{1}{a})}^2=a^2+2+{\displaystyle \frac{1}{a^2},\text{syarat}a\ne 0}\end{array}$

$\text{ CONTOH SOAL}$

$\begin{array}{r}\\ (1).& 2^6\times 2^4\times 2^7=2^{6+4+7}=2^{17}\\ (2).& 2^5\times 3^5\times 7^5={(2.3.7)}^5={\left(42\right)}^5\\ (3).& {\displaystyle \frac{a^3.a^7.a^6}{a^9}={\displaystyle \frac{a^{3+7+6}}{a^9}=\frac{a^{16}}{a^9}=a^{16-9}=a^7,\text{syarat}a\ne 0}}\end{array}$

$\begin{array}{rl}(4).{\displaystyle \frac{3^7{.7}^3.2}{{\left(42\right)}^3}}& =\frac{2^1{.3}^7{.7}^3}{{\left(2.3.7\right)}^3}=\frac{2^1{.3}^7{.7}^3}{2^3{.3}^3{.7}^3}\\ & =2^{1-3}{.3}^{7-3}{.7}^{3-3}=2^{-2}{.3}^4{.7}^0\\ & =\frac{1}{2^2}{.3}^4.1=\frac{3^4}{2^2}\end{array}$

$\begin{array}{rl}(5).{\displaystyle \frac{2^{2020}+2^{2021}+2^{2022}}{7}}& ={\displaystyle \frac{{1.2}^{2020}+2^1{.2}^{2020}+2^2{.2}^{2020}}{7}}\\ & =\frac{(1+2+4){.2}^{2020}}{7}\\ & =\frac{{7.2}^{2020}}{7}\\ & =2^{2020}\end{array}$

$\begin{array}{rl}(6){\displaystyle \frac{{\left(2^{n+2}\right)}^2-2^2{.2}^{2n}}{2^n{.2}^{n+2}}}& ={\displaystyle \frac{2^{2(n+2)}-2^2{.2}^{2n}}{2^n{.2}^n{.2}^2}}\\ & ={\displaystyle \frac{2^{2n}{.2}^{2.2}-2^2{.2}^{2n}}{2^{n+n}{.2}^2}}\\ & ={\displaystyle \frac{2^{2n}(2^4-2^2)}{2^{2n}{.2}^2}}\\ & ={\displaystyle \frac{(2^4-2^2)}{2^2}=\frac{16-4}{4}}\\ & ={\displaystyle \frac{12}{4}=3}\end{array}$

$\text{C. Bentuk Akar}$

Bilangan bentuk akar di sini adalah kebalikan dari bilangan bentuk pangkat. Bilangan bentuk akar selanjutnya disebut bilangan irasional. Sebagai contoh $\sqrt{2}$, $\sqrt{3}$, $\sqrt{8}$, $\sqrt[3]{3}$, $\sqrt[3]{4}$, $\sqrt[3]{7}$ dan tapi ingat $\sqrt{4}$ dan  $\sqrt[3]{8}$ serta  $\sqrt[3]{27}$ adalah bukan bentuk akar, karena nantinya akan menghasilkan masing-masing 2 dan 3 serta 3.

$\begin{array}{rl}& \\ 1.& a^{\frac{1}{n}}=\sqrt[n]{a}\\ 2.& a^{\frac{m}{n}}=\sqrt[n]{a^m}\\ 3.& a^{\frac{1}{2}}=\sqrt[2]{a^1}=\sqrt{a}\end{array}$.

$\text{Cara membaca}$.

$\begin{array}{rl}1.& \sqrt[n]{p}\mathbf{\text{dibaca}}\text{akar pangkat n dari p}\\ 2.& \sqrt[n]{p^2}\mathbf{\text{dibaca}}\text{akar pangkat n dari p kuadrat}\\ 3.& \sqrt[n]{p^3}\mathbf{\text{dibaca}}\text{akar pangkat n dari p pangkat tiga}\\ 4.& \sqrt{p}\mathbf{\text{dibaca}}\text{akar dari p}\text{atau}\\ & \text{akar kuadrat dari p}\\ & \text{ingat bahwa}:\sqrt{p}=\sqrt[2]{p}\end{array}$.

$\begin{array}{rl}\text{Defini}& \text{si}\\ \text{Jika}& a\text{dan}b\text{bilangan real dan}\\ & n\text{bilangan bulat positif, maka}:\\ & a^n=b\iff \sqrt[n]{b}=a\\ \text{keter}& \text{angan}:\\ \sqrt[n]{b}& \text{disebut}\mathbf{\text{akar (radikal)}}\\ b& \text{disebut}\mathbf{\text{radikan}}\\ & \text{(bilangan pokok yang ditarik akarnya)}\\ n& \text{disebut}\mathbf{\text{indeks}}\\ & (\text{pangkat akar})\end{array}$.

$\text{C.1 Bilangan Pangkat Pecahan}$.

Operasi Bilangan pangkat pecahan sama dengan operasi pangkat bilangan bulat.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& a^{{.}^{\frac{1}{2}}}\times a^{{.}^{\frac{1}{3}}}=a^{{.}^{\frac{1}{2}+\frac{1}{3}}}=a^{{.}^{\frac{5}{6}}}\\ 2.& a^{{.}^{\frac{1}{5}}}:a^{{.}^{\frac{1}{3}}}=a^{{.}^{\frac{1}{5}-\frac{1}{3}}}=a^{{.}^{-\frac{2}{15}}}\\ 3.& {\left(a^{{.}^{\frac{2}{5}}}\right)}^{\frac{4}{7}}=a^{{.}^{\frac{8}{35}}}\\ 4.& {81}^{{.}^{\frac{1}{2}}}={\left(9^2\right)}^{{.}^{\frac{1}{2}}}=9^1=9\\ 5.& {27}^{{.}^{-\frac{2}{3}}}={\left(3^3\right)}^{{.}^{-\frac{2}{3}}}={\left(3\right)}^{-2}={\displaystyle \frac{1}{3^2}=\frac{1}{9}}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakanlah bentuk berikut dan}\\ & \text{nyatakan hasilnya dalam pangkat positif}\\ \\ & \text{a}.\left(3p^{{.}^{\frac{5}{3}}}{q}^{{.}^{-\frac{3}{4}}}\right)\left(2p^{{.}^{-\frac{2}{3}}}{q}^{{.}^{\frac{5}{4}}}\right)\\ \\ & \text{b}.{\displaystyle \frac{\left(8p^{{.}^{\frac{2}{3}}}{q}^0{r}^{{.}^{-\frac{1}{2}}}\right)}{\left(4p^{{.}^{-\frac{1}{2}}}{q}^{{.}^{-\frac{1}{3}}}r\right)}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \left(3p^{{.}^{\frac{5}{3}}}{q}^{{.}^{-\frac{3}{4}}}\right)\left(2p^{{.}^{-\frac{2}{3}}}{q}^{{.}^{\frac{5}{4}}}\right)\\ & =3.2.p^{{.}^{\frac{5}{3}+(-\frac{2}{3})}}.q^{{.}^{-\frac{3}{4}+\frac{5}{4}}}\\ & =6.p^{{.}^{\frac{3}{3}}}{q}^{{.}^{\frac{2}{4}}}\\ & =6p{q}^{{.}^{\frac{1}{2}}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\displaystyle \frac{\left(8p^{{.}^{\frac{2}{3}}}{q}^0{r}^{{.}^{-\frac{1}{2}}}\right)}{\left(4p^{{.}^{-\frac{1}{2}}}{q}^{{.}^{-\frac{1}{3}}}r\right)}}\\ & =2.p^{{.}^{\frac{2}{3}-(-\frac{1}{2})}}{q}^{{.}^0-(-\frac{1}{3})}{r}^{{.}^{-\frac{1}{2}-1}}\\ & =2p^{{.}^{\frac{2}{3}+\frac{1}{2}}}{q}^{{.}^{\frac{1}{3}}}{r}^{{.}^{-\frac{3}{2}}}\\ & =2p^{{.}^{\frac{4+3}{6}}}{q}^{{.}^{\frac{1}{3}}}.r^{{.}^{-\frac{3}{2}}}\\ & =2p^{{.}^{\frac{7}{6}}}{q}^{{.}^{\frac{1}{3}}}.r^{{.}^{-\frac{3}{2}}}\\ & ={\displaystyle \frac{2p^{{.}^{\frac{7}{6}}}{q}^{{.}^{\frac{1}{3}}}}{r^{{.}^{\frac{3}{2}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Sederhanakanlah bentuk berikut dan}\\ & \text{nyatakan hasilnya dalam pangkat positif}\\ & \text{a}.{\left({\displaystyle \frac{p^{3n+1}{q}^n}{p^{3n+4}{q}^{4n}}}\right)}^{\frac{1}{3}}\\ & \text{b}.{\left({\displaystyle \frac{p^{-2}{q}^3}{p^4{q}^{-3}}}\right)}^{-\frac{1}{2}}{\left({\displaystyle \frac{p^4{q}^{-5}}{pq}}\right)}^{-\frac{1}{3}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\left({\displaystyle \frac{p^{3n+1}{q}^n}{p^{3n+4}{q}^{4n}}}\right)}^{\frac{1}{3}}\\ & ={\left(p^{(3n+1)-(3n+4)}{q}^{n-4n}\right)}^{\frac{1}{3}}\\ & ={\left(p^{-3}{q}^{-3n}\right)}^{\frac{1}{3}}\\ & =p^{-3.\frac{1}{3}}{q}^{-3n.\frac{1}{3}}\\ & =p^{-1}{q}^{-n}\\ & ={\displaystyle \frac{1}{p{q}^n}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\left({\displaystyle \frac{p^{-2}{q}^3}{p^4{q}^{-3}}}\right)}^{-\frac{1}{2}}{\left({\displaystyle \frac{p^4{q}^{-5}}{pq}}\right)}^{-\frac{1}{3}}\\ & =\left({\displaystyle \frac{p^{-2.(-\frac{1}{2})}{q}^{3.(-\frac{1}{2})}}{p^{4.(-\frac{1}{2})}{q}^{-3.(-\frac{1}{2})}}}\right)\left({\displaystyle \frac{p^{4.(-\frac{1}{3})}{q}^{-5.(-\frac{1}{3})}}{p^{{.}^{-\frac{1}{3}}}{q}^{{.}^{-\frac{1}{3}}}}}\right)\\ & ={\displaystyle \frac{p^1{q}^{{.}^{-\frac{3}{2}}}}{p^{-2}{q}^{{.}^{\frac{3}{2}}}}\times \frac{p^{{.}^{-\frac{4}{3}}}{q}^{{.}^{\frac{5}{3}}}}{p^{{.}^{-\frac{1}{3}}}{q}^{{.}^{-\frac{1}{3}}}}}\\ & =p^{1-(-2)+(-\frac{4}{3})-(-\frac{1}{3})}{q}^{-\frac{3}{2}-\frac{3}{2}+\frac{5}{3}-(-\frac{1}{3})}\\ & =p^{3-\frac{3}{3}}{q}^{-\frac{6}{2}+\frac{6}{3}}\\ & =p^{3-1}{q}^{-3+2}\\ & =p^2{q}^{-1}\\ & ={\displaystyle \frac{p^2}{q}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jabarkanlah bentuk}\\ & {(2m^{{.}^{\frac{3}{2}}}+n^{{.}^{\frac{3}{4}}})}^2\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {(2m^{{.}^{\frac{3}{2}}}+n^{{.}^{\frac{3}{4}}})}^2\\ & ={\left(2m^{{.}^{\frac{3}{2}}}\right)}^2+2\left(2m^{{.}^{\frac{3}{2}}}\right)\left(n^{{.}^{\frac{3}{4}}}\right)+{\left(n^{{.}^{\frac{3}{4}}}\right)}^2\\ & \text{INGAT}:{(A+B)}^2=A^2+2AB+B^2\\ & =2^2{m}^{{.}^{\frac{3.2}{2}}}+2.2.m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3}{4}}}+n^{{.}^{\frac{3.2}{4}}}\\ & =4m^3+4m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3}{4}}}+n^{{.}^{\frac{3}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jabarkanlah bentuk}\\ & {(2m^{{.}^{\frac{3}{2}}}-n^{{.}^{\frac{3}{4}}})}^3\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {(2m^{{.}^{\frac{3}{2}}}-n^{{.}^{\frac{3}{4}}})}^3\\ & ={\left(2m^{{.}^{\frac{3}{2}}}\right)}^3-3{\left(2m^{{.}^{\frac{3}{2}}}\right)}^2\left(n^{{.}^{\frac{3}{4}}}\right)+3\left(2m^{{.}^{\frac{3}{2}}}\right){\left(n^{{.}^{\frac{3}{4}}}\right)}^2-{\left(n^{{.}^{\frac{3}{4}}}\right)}^3\\ & \text{INGAT}:{(A-B)}^3=A^3-3A^{2}B+3A{B}^2-B^3\\ & =2^3{m}^{{.}^{\frac{3.3}{2}}}-{3.2}^2.m^{{.}^{\frac{3.2}{2}}}{n}^{{.}^{\frac{3}{4}}}+3.2.m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3.2}{4}}}-n^{{.}^{\frac{3.3}{4}}}\\ & =8m^{{.}^{\frac{9}{2}}}-12m^3{n}^{{.}^{\frac{3}{4}}}+6m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3}{2}}}-n^{{.}^{\frac{9}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jabarkanlah bentuk berikut}\\ \\ & \text{a}.(2p^{{.}^{\frac{1}{2}}}-3q^{{.}^{\frac{1}{2}}})(p^{{.}^{\frac{1}{2}}}+4q^{{.}^{\frac{1}{2}}})\\ \\ & \text{b}.(p^{{.}^{\frac{1}{3}}}-q^{{.}^{\frac{1}{3}}})(p^{{.}^{\frac{2}{3}}}+p^{{.}^{\frac{1}{3}}}{q}^{{.}^{\frac{1}{3}}}+q^{{.}^{\frac{2}{3}}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& (2p^{{.}^{\frac{1}{2}}}-3q^{{.}^{\frac{1}{2}}})(p^{{.}^{\frac{1}{2}}}+4q^{{.}^{\frac{1}{2}}})\\ & =2{\left(p^{{.}^{\frac{1}{2}}}\right)}^2+2.4.p^{{.}^{\frac{1}{2}}}{q}^{{.}^{\frac{1}{2}}}-3q^{{.}^{\frac{1}{2}}}.p^{{.}^{\frac{1}{2}}}-3.4.{\left(q^{{.}^{\frac{1}{2}}}\right)}^2\\ & =2p^1+8q^{{.}^{\frac{1}{2}}}{q}^{{.}^{\frac{1}{2}}}-3p^{{.}^{\frac{1}{2}}}{q}^{{.}^{\frac{1}{2}}}-12.q^1\\ & =2p+5(pq{)}^{{.}^{\frac{1}{2}}}-12q\end{array}\\ & \begin{array}{rl}\text{b}.& (p^{{.}^{\frac{1}{3}}}-q^{{.}^{\frac{1}{3}}})(p^{{.}^{\frac{2}{3}}}+p^{{.}^{\frac{1}{3}}}{q}^{{.}^{\frac{1}{3}}}+q^{{.}^{\frac{2}{3}}})\\ & =p^{{.}^{\frac{1+2}{3}}}+{\left(p^{{.}^{\frac{1}{3}}}\right)}^2{q}^{{.}^{\frac{1}{3}}}+p^{{.}^{\frac{1}{3}}}{q}^{{.}^{\frac{2}{3}}}-p^{{.}^{\frac{2}{3}}}{q}^{{.}^{\frac{1}{3}}}-p^{{.}^{\frac{1}{3}}}{\left(q^{{.}^{\frac{1}{3}}}\right)}^2-q^{{.}^{\frac{1+2}{3}}}\\ & =p^1+0+0-q^1\\ & =p-q\end{array}\end{array}$

DAFTAR PUSTAKA

1. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.

Persamaan Garis Singgung Lingkaran (PGSL)

$\text{A. PGS melalui titik pada lingkaran pusat (0,0)}$

Misalkan titik  $P(x_1,y_1)$ yang terletak pada lingkaran  $x^2+y^2=r^2$. Gradien dari garis OP adalah  ${\displaystyle \frac{y_1}{x_1}}$

Perhatikanlah ilustrasi gambar berikut

$\text{B. PGS melalui titik pada lingkaran pusat (p,q)}$

Kurang lebih dengan penjelasan yang sama dan persamaan garisnya di rumuskan 

$\{\begin{array}{ll}(p,q)& \equiv (x_1-p)(x-p)+(y_1-q)(y-q)=r^2\\ (p,q)& \equiv px+qy+{\displaystyle \frac{1}{2}A(p+x)+{\displaystyle \frac{1}{2}B(q+y)+C=0}}\end{array}$.

$\text{C. PGSL dengan gradien m}$

$\text{D. PGS melalui titik di luar lingkaran pusat (0,0)}$

Contoh 9 Vektor

$\begin{array}{l}\\ 41.& \text{Diketahui segi empat ABCD dengan}\overrightarrow{DA}=\overrightarrow{a},\\ & \overrightarrow{DB}=\overrightarrow{b},\text{dan}\overrightarrow{DC}=\overrightarrow{c}.\text{Jika titik H pada AB}\\ & \text{dengan}\overrightarrow{AH}:\overrightarrow{HB}=1:2,\text{dan titik J pada BC}\\ & \text{dengan}\overrightarrow{BJ}:\overrightarrow{JC}=1:2\text{maka}\overrightarrow{HJ}=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{-\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}}\\ \text{b}.{\displaystyle \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}}\\ \text{c}.{\displaystyle \frac{-2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}}& \\ \text{d}.{\displaystyle \frac{-2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}}{3}}\\ \text{e}.{\displaystyle \frac{-2\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c}}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Coba perhatikanlah ilustrasi berikut}\end{array}$

$.\begin{array}{rl}& \begin{array}{|cc|}\hline \text{Diketahui 1}& \text{Diketahui 2}\\ \begin{array}{rl}\overrightarrow{AH}:\overrightarrow{HB}& =1:2\\ \overrightarrow{h}& ={\displaystyle \frac{2\overrightarrow{a}+\overrightarrow{b}}{3}}\\ & \end{array}& \begin{array}{rl}\overrightarrow{BJ}:\overrightarrow{JC}& =1:2\\ \overrightarrow{j}& ={\displaystyle \frac{2\overrightarrow{b}+\overrightarrow{c}}{3}}\\ & \end{array}\\ \hline\end{array}\\ & \text{Proses Penyelesaian}\\ & \begin{array}{rl}\overrightarrow{HJ}& =\overrightarrow{j}-\overrightarrow{h}\\ & =\left({\displaystyle \frac{2\overrightarrow{b}+\overrightarrow{c}}{3}}\right)-\left({\displaystyle \frac{2\overrightarrow{a}+\overrightarrow{b}}{3}}\right)\\ & ={\displaystyle \frac{-2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 42.& \text{Supaya vektor}\overrightarrow{a}=\left(\begin{array}{c}x\\ 4\\ 7\end{array}\right),\text{dan}\overrightarrow{b}=\left(\begin{array}{c}6\\ y\\ 14\end{array}\right),\\ & \text{segaris, harga}x-y=....\\ & \begin{array}{l}\\ \text{a}.-5& & \text{d}.4\\ \text{b}.-2& \text{c}.3& \text{e}.6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Dikethui}& \text{bahwa}:\\ \text{vektor}& \overrightarrow{a}\text{dan}\overrightarrow{b}\text{segaris},\text{maka}\\ m\overrightarrow{a}& =\overrightarrow{b}\\ \text{dengan}& m\text{adalah skalar/faktor pengali}\\ m\left(\begin{array}{c}x\\ 4\\ 7\end{array}\right)& =\left(\begin{array}{c}6\\ y\\ 14\end{array}\right)\\ \text{di}\text{dapat}& \text{kan}\{\begin{array}{ll}mx& =6...................(1)\\ 4m& =y....................(2)\\ 7m& =14...................(3)\end{array}\\ \text{Dari per}& \text{samaan}(3)\text{akan}\\ \text{didapat}& \text{kan nilai}m=2\\ \text{maka}& \text{akan didapatkan juga}\{\begin{array}{ll}x& =3\\ y& =8\end{array}\\ \text{sehin}& \text{gga nilai dari}\\ x-y& =3-8\\ & =-5\end{array}\end{array}$.

$\begin{array}{l}\\ 43.& \text{Diketahui}O\text{titik pangkal}A(0,1,2)\\ & \text{dan}B(3,4,5),\text{maka luas segitiga}\\ & OAB\text{sama dengan}....\\ & \begin{array}{l}\\ \text{a}.3\sqrt{6}\\ \text{b}.{\displaystyle \frac{2}{3}\sqrt{6}}\\ \text{c}.{\displaystyle \frac{4}{3}\sqrt{6}}& \\ \text{d}.{\displaystyle \frac{3}{2}\sqrt{6}}\\ \text{e}.2\sqrt{6}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Coba perhatikanlah ilustrasi berikut}\\ & \begin{array}{rl}& \text{Misalkan luas segitiga}{\displaystyle \frac{1}{2}|\overrightarrow{p}\times \overrightarrow{q}|,}\\ & \text{dengan}\\ & \{\begin{array}{ll}\overrightarrow{p}& =\overline{OA}=\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)\\ \overrightarrow{q}& =\overline{OB}=\left(\begin{array}{c}3\\ 4\\ 5\end{array}\right)\end{array}\end{array}\\ & \begin{array}{rl}\overrightarrow{p}& \times \overrightarrow{q}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 0& 1& 2\\ 3& 4& 5\end{array}\right|\\ & =(5-8)\overrightarrow{i}-(0-6)\overrightarrow{j}+(0-3)\overrightarrow{k}\\ & =-3\overrightarrow{i}+6\overrightarrow{j}-3\overrightarrow{k}\\ & \text{Sehingga}\\ & |\overrightarrow{p}\times \overrightarrow{q}|=\sqrt{(-3{)}^2+6^2+(-3{)}^2}\\ & =\sqrt{9+36+9}=\sqrt{54}=3\sqrt{6}\\ & \text{Maka luas segi tiganya adalah}:\\ & \text{luas}\mathrm{△}ABC={\displaystyle \frac{1}{2}|\overrightarrow{p}\times \overrightarrow{q}|={\displaystyle \frac{1}{2}\left(3\sqrt{6}\right)={\displaystyle \frac{3}{2}\sqrt{6}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 44.& \text{Proyeksi skalar ortogonal}\\ & \overrightarrow{a}=2\overrightarrow{i}-3\overrightarrow{j}+6\overrightarrow{k},\text{pada}\\ & \overrightarrow{b}=\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{4}{3}}& & & & & \text{d}.& {\displaystyle \frac{16}{3}}\\ \\ \text{b}.{\displaystyle \frac{8}{3}}& & \text{c}& {\displaystyle \frac{10}{3}}& & \text{e}.& {\displaystyle \frac{20}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \{\begin{array}{ll}\overrightarrow{a}& =(2,-3,6)\\ \overrightarrow{b}& =(1,2,2)\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{r}\\ \left|\overrightarrow{c}\right|& =\left|{\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right|\\ & ={\displaystyle \frac{\left(\begin{array}{c}2\\ -3\\ 6\end{array}\right)\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)}{\sqrt{1^2+2^2+2^2}}}\\ & ={\displaystyle \frac{2-6+12}{\sqrt{9}}=\frac{8}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 45.& \text{Diketahui vektor}\overrightarrow{a}=3\overrightarrow{i}-4\overrightarrow{j}+p\overrightarrow{k}\\ & \text{dan}\overrightarrow{b}=2\overrightarrow{i}+2\overrightarrow{j}-3\overrightarrow{k}.\text{Jika panjang}\\ & \text{proyeksi vektor}\overrightarrow{a}\text{pada}\overrightarrow{b}\text{adalah}\\ & {\displaystyle \frac{4}{\sqrt{17}},\text{maka nilai}p\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -2}& & & & & \text{d}.& {\displaystyle 2}\\ \\ \text{b}.{\displaystyle -1}& & \text{c}& {\displaystyle 1}& & \text{e}.& {\displaystyle 3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Panjang proyeksi skalar vektor}\overrightarrow{a}\text{pada}\overrightarrow{b}\\ & \begin{array}{r}\\ \left|\overrightarrow{c}\right|& ={\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\\ {\displaystyle \frac{4}{\sqrt{17}}}& ={\displaystyle \frac{\left(\begin{array}{c}3\\ -4\\ p\end{array}\right)\left(\begin{array}{c}2\\ 2\\ -3\end{array}\right)}{\sqrt{2^2+2^2+(-3{)}^2}}}\\ {\displaystyle \frac{4}{\sqrt{17}}}& ={\displaystyle \frac{6-8-3p}{\sqrt{17}}}\\ 4& =-2-3p\\ 6& =-3p\\ -3p& =6\\ p& =-2\end{array}\end{array}$

Contoh 8 Vektor

Contoh soal sebelumnya di sini Contoh Soal 7

$\begin{array}{l}\\ 34.& \text{Vektor satuan untuk}\overrightarrow{a}=\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}\sqrt{3}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{c}.{\displaystyle \frac{1}{5}\sqrt{5}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{d}.{\displaystyle \frac{1}{7}\sqrt{7}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\\ \text{e}.{\displaystyle \frac{1}{21}\sqrt{21}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Vektor satuan}\overrightarrow{a}& \text{adalah}{\overrightarrow{e}}_{\overrightarrow{a}},\text{yaitu}:\\ {\overrightarrow{e}}_{\overrightarrow{a}}& ={\displaystyle \frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}{\sqrt{2^2+(-1{)}^2+4^2}}}\\ & ={\displaystyle \frac{1}{\sqrt{21}}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\\ & ={\displaystyle \frac{1}{21}\sqrt{21}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Posisi suatu titik dalam ruang saat }\\ & \text{waktu}t\text{ditunjukkan oleh vektor}\\ & \left(\begin{array}{c}t\\ t^2\\ -t\end{array}\right).\text{Jika pada saat}t=1\text{titik }\\ & \text{tersebut berada di titik P dan pada}\\ & \text{saat}t=2\text{titik tersebut berada }\\ & \text{di titik Q, maka jarak titik P dari Q}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.\sqrt{24}-\sqrt{3}& & \text{d}.\sqrt{11}\\ \text{b}.2-\sqrt{2}& \text{c}.3& \text{e}.\sqrt{43}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left|\overrightarrow{PQ}\right|& =\sqrt{(x_q-x_p{)}^2+(y_q-y_p{)}^2+(y_q-y_p{)}^2}\\ & =\sqrt{(2-1{)}^2+(2^2-1^2{)}^2+((-2)-(-1){)}^2}\\ & =\sqrt{1^2+3^2+(-1{)}^2}\\ & =\sqrt{1+9+1}\\ & =\sqrt{11}\end{array}\end{array}$

$\begin{array}{l}\\ 36.& \text{Jika diketahui}\left|\overrightarrow{a}\right|=4\sqrt{3},\left|\overrightarrow{b}\right|=5,\\ & \text{ dan}(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})=13,\\ & \text{maka}\mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle {30}^{\circ }}& & \text{d}.{\displaystyle {135}^{\circ }}\\ \text{b}.{\displaystyle {60}^{\circ }}& \text{c}.{\displaystyle {120}^{\circ }}& \text{e}.{\displaystyle {150}^{\circ }}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})& =13\\ \overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}& =13\\ {\left|\overrightarrow{a}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}+{\left|\overrightarrow{b}\right|}^2& =13,\\ \text{ingat bahwa}\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a}& \\ {\left(4\sqrt{3}\right)}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})+5^2& =13\\ 48+2.(4\sqrt{3}).5.\cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})+25& =13\\ 40\sqrt{3}\cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})& =13-25-48\\ \cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})& ={\displaystyle \frac{-60}{40\sqrt{3}}}\\ & =-{\displaystyle \frac{1}{2}\sqrt{3}}\\ & =-\cos 30\\ & =\cos ({180}^{\circ }-{30}^{\circ })\\ \cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})& =\cos {150}^{\circ }\\ \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})& ={150}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 37.& \text{Jika diketahui titik}A(2,-1,4),B(4,1,3),\\ & \text{ dan}C(2,0,5),\text{maka}\sin \mathrm{\angle }(\overline{AB},\overline{AC})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{7}\sqrt{5}}& & \text{d}.{\displaystyle \frac{1}{6}\sqrt{3}}\\ \\ \text{b}.{\displaystyle \frac{1}{6}\sqrt{34}}& \text{c}.{\displaystyle \frac{2}{3}\sqrt{2}}& \text{e}.{\displaystyle \frac{1}{6}\sqrt{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \cos \mathrm{\angle }(\overline{AB},\overline{AC})={\displaystyle \frac{\overline{AB}.\overline{AC}}{\left|\overline{AB}\right|.\left|\overline{AC}\right|}}\\ & ={\displaystyle \frac{(\overrightarrow{b}-\overrightarrow{a}).(\overrightarrow{c}-\overrightarrow{a})}{\sqrt{x_{(\overrightarrow{b}-\overrightarrow{a})}^2+y_{(\overrightarrow{b}-\overrightarrow{a})}^2+z_{(\overrightarrow{b}-\overrightarrow{a})}^2}.\sqrt{x_{(\overrightarrow{c}-\overrightarrow{a})}^2+y_{(\overrightarrow{c}-\overrightarrow{a})}^2+z_{(\overrightarrow{c}-\overrightarrow{a})}^2}}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}4-2\\ 1+1\\ 3-4\end{array}\right).\left(\begin{array}{c}2-2\\ 0+1\\ 5-4\end{array}\right)}{\sqrt{(4-2{)}^2+(1+1{)}^2+(3-4{)}^2}.\sqrt{(2-2{)}^2+(0+1{)}^2+(5-4{)}^2}}}\\ & ={\displaystyle \frac{2.0+2.1+-1.1}{\sqrt{4+4+1}.\sqrt{0+1+1}}}\\ & ={\displaystyle \frac{1}{3\sqrt{2}}}\\ & ={\displaystyle \frac{1}{6}\sqrt{2}}\end{array}\\ & \begin{array}{rl}& \text{Sehingga},\\ & \sin \mathrm{\angle }(\overline{AB},\overline{AC})\\ & =\sqrt{1-{\cos }^2\mathrm{\angle }(\overline{AB},\overline{AC})}\\ & =\sqrt{1-{\left({\displaystyle \frac{1}{6}\sqrt{2}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{2}{36}}}\\ & =\sqrt{{\displaystyle \frac{34}{36}}}\\ & ={\displaystyle \frac{1}{6}\sqrt{34}}\end{array}\end{array}$.

$\begin{array}{l}\\ 38.& \text{Diketahui segitiga ABC. Titik M }\\ & \text{di tengah AC, dan titik N pada BC}\\ & \text{Jika}\overrightarrow{AB}=\overrightarrow{c},\overrightarrow{AC}=\overrightarrow{b},\overrightarrow{BC}=\overrightarrow{a},\\ & \text{maka}\overrightarrow{MN}=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}(\overrightarrow{b}-\overrightarrow{c})}& \\ \text{b}.{\displaystyle \frac{1}{2}(-\overrightarrow{b}+\overrightarrow{c})}& \\ \text{c}.{\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{c})}\\ \text{d}.{\displaystyle \frac{1}{2}(\overrightarrow{a}-\overrightarrow{b})}\\ \text{e}.{\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{b})}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\overrightarrow{MN}& =\overrightarrow{MC}+\overrightarrow{CN}\\ & ={\displaystyle \frac{1}{2}\overrightarrow{AC}+{\displaystyle \frac{1}{2}(-\overrightarrow{BC})}}\\ & ={\displaystyle \frac{1}{2}(\overrightarrow{AC}-\overrightarrow{BC})}\\ & ={\displaystyle \frac{1}{2}(\overrightarrow{b}-\overrightarrow{a})\mathbf{\text{atau}}}\\ & ={\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{b})}\end{array}\end{array}$.

$\begin{array}{l}\\ 39.& \text{Jika titik berat segitiga ABC adalah Z}\\ & \text{dengan}\text{A}(1,0,2),\text{B}(5,4,10),\text{C}(0,-1,6),\\ & \text{maka koordinat titik Z tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.(-2,-1,6)& & \text{d}.(3,2,6)\\ \text{b}.(2,1,6)& \text{c}.(3,-1,6)& \text{e}.(6,4,12)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \text{Coba perhatikanlah ilustrasi berikut}\end{array}$

$.\begin{array}{r}\\ & \begin{array}{rl}\text{koordinat}& {\text{titik A}}^{\prime }\\ & ={\displaystyle \frac{1}{2}(5+0,4-1,10+6)}\\ & =(\frac{5}{2},\frac{3}{2},8)\end{array}\\ & \begin{array}{rl}\text{Dalam se}& \text{gitiga ABC untuk titik berat Z }\\ \text{berlaku}& \text{ketentuan sebagai berikut}\\ AZ:Z{A}^{\prime }& =2:1\\ \overrightarrow{AZ}:\overrightarrow{Z{A}^{\prime }}& =2:1\\ \overrightarrow{OZ}& ={\displaystyle \frac{\overrightarrow{OA}+2\overrightarrow{O{A}^{\prime }}}{3}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right)+2\left(\begin{array}{c}\frac{5}{2}\\ \frac{3}{2}\\ 8\end{array}\right)}{3}}\\ & =\left(\begin{array}{c}2\\ 1\\ 6\end{array}\right)\\ \text{Jadi},& \text{koordinat titik Z adalah}(2,1,6)\end{array}\end{array}$.

$\begin{array}{l}\\ 40.& \text{Diketahui titik}\text{A}(-4,-1,-2),\text{B}(-6,4,3),\\ & \text{C}(2,3,5).\text{Jika titik M membagi}\overrightarrow{AB}\\ & \text{sehingga}\overrightarrow{AM}:\overrightarrow{MB}=3:2\\ & \text{maka vektor yang diwakili oleh}\overrightarrow{MC}=....\\ & \begin{array}{l}\\ \text{a}.(-4,1,4)& & \text{d}.(6,4,1)\\ \text{b}.(-2,2,1)& \text{c}.(0,5,6)& \text{e}.(4,1,4)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Coba perhatikanlah ilustrasi berikut}\end{array}$

$.\begin{array}{|cc|}\hline \text{Diketahui}& \text{Proses Penyelesaian}\\ \begin{array}{rl}\overrightarrow{AM}:\overrightarrow{MB}& =3:2\\ \overrightarrow{m}& ={\displaystyle \frac{2\overrightarrow{a}+3\overrightarrow{b}}{5}}\\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}\overrightarrow{MC}& =\overrightarrow{c}-\overrightarrow{m}\\ & =\overrightarrow{c}-{\displaystyle \frac{2\overrightarrow{a}+3\overrightarrow{b}}{5}}\\ & ={\displaystyle \frac{5\overrightarrow{c}-2\overrightarrow{a}-3\overrightarrow{b}}{5}}\\ & ={\displaystyle \frac{5\left(\begin{array}{c}2\\ 3\\ 5\end{array}\right)-2\left(\begin{array}{c}4\\ -1\\ -2\end{array}\right)-3\left(\begin{array}{c}-6\\ 4\\ 3\end{array}\right)}{5}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}20\\ 5\\ 20\end{array}\right)}{5}}\\ & =\left(\begin{array}{c}4\\ 1\\ 4\end{array}\right)\end{array}\\ \hline\end{array}$