Lanjutan 3 Persamaan Trigonometri

 B. 1 Persamaan Trigonometri Sederhana

Dalam penyelesaian persamaan trigonometri sederhana dapat digunakan salah satu rumus berikut, yaitu:

$\begin{array}{rl}(1).\sin x& =\sin {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k{.360}^{\circ }\\ \text{atau}\\ x=({180}^{\circ }-{\alpha }^{\circ })+k{.360}^{\circ }\end{array}\\ (2).\cos x& =\cos {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k{.360}^{\circ }\\ \text{atau}\\ x=-{\alpha }^{\circ }+k{.360}^{\circ }\end{array}\\ (3).\tan x& =\tan {\alpha }^{\circ }\\ & x={\alpha }^{\circ }+k{.180}^{\circ }\end{array}$.

Jika sudutnya dinyatakan dalam phi radian $(\pi \text{dibaca}:phi)$, maka persamaan trigonometri sederhananya adalah:

$\begin{array}{rl}(1).\sin x& =\sin {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k.2\pi \\ \text{atau}\\ x=(\pi -{\alpha }^{\circ })+k.2\pi \end{array}\\ (2).\cos x& =\cos {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k.2\pi \\ \text{atau}\\ x=-{\alpha }^{\circ }+k.2\pi \end{array}\\ (3).\tan x& =\tan {\alpha }^{\circ }\\ & x={\alpha }^{\circ }+k.\pi \end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah akar-akar persamaan trigonometri}\\ & \text{berikut dan tentukan pula himpunan}\\ & \text{penyelesaiannya untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \text{a}.\sin x=\sin {50}^{\circ }\\ & \text{b}.\cos x=\cos {50}^{\circ }\\ & \text{c}.\tan x=\tan {50}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}.\text{a}.\sin x& =\sin {50}^{\circ }\\ x& =\{\begin{array}{ll}{50}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{50}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{50}^{\circ }(\text{memenuhi})& \\ {130}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{50}^{\circ },{130}^{\circ }\}\end{array}\\ & \begin{array}{rl}.\text{b}.\cos x& =\cos {50}^{\circ }\\ x& =\{\begin{array}{ll}{50}^{\circ }& +k{.360}^{\circ }\\ -{50}^{\circ }& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{50}^{\circ }(\text{memenuhi})& \\ -{50}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{50}^{\circ }+{360}^{\circ }={410}^{\circ }(\text{tidak memenuhi})& \\ -{50}^{\circ }+{360}^{\circ }={310}^{\circ }(\text{memenuhi})& \end{array}\\ \text{HP}& =\{{50}^{\circ },{310}^{\circ }\}\end{array}\\ & \begin{array}{rl}.\text{c}.\tan x& =\tan {50}^{\circ }\\ x& ={50}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={50}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={50}^{\circ }+{180}^{\circ }={230}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={50}^{\circ }+{360}^{\circ }={410}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{50}^{\circ },{230}^{\circ }\}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah himpunan penyelesaian dari }\\ & \text{persamaan-persamaan trigonometri berikut}\\ & \text{ini untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \begin{array}{l}\\ \text{a}.& \sin x={\displaystyle \frac{1}{2}}& \text{f}.& \tan x=-{\displaystyle \frac{1}{3}\sqrt{3}}& \text{k}.& \sin 2x={\displaystyle \frac{1}{2}}\\ \text{b}.& \cos x={\displaystyle \frac{1}{2}\sqrt{3}}& \text{g}& 2\cos x=-\sqrt{3}& \text{l}.& \cos 2x=-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{c}.& \tan x=\sqrt{3}& \text{h}& 3\tan x=\sqrt{3}& \text{m}.& \tan 2x=\sqrt{3}\\ \text{d}.& \sin x=-1& \text{i}.& \sin x=\sin {46}^{\circ }& \text{n}.& \sin (2x-{30}^{\circ })=\sin {45}^{\circ }\\ \text{e}.& \cos x=-{\displaystyle \frac{1}{2}\sqrt{2}}& \text{j}.& \cos x=\cos {93}^{\circ }& \text{o}.& \sin (2x+{60}^{\circ })=\sin {90}^{\circ }\end{array}\end{array}$

$.\text{Jawab}:$

$\begin{array}{rl}.\text{a}.\sin x& ={\displaystyle \frac{1}{2}}\\ \sin x& =\sin {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{30}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ {150}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{30}^{\circ },{150}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{b}.\cos x& ={\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos x& =\cos {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ -{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ -{30}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tidak memenuhi})& \\ -{30}^{\circ }+{360}^{\circ }={330}^{\circ }(\text{memenuhi})& \end{array}\\ \text{HP}& =\{{30}^{\circ },{330}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{c}.\tan x& =\sqrt{3}\\ \tan x& =\tan {60}^{\circ }\\ x& ={60}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={60}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={60}^{\circ }+{180}^{\circ }={240}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={60}^{\circ }+{360}^{\circ }={420}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{60}^{\circ },{240}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{d}.\sin x& =-1\\ \sin x& =\sin {270}^{\circ }\\ x& =\{\begin{array}{ll}{270}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{270}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}{270}^{\circ }& \text{memenuhi}\\ -{90}^{\circ }& \text{tidak memenuhi}\end{array}\\ k=1& \text{tidak memenuhi semuanya}\\ \text{HP}& =\left\{{270}^{\circ }\right\}\end{array}$.

$\begin{array}{rl}.\text{k}.\sin 2x& ={\displaystyle \frac{1}{2}}\\ \sin 2x& =\sin {30}^{\circ }\\ 2x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{30}^{\circ })& +k{.360}^{\circ }\end{array}\\ \text{sehin}& \text{gga}\\ x& =\{\begin{array}{ll}{15}^{\circ }& +k{.180}^{\circ }\\ ({90}^{\circ }-{15}^{\circ })& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{15}^{\circ }(\text{memenuhi})& \\ {75}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{15}^{\circ }+{180}^{\circ }={195}^{\circ }(\text{memenuhi})& \\ {75}^{\circ }+{180}^{\circ }={255}^{\circ }(\text{memenuhi})& \end{array}\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{15}^{\circ },{75}^{\circ },{195}^{\circ },{255}^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{l}.\cos 2x& =-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos 2x& =-\cos {30}^{\circ }=\cos ({180}^{\circ }-{30}^{\circ })=\cos {150}^{\circ }\\ 2x& =\{\begin{array}{ll}{150}^{\circ }& +k{.360}^{\circ }\\ -{150}^{\circ }& +k{.360}^{\circ }\end{array}\\ \text{sehin}& \text{gga}\\ x& =\{\begin{array}{ll}{75}^{\circ }& +k{.180}^{\circ }\\ -{75}^{\circ }& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{75}^{\circ }(\text{memenuhi})& \\ -{75}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{75}^{\circ }+{180}^{\circ }={255}^{\circ }(\text{memenuhi})& \\ -{75}^{\circ }+{180}^{\circ }={105}^{\circ }(\text{memenuhi})& \end{array}\\ k=2& \\ x& =\{\begin{array}{ll}{75}^{\circ }+{360}^{\circ }={435}^{\circ }(\text{tidak memenuhi})& \\ -{75}^{\circ }+{360}^{\circ }={285}^{\circ }(\text{memenuhi})& \end{array}\\ k=3& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{75}^{\circ },{105}^{\circ },{255}^{\circ },{285}^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{m}.\tan 2x& =\sqrt{3}\\ \tan 2x& =\tan {60}^{\circ }\\ 2x& ={60}^{\circ }+k{.180}^{\circ }\\ \text{sehin}& \text{gga}\\ x& ={30}^{\circ }+k{.90}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={30}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={30}^{\circ }+{90}^{\circ }={120}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={30}^{\circ }+{180}^{\circ }={210}^{\circ }\text{memenuhi}\\ k=3& \\ x& ={30}^{\circ }+{270}^{\circ }={300}^{\circ }\text{memenuhi}\\ k=4& \\ x& ={30}^{\circ }+{360}^{\circ }={390}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{30}^{\circ },{120}^{\circ },{210}^{\circ },{300}^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{n}.\sin (2x-{30}^{\circ })& =\sin {45}^{\circ }\\ (2x-{30}^{\circ })& =\{\begin{array}{ll}{45}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{45}^{\circ })& +k{.360}^{\circ }\end{array}\\ 2x& =\{\begin{array}{ll}{45}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\\ {135}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ x& =\{\begin{array}{ll}37,5^{\circ }& +k{.180}^{\circ }\\ 82,5^{\circ }& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }& \\ 82,5^{\circ }& \end{array}\\ k=1& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }+{180}^{\circ }& =217,5^{\circ }\\ 82,5^{\circ }+{180}^{\circ }& =262,5^{\circ }\end{array}\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{37,5^{\circ },82,5^{\circ },217,5^{\circ },262,5^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{o}.\sin (2x+{60}^{\circ })& =\sin {90}^{\circ }\\ (2x+{60}^{\circ })& =\{\begin{array}{ll}{90}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{90}^{\circ })& +k{.360}^{\circ }\end{array}\\ 2x& =\{\begin{array}{ll}{90}^{\circ }-{60}^{\circ }& +k{.360}^{\circ }\\ {90}^{\circ }-{60}^{\circ }& +k{.360}^{\circ }\end{array}\\ x& ={15}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}\\ x& ={15}^{\circ }\\ k=1& \text{diperoleh}\\ x& ={15}^{\circ }+{180}^{\circ }={195}^{\circ }\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{15}^{\circ },{195}^{\circ }\}\end{array}$

Lanjutan 2 Persamaan Trigonometri

 A. 2  Relasi Sudut

Mengingatkan kembali materi tentang nilai sudut diberbagai kuadran yang selanjutnya berkaitan erat dengan relasi sudutnya dari kuadran selain satu diubah ke kuadran satu supaya mudah menentukan nilai trigonometri.

Untuk tanda perbandingan trigonometrinya berkaitan dengan relasi sudutnya adalah disajikan sebagaimana dalam bagan berikut

$\begin{array}{cccccc}\text{Nilai yang positif}& & & & & \\ \text{hanya}\mathbf{\text{sinus}}& & & & \text{Semua nilai trigon}& \mathbf{\text{positif}}\\ & & & & & \\ \text{Nilai yang positif}& & & & \text{Nilai yang positif}& \\ \text{hanya}\mathbf{\text{tangen}}& & & & \text{hanya}\mathbf{\text{cosinus}}& \end{array}$.

atau

$\begin{array}{cccccc}\begin{array}{l}\\ \{\begin{array}{ll}\sin & =+\\ \cos & =-\\ \tan & =-\\ \csc & =+\\ \sec & =-\\ \cot & =-\end{array}& \end{array}& & & & & \begin{array}{l}\\ \{\begin{array}{ll}\sin & =+\\ \cos & =+\\ \tan & =+\\ \csc & =+\\ \sec & =+\\ \cot & =+\end{array}& \end{array}\\ & & & & \\ & & & & & \\ & & & & & \\ \begin{array}{l}\\ \{\begin{array}{ll}\sin & =-\\ \cos & =-\\ \tan & =+\\ \csc & =-\\ \sec & =-\\ \cot & =+\end{array}& \end{array}& & & & & \begin{array}{l}\\ \{\begin{array}{ll}\sin & =-\\ \cos & =+\\ \tan & =-\\ \csc & =-\\ \sec & =+\\ \cot & =-\end{array}& \end{array}\end{array}$.

Adapun penjabaran sudut-sudut yang berelasi sebagaimana ilustrasi bagan berikut, yaitu:

$\begin{array}{cccccc}\text{Kuadran II}& & & & \text{Kuadran I}& \\ ({180}^{\circ }-\alpha )& & & & \text{Semua nilai trigon}& \mathbf{\text{positif}}\\ & & & & & \\ \text{Kuadran III}& & & & \text{Kuadran IV}& \\ ({180}^{\circ }+\alpha )& & & & ({360}^{\circ }-\alpha )& \end{array}$

Ketentuan perubahan trigonometri berkaitan dengan sudut berelasi adalah sebagaimana tabel berikut:

KUADRAN PERTAMA

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran I}\\ & 0^{\circ }<\alpha <{90}^{\circ }\\ & =({90}^{\circ }-\alpha )\end{array}& \{\begin{array}{ll}\sin & =\cos \\ \cos & =\sin \\ \tan & =\cot \\ \csc & =\sec \\ \sec & =\csc \\ \cot & =\tan \end{array}& \begin{array}{rl}\sin ({90}^{\circ }-\alpha )& =\cos \alpha \\ \cos ({90}^{\circ }-\alpha )& =\sin \alpha \\ \tan ({90}^{\circ }-\alpha )& =\cot \alpha \\ \csc ({90}^{\circ }-\alpha )& =\sec \alpha \\ \sec ({90}^{\circ }-\alpha )& =\csc \alpha \\ \cot ({90}^{\circ }-\alpha )& =\tan \alpha \end{array}\\ \hline\end{array}$.

KUADRAN KEDUA

ada 2 pilihan yaitu:

pertama

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran II}\\ & {90}^{\circ }<\alpha <{180}^{\circ }\\ & =({90}^{\circ }+\alpha )\end{array}& \{\begin{array}{ll}\sin & =\cos \\ \cos & =\sin \\ \tan & =\cot \\ \csc & =\sec \\ \sec & =\csc \\ \cot & =\tan \end{array}& \begin{array}{rl}\sin ({90}^{\circ }+\alpha )& =\cos \alpha \\ \cos ({90}^{\circ }+\alpha )& =-\sin \alpha \\ \tan ({90}^{\circ }+\alpha )& =-\cot \alpha \\ \csc ({90}^{\circ }+\alpha )& =\sec \alpha \\ \sec ({90}^{\circ }+\alpha )& =-\csc \alpha \\ \cot ({90}^{\circ }+\alpha )& =-\tan \alpha \end{array}\\ \hline\end{array}$.

kedua

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Tidak Ada Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran II}\\ & {90}^{\circ }<\alpha <{180}^{\circ }\\ & =({180}^{\circ }-\alpha )\end{array}& \{\begin{array}{ll}\sin & =\sin \\ \cos & =\cos \\ \tan & =\tan \\ \csc & =\csc \\ \sec & =\sec \\ \cot & =\cot \end{array}& \begin{array}{rl}\sin ({180}^{\circ }-\alpha )& =\sin \alpha \\ \cos ({180}^{\circ }-\alpha )& =-\cos \alpha \\ \tan ({180}^{\circ }-\alpha )& =-\tan \alpha \\ \csc ({180}^{\circ }-\alpha )& =\csc \alpha \\ \sec ({180}^{\circ }-\alpha )& =-\sec \alpha \\ \cot ({180}^{\circ }-\alpha )& =-\cot \alpha \end{array}\\ \hline\end{array}$.

KUADRAN KETIGA

ada 2 pilihan juga yaitu:

pertama

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Tidak Ada Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran III}\\ & {180}^{\circ }<\alpha <{270}^{\circ }\\ & =({180}^{\circ }+\alpha )\end{array}& \{\begin{array}{ll}\sin & =\sin \\ \cos & =\cos \\ \tan & =\tan \\ \csc & =\csc \\ \sec & =\sec \\ \cot & =\cot \end{array}& \begin{array}{rl}\sin ({180}^{\circ }+\alpha )& =-\sin \alpha \\ \cos ({180}^{\circ }+\alpha )& =-\cos \alpha \\ \tan ({180}^{\circ }+\alpha )& =\tan \alpha \\ \csc ({180}^{\circ }+\alpha )& =-\csc \alpha \\ \sec ({180}^{\circ }+\alpha )& =-\sec \alpha \\ \cot ({180}^{\circ }+\alpha )& =\cot \alpha \end{array}\\ \hline\end{array}$.

kedua

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran III}\\ & {180}^{\circ }<\alpha <{270}^{\circ }\\ & =({270}^{\circ }-\alpha )\end{array}& \{\begin{array}{ll}\sin & =\cos \\ \cos & =\sin \\ \tan & =\cot \\ \csc & =\sec \\ \sec & =\csc \\ \cot & =\tan \end{array}& \begin{array}{rl}\sin ({270}^{\circ }-\alpha )& =-\cos \alpha \\ \cos ({270}^{\circ }-\alpha )& =-\sin \alpha \\ \tan ({270}^{\circ }-\alpha )& =\cot \alpha \\ \csc ({270}^{\circ }-\alpha )& =-\sec \alpha \\ \sec ({270}^{\circ }-\alpha )& =-\csc \alpha \\ \cot ({270}^{\circ }-\alpha )& =\tan \alpha \end{array}\\ \hline\end{array}$.

KUADRAN KEEMPAT

ada 2 pilihan juga yaitu:

pertama

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran IV}\\ & {270}^{\circ }<\alpha <{360}^{\circ }\\ & =({270}^{\circ }+\alpha )\end{array}& \{\begin{array}{ll}\sin & =\cos \\ \cos & =\sin \\ \tan & =\cot \\ \csc & =\sec \\ \sec & =\csc \\ \cot & =\tan \end{array}& \begin{array}{rl}\sin ({270}^{\circ }+\alpha )& =-\cos \alpha \\ \cos ({270}^{\circ }+\alpha )& =\sin \alpha \\ \tan ({270}^{\circ }+\alpha )& =-\cot \alpha \\ \csc ({270}^{\circ }+\alpha )& =-\sec \alpha \\ \sec ({270}^{\circ }+\alpha )& =\csc \alpha \\ \cot ({270}^{\circ }+\alpha )& =-\tan \alpha \end{array}\\ \hline\end{array}$.

kedua

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Tidak Ada Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran IV}\\ & {270}^{\circ }<\alpha <{360}^{\circ }\\ & =({360}^{\circ }-\alpha )\end{array}& \{\begin{array}{ll}\sin & =\sin \\ \cos & =\cos \\ \tan & =\tan \\ \csc & =\csc \\ \sec & =\sec \\ \cot & =\cot \end{array}& \begin{array}{rl}\sin ({360}^{\circ }-\alpha )& =-\sin \alpha \\ \cos ({360}^{\circ }-\alpha )& =\cos \alpha \\ \tan ({360}^{\circ }-\alpha )& =-\tan \alpha \\ \csc ({360}^{\circ }-\alpha )& =-\csc \alpha \\ \sec ({360}^{\circ }-\alpha )& =\sec \alpha \\ \cot ({360}^{\circ }-\alpha )& =-\cot \alpha \end{array}\\ \hline\end{array}$.

 A. 3  Sudut Negatif dan Sudut lebih Besar dari  ${360}^{\circ }$

$\begin{array}{rl}\text{a}.& \{\begin{array}{ll}\sin (-A)& =-\sin A\\ \cos (-A)& =\cos A\\ \tan (-A)& =-\tan A\end{array}\\ \text{b}.& \{\begin{array}{ll}\csc (-A)& =-\csc A\\ \sec (-A)& =\sec A\\ \cot (-A)& =-\cot A\end{array}\\ \text{c}.& \{\begin{array}{ll}\sin (n{.360}^{\circ }+A)& =\sin A\\ \cos (n{.360}^{\circ }+A)& =\cos A\\ \tan (n{.360}^{\circ }+A)& =\tan A\end{array},n\in \mathbb{N}\end{array}$.

Catatan : $0^{\circ }$=${360}^{\circ }$=${720}^{\circ }$=${1080}^{\circ }$=$n{.360}^{\circ }$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai}\\ & \text{a}.\sin {120}^{\circ }\\ & \text{b}.\cos {240}^{\circ }\\ & \text{c}.\tan {315}^{\circ }\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.\sin {120}^{\circ }& =\sin ({180}^{\circ }-{60}^{\circ })=\sin {60}^{\circ }\\ & ={\displaystyle \frac{1}{2}\sqrt{3},\text{atau}}\\ & =\sin ({90}^{\circ }+{30}^{\circ })=\cos {30}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{b}.\cos {240}^{\circ }& =\cos ({180}^{\circ }+{60}^{\circ })=-\cos {60}^{\circ }\\ & =-{\displaystyle \frac{1}{2},\text{atau}}\\ & =\cos ({270}^{\circ }-{30}^{\circ })=-\sin {30}^{\circ }=-\frac{1}{2}\\ \text{c}.\tan {315}^{\circ }& =\tan ({360}^{\circ }-{45}^{\circ })=-\tan {45}^{\circ }\\ & =-1,\text{atau}\\ & =\tan ({270}^{\circ }+{45}^{\circ })=-\cot {45}^{\circ }=-1\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Buktikan bahwa}\\ \\ & \text{a}.{\displaystyle \frac{\cos ({90}^{\circ }-B)}{\sec B}+\frac{\sin ({90}^{\circ }-B)}{\csc B}=2\sin B\cos B}\\ \\ & \text{b}.\tan C+\tan ({90}^{\circ }-C)=\sec C.\sec ({90}^{\circ }-C)\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{\cos ({90}^{\circ }-B)}{\sec B}+\frac{\sin ({90}^{\circ }-B)}{\csc B}}\\ & ={\displaystyle \frac{\sin B}{\sec B}+\frac{\cos B}{\csc B}}\\ & ={\displaystyle \frac{\sin B}{{\displaystyle \frac{1}{\cos B}}}+\frac{\cos B}{{\displaystyle \frac{1}{\sin B}}}}\\ & =\sin B\cos B+\sin B\cos B\\ & =2\sin B\cos B◼\end{array}\\ & \begin{array}{rl}\text{b}.& \tan C+\tan ({90}^{\circ }-C)\\ & =\tan C+\cot C\\ & ={\displaystyle \frac{\sin C}{\cos C}+\frac{\cos C}{\sin C}}\\ & ={\displaystyle \frac{{\sin }^{2}C+{\cos }^{2}C}{\sin C\cos C}={\displaystyle \frac{1}{\sin C\cos C}}}\\ & ={\displaystyle \frac{1}{\cos C}.\frac{1}{\sin C}}\\ & =\sec C.\csc C\\ & =\sec C.\sec ({90}^{\circ }-C)◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai}\\ & \text{a}.\tan (A-{90}^{\circ })\sin (-A)\\ & \text{b}.\cos {540}^{\circ }+\sin {690}^{\circ }\\ & \text{c}.\sin {2021}^{\circ }+\cos {2021}^{\circ }\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \tan (A-{90}^{\circ })\sin (-A)\\ & =\tan (-({90}^{\circ }-A))(-\sin A)\\ & =-\tan ({90}^{\circ }-A)(-\sin A)\\ & =\tan ({90}^{\circ }-A)(\sin A)\\ & =\cot A.\sin A\\ & ={\displaystyle \frac{\cos A}{\sin A}.\sin A}\\ & =\cos A\end{array}\\ & \begin{array}{rl}\text{b}.& \cos {540}^{\circ }+\sin {690}^{\circ }\\ & =\cos ({360}^{\circ }+{180}^{\circ })+\sin ({720}^{\circ }-{30}^{\circ })\\ & =\cos (0^{\circ }+{180}^{\circ })+\sin (0^{\circ }-{30}^{\circ })\\ & =\cos {180}^{\circ }+\sin (-{30}^{\circ })\\ & =\cos {180}^{\circ }-\sin {30}^{\circ }\\ & =-1-{\displaystyle \frac{1}{2}}\\ & =-{\displaystyle \frac{3}{2}}\end{array}\\ & \begin{array}{rl}\text{c}.& \sin {2021}^{\circ }+\cos {2021}^{\circ }\\ & =\sin ({5.360}^{\circ }+{221}^{\circ })+\cos ({5.360}^{\circ }+{221}^{\circ })\\ & =\sin (0^{\circ }+{221}^{\circ })+\cos (0^{\circ }+{221}^{\circ })\\ & =\sin {221}^{\circ }+\cos {221}^{\circ }\\ & =\sin ({180}^{\circ }+{41}^{\circ })+\cos ({180}^{\circ }+{41}^{\circ })\\ & =-\sin {41}^{\circ }-\cos {41}^{\circ }\end{array}\end{array}$

Lanjutan : Fungsi Eksponensial

1. Pengertian Fungsi Eksponen

Sebuah fungsi adalah relasi khusus dengan aturan tertentu. Fungsi adalah sebuah pemetaan yang memetakan setiap anggota domoain dengan tepat satu anggota kodomain. Jika suatu himpunan A sebagai domain yang setiap anggota himpunannya dipetakan ke tepat satu anggota himpunan B sebagai kodomain selanjutnya disebut fungsi dari himpunan A ke B atau  $f:A\to B$.

Perhatikanlah gambar berikut

Pada gambar di atas terlihat jelas bahwa setiap bilangan riil  $x$ dipetakan dengan tepat ke bilangan riil  $y$. Sehingga fungsi $f$  memtakan  $x\in A$  ke  $y$  atau  $f:x\to y$ dan aturan dari fungsi  $f$ ini sendiri ini biasanya sering dituliskan dalam notasi  $y=f(x)$. Selanjutnya untuk ilustrasi fungsi eksponensial adalah sebagai berikut:

Tampak jelas bahwa  $y\in B$  adalah  $y=f(x)=k.a^x$,  dengan  $k$ konstanta,  $x$ sebagai variabel bebas, serta  $a$ adalah bilangan basis atau bilangan pokok, dengan  $a>0$  dan   $a\ne 1$.

2. Garfik Fungsi Eksponen

a. Grafik fungsi eksponen  $y=f(x)=k.a^x$, dengan  $a>1$  dan   $x\in \mathbb{R}$.

Perhatikan ilustrasi berikut

a. Grafik fungsi eksponen  $y=f(x)=k.a^x$, dengan  $0<a<1$,  $a\in \mathbb{Q}$  dan   $x\in \mathbb{R}$.

Ilustrasi lain dari garfik fungsi eksponen adalah sebagai berikut

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Lengkapilah tabel berikut}\\ & \begin{array}{|ccccccccc|}\hline \text{fungsi}& -3& -2& -1& 0& 1& 2& 3& 4\\ f(x)=2^x& & & & & & & & \\ f(x)=2^{-x}& & & & & & & & \\ f(x)=3^x& & & & & & & & \\ f(x)=3^{-x}& & & & & & & & \\ \hline\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{|ccccccccc|}\hline \text{fungsi}& -3& -2& -1& 0& 1& 2& 3& 4\\ f(x)=2^x& {\displaystyle \frac{1}{8}}& {\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{2}}& 1& 2& 4& 8& 16\\ f(x)=2^{-x}& 8& 4& 2& 1& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{8}}& {\displaystyle \frac{1}{16}}\\ f(x)=3^x& {\displaystyle \frac{1}{27}}& {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{3}}& 1& 3& 9& 27& 81\\ f(x)=3^{-x}& 27& 9& 3& 1& {\displaystyle \frac{1}{3}}& {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{27}}& {\displaystyle \frac{1}{81}}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Gambarlah grafik fungsi eksponen berikut}\\ & \text{a}.f(x)=3^{x+1}\\ & \text{b}.f(x)=3^x+1\\ \\ & \mathbf{\text{Jawab}}\\ & \text{a. Untuk fungsi}f(x)=3^{x+1}\text{sebagai berikut}:\\ & \begin{array}{|cccccccccccc|}\hline \text{fungsi/titik}& -\mathrm{\infty }& \cdots & -3& -2& -1& 0& 1& 2& 3& \cdots & \mathrm{\infty }\\ f(x)=3^x& 0& \cdots & {\displaystyle \frac{1}{27}}& {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{3}}& 1& 3& 9& 27& \cdots & \text{TD}\\ f(x)=3^{x+1}& 0& \cdots & {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{3}}& 1& 3& 9& 27& 81& \cdots & \text{TD}\\ (x,f(x))& (-\mathrm{\infty },0)& \cdots & (-3,{\displaystyle \frac{1}{9}})& (-2,{\displaystyle \frac{1}{3}})& (-1,1)& (0,3)& (1,9)& (2,27)& (3,81)& \cdots & \\ \hline\end{array}\\ & \text{b. Dan untuk fungsi}f(x)=3^x+1\text{sebagai berikut}:\\ & \begin{array}{|cccccccccccc|}\hline \text{fungsi/titik}& -\mathrm{\infty }& \cdots & -3& -2& -1& 0& 1& 2& 3& \cdots & \mathrm{\infty }\\ f(x)=3^x& 0& \cdots & {\displaystyle \frac{1}{27}}& {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{3}}& 1& 3& 9& 27& \cdots & \text{TD}\\ f(x)=3^x+1& 1& \cdots & 1{\displaystyle \frac{1}{27}}& 1{\displaystyle \frac{1}{9}}& 1{\displaystyle \frac{1}{3}}& 2& 4& 10& 28& \cdots & \text{TD}\\ (x,f(x))& (-\mathrm{\infty },1)& \cdots & (-3,1{\displaystyle \frac{1}{27}})& (-2,1{\displaystyle \frac{1}{9}})& (-1,1{\displaystyle \frac{1}{3}})& (0,2)& (1,4)& (2,10)& (3,28)& \cdots & \\ \hline\end{array}\end{array}$

.

Limit Fungsi di tak Berhingga

Lanjutan 2 Limit Fungsi Trigonometri

 A. Teorema Apit

Misalkan  $f$,  $g$, dan   $h$  adalah fungsi yang memenuhi  $f(x)\le g(x)\le h(x)$ untuk seluruh titik di sekitar  $c$.  

Jika  $\underset{x\to c}{\text{lim}}f(x)=\underset{x\to c}{\text{lim}}g(x)=\text{L}$,  maka   $\underset{x\to c}{\text{lim}}g(x)=\text{L}$.

B. Penentuan nilai  $\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}}$  dan  $\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}}$.

Untuk bukti dari 

$\begin{array}{rl}1.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}=1}}\\ 2.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}=1}}\end{array}$.

Berikut penjabaran buktinya

Perhatikan ilustrasi gambar berikut ini

$\begin{array}{|cll|}\hline \text{No}& \text{Nama}& \text{Luas Bangun}\\ 1& \mathrm{△}\text{AOC}& \begin{array}{rl}{\text{L}}_{{}_{\mathrm{△}\text{AOC}}}& ={\displaystyle \frac{1}{2}.OA.CD}\\ & ={\displaystyle \frac{1}{2}.OA.OC.\sin x}\\ & ={\displaystyle \frac{1}{2}.r.r.\sin x}\\ & ={\displaystyle \frac{1}{2}{r}^2\sin x}\end{array}\\ 2& \text{Juring AOC}& \begin{array}{rl}{\text{L}}_{{}_{\text{Juring AOC}}}& ={\displaystyle \frac{x}{2\pi }.\pi r^2}\\ & ={\displaystyle \frac{x}{2\pi }.\pi .r^2}\\ & ={\displaystyle \frac{1}{2}x.r^2}\end{array}\\ 3& \mathrm{△}\text{AOB}& \begin{array}{rl}{\text{L}}_{{}_{\mathrm{△}\text{AOB}}}& ={\displaystyle \frac{1}{2}.OA.AB}\\ & ={\displaystyle \frac{1}{2}.OA.OA.\tan x}\\ & ={\displaystyle \frac{1}{2}.r.r.\tan x}\\ & ={\displaystyle \frac{1}{2}{r}^2\tan x}\end{array}\\ \hline\end{array}$.

Selanjutnya perhatikan pula bahwa dari fakta di atas dapat dituliskan sebagai berikut, yaitu:

$\begin{array}{|cc|}\hline \mathbf{\text{Bagian Pertama}}& \text{Bagian Kedua}\\ \begin{array}{r}\\ {\displaystyle \frac{1}{2}{r}^2\sin x}& \le & {\displaystyle \frac{1}{2}x{r}^2}& \le & {\displaystyle \frac{1}{2}{r}^2\tan x}\\ \\ \sin x& \le & x& \le & {\displaystyle \frac{\sin x}{\cos x}}\\ \\ 1& \le & {\displaystyle \frac{x}{\sin x}}& \le & {\displaystyle \frac{1}{\cos x}}\\ \\ 1& \ge & {\displaystyle \frac{\sin x}{x}}& \ge & \cos x\\ \\ \cos x& \le & {\displaystyle \frac{\sin x}{x}}& \le & 1\\ \\ & & & & \end{array}& \begin{array}{r}\\ {\displaystyle \frac{1}{2}{r}^2\sin x}& \le & {\displaystyle \frac{1}{2}x{r}^2}& \le & {\displaystyle \frac{1}{2}{r}^2\tan x}\\ \\ {\displaystyle \sin x}& \le & x& \le & {\displaystyle \tan x}\\ \\ {\displaystyle \frac{\sin x}{\tan x}}& \le & {\displaystyle \frac{x}{\tan x}}& \le & 1\\ \\ \cos x& \le & {\displaystyle \frac{x}{\tan x}}& \le & 1\\ \\ {\displaystyle \frac{1}{\cos x}}& \ge & {\displaystyle \frac{\tan x}{x}}& \ge & 1\\ 1& \le & {\displaystyle \frac{\tan x}{x}}& \le & {\displaystyle \frac{1}{\cos x}}\end{array}\\ \hline\end{array}$

Dan untuk mendapatkan nilai  yang diinginkan adalah:

$\begin{array}{rl}\mathbf{\text{Bag}}& \mathbf{\text{ian pertama}}:\\ & \underset{x\to 0}{\text{lim}}\cos x\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\le \underset{x\to 0}{\text{lim}}1}\\ & \iff 1\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\le 1}\\ & \text{Dengan teorema apit},\text{maka}\\ & \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=1.}\\ \mathbf{\text{Bag}}& \mathbf{\text{ian kedua}}:\\ & \underset{x\to 0}{\text{lim}}1\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan x}{x}\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1}{\cos x}}}\\ & \iff 1\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\le 1}\\ & \text{Dengan teorema apit},\text{maka}\\ & \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=1.}\end{array}$.

Selanjutnya untuk mendapatkan nilai $\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan x}{x}}$ dan  $\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan x}}$ dapat diperoleh dari bagian pertama dan kedua di atas, yaitu:

$\begin{array}{rl}\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan x}{x}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\times \frac{1}{\cos x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1}{\cos x}}}\\ & =1\times 1\\ & =1\\ \text{Demikia}& \text{n juga}\\ \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan x}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{1}{\cos x}\times \frac{x}{\sin x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{1}{\cos x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}}}\\ & =1\times 1\\ & =1\end{array}$.

Dari uraian panjang di atas telah ditunjukkan dengan bukti-buktinya bahwa

$\begin{array}{rl}1.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}=1}}\\ 2.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}=1}}\end{array}$

DAFTAR PUSTAKA

1. Kartini, Suprapto, Subandi, Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
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.

Lanjutan 2 Fungsi Eksponen

 C. 2. 2  Merasionalkan penyebut

Jika suatu pecahan penyebutnya mengandung bilangan irasional atau bentuk akar, maka penyebut ini dapat dibuat menjadi bilangan rasional. Perhatikanlah langkah berikut

$\begin{array}{rl}1.& {\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{\left(\sqrt{b^2}\right)}=\frac{a}{b}\sqrt{b}}\\ 2.& {\displaystyle \frac{a}{\sqrt[3]{b}}=\frac{a}{\sqrt[3]{b}}\times \frac{\sqrt[3]{b^2}}{\sqrt[3]{b^2}}=\frac{a\sqrt[3]{b^2}}{\left(\sqrt[3]{b^3}\right)}=\frac{a}{b}\sqrt[3]{b^2}}\\ 3.& {\displaystyle \frac{a}{\sqrt[5]{b^3}}={\displaystyle \frac{a}{\sqrt[5]{b^3}}\times \frac{\sqrt[5]{b^2}}{\sqrt[5]{b^2}}=\frac{a\sqrt[5]{b^2}}{\sqrt[5]{b^5}}=\frac{a}{b}\sqrt[5]{b^2}}}\end{array}$

Merasionalkan di atas adalah contoh bebrapa contoh model merasionalkan jika berjenis tunggal tetapi jika nanti jenisnya lebih dari itu, maka perhatikanlah simulasi contoh berikut

$\begin{array}{rl}& \\ 1.& {\displaystyle \frac{c}{a+\sqrt{b}}=\frac{c}{a+\sqrt{b}}.\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{c(a-\sqrt{b})}{a^2-b}}\\ 2.& {\displaystyle \frac{c}{a-\sqrt{b}}=\frac{c}{a-\sqrt{b}}.\frac{a+\sqrt{b}}{a+\sqrt{b}}=\frac{c(a+\sqrt{b})}{a^2-b}}\\ 3.& {\displaystyle \frac{c}{\sqrt{a}+\sqrt{b}}=\frac{c}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{c(\sqrt{a}-\sqrt{b})}{a-b}}\end{array}$

Perhatikanlah simulasi contoh di atas, bentuk $a+\sqrt{b}$ memiliki bentuk sekawan (irasional juga) $a-\sqrt{b}$, demikian juga bentuk $\sqrt{a}+\sqrt{b}$ memiliki sekawan $\sqrt{a}-\sqrt{b}$. Disamping itu ada bentuk khusus yatu bentuk  $\sqrt[3]{a}+\sqrt[3]{b}$ memiliki bentuk sekawan $\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Rasionalkanlah penyebut pecahan berikut}\\ & \text{dan serderhankanlah hasilnya}\\ & \text{a}.{\displaystyle \frac{2}{\sqrt{5}}\text{d}.{\displaystyle \frac{\sqrt{2}}{\sqrt{5}}}}\\ & \text{b}.{\displaystyle \frac{2}{5\sqrt{2}}\text{e}.{\displaystyle \frac{p}{\sqrt{q}}}}\\ & \text{c}.{\displaystyle \frac{6}{3\sqrt{5}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{2}{\sqrt{5}}={\displaystyle \frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}={\displaystyle \frac{2\sqrt{5}}{\sqrt{25}}={\displaystyle \frac{2}{5}\sqrt{5}}}}}\\ \text{b}.& {\displaystyle \frac{2}{5\sqrt{2}}={\displaystyle \frac{2}{5\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}={\displaystyle \frac{2\sqrt{2}}{5\sqrt{4}}=\frac{2\sqrt{2}}{5.2}={\displaystyle \frac{1}{5}\sqrt{2}}}}}\\ \text{c}.& {\displaystyle \frac{6}{3\sqrt{5}}={\displaystyle \frac{6}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}={\displaystyle \frac{6\sqrt{5}}{3\sqrt{25}}=\frac{6\sqrt{5}}{3.5}={\displaystyle \frac{2}{5}\sqrt{5}}}}}\\ \text{d}.& {\displaystyle \frac{\sqrt{2}}{\sqrt{5}}={\displaystyle \frac{\sqrt{2}}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{10}}{\sqrt{25}}=\frac{\sqrt{10}}{5}={\displaystyle \frac{1}{5}\sqrt{10}}}}\\ \text{e}.& {\displaystyle \frac{p}{\sqrt{q}}={\displaystyle \frac{p}{\sqrt{q}}\times \frac{\sqrt{q}}{\sqrt{q}}={\displaystyle \frac{p\sqrt{q}}{\sqrt{q^2}}=\frac{p\sqrt{q}}{q}={\displaystyle \frac{p}{q}\sqrt{q}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Rasionalkanlah penyebut pecahan berikut}\\ & \text{dan serderhankanlah hasilnya}\\ & \text{a}.{\displaystyle \frac{3}{6-\sqrt{5}}\text{f}.{\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}}}\\ & \text{b}.{\displaystyle \frac{3}{6+\sqrt{5}}\text{g}.\frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}}\\ & \text{c}.{\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}\text{h}.\frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}}\\ & \text{d}.{\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}\text{i}.\frac{\sqrt{3}}{\sqrt{6-2\sqrt{5}}}}\\ & \text{e}.{\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}\text{j}.\frac{\sqrt{3}}{\sqrt{6+2\sqrt{5}}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{3}{6-\sqrt{5}}={\displaystyle \frac{3}{6-\sqrt{5}}\times \frac{6+\sqrt{5}}{6+\sqrt{5}}={\displaystyle \frac{3(6+\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{18+3\sqrt{5}}{36-5}=\frac{18+3\sqrt{5}}{31}={\displaystyle \frac{1}{31}(18+3\sqrt{5})}}\\ \text{b}.& {\displaystyle \frac{3}{6+\sqrt{5}}={\displaystyle \frac{3}{6+\sqrt{5}}\times \frac{6-\sqrt{5}}{6-\sqrt{5}}={\displaystyle \frac{3(6-\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{18-3\sqrt{5}}{36-5}=\frac{18-3\sqrt{5}}{31}={\displaystyle \frac{1}{31}(18-3\sqrt{5})}}\\ \text{c}.& {\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}={\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}\times \frac{6+\sqrt{5}}{6+\sqrt{5}}={\displaystyle \frac{\sqrt{3}(6+\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{6\sqrt{3}+\sqrt{15}}{36-5}=\frac{6\sqrt{3}+\sqrt{15}}{31}={\displaystyle \frac{1}{31}(6\sqrt{3}+\sqrt{15})}}\\ \text{d}.& {\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}={\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}\times \frac{6-\sqrt{5}}{6-\sqrt{5}}={\displaystyle \frac{\sqrt{3}(6-\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{6\sqrt{3}-\sqrt{15}}{36-5}=\frac{6\sqrt{3}-\sqrt{15}}{31}={\displaystyle \frac{1}{31}(6\sqrt{3}-\sqrt{15})}}\\ \text{e}.& {\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}={\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}={\displaystyle \frac{3(\sqrt{6}+\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{3(\sqrt{6}+\sqrt{5})}{6-5}=\frac{3(\sqrt{6}+\sqrt{5})}{1}=3(\sqrt{6}+\sqrt{5})}\\ \text{f}.& {\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}={\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}={\displaystyle \frac{3(\sqrt{6}-\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{3(\sqrt{6}-\sqrt{5})}{6-5}=\frac{3(\sqrt{6}-\sqrt{5})}{1}=3(\sqrt{6}-\sqrt{5})}\\ \text{g}.& {\displaystyle \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}={\displaystyle \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}={\displaystyle \frac{\sqrt{3}(\sqrt{6}+\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{\sqrt{18}+\sqrt{15}}{6-5}=\frac{\sqrt{9.2}+\sqrt{15}}{1}=(3\sqrt{2}+\sqrt{15})}\\ \text{h}.& {\displaystyle \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}={\displaystyle \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}={\displaystyle \frac{\sqrt{3}(\sqrt{6}-\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{\sqrt{18}-\sqrt{15}}{6-5}=\frac{\sqrt{9.2}-\sqrt{15}}{1}=(3\sqrt{2}-\sqrt{15})}\\ \text{i}.& \frac{\sqrt{3}}{\sqrt{6-2\sqrt{5}}}=\frac{\sqrt{3}}{\sqrt{5+1-2\sqrt{5.1}}}={\displaystyle \frac{\sqrt{3}}{\sqrt{5}-\sqrt{1}}=\frac{\sqrt{3}}{\sqrt{5}-1}}\\ & =\frac{\sqrt{3}}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1}={\displaystyle \frac{\sqrt{3.5}+\sqrt{3.1}}{{\sqrt{5}}^2-1^2}=\frac{\sqrt{15}+\sqrt{3}}{5-1}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{15}+\sqrt{3})}\\ \text{j}.& \frac{\sqrt{3}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{3}}{\sqrt{5+1+2\sqrt{5.1}}}={\displaystyle \frac{\sqrt{3}}{\sqrt{5}+\sqrt{1}}=\frac{\sqrt{3}}{\sqrt{5}+1}}\\ & =\frac{\sqrt{3}}{\sqrt{5}+1}\times \frac{\sqrt{5}-1}{\sqrt{5}-1}={\displaystyle \frac{\sqrt{3.5}-\sqrt{3.1}}{{\sqrt{5}}^2-1^2}=\frac{\sqrt{15}-\sqrt{3}}{5-1}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{15}-\sqrt{3})}\end{array}\end{array}$.

$\begin{array}{l}\\ 3& \text{Rasionalkan penyebut dan sederhanakanlah}\\ & \text{a}.{\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}}\\ & \text{b}.{\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}}\\ & ={\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}\times {\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{\sqrt{2}+\sqrt{5}-\sqrt{7}}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{{(\sqrt{2}+\sqrt{5})}^2-{\left(\sqrt{7}\right)}^2}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{(2+2\sqrt{10}+5)-7}={\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{2\sqrt{10}}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{2\sqrt{10}}\times {\displaystyle \frac{\sqrt{10}}{\sqrt{10}}={\displaystyle \frac{\sqrt{20}+\sqrt{50}-\sqrt{70}}{2\times 10}}}}\\ & ={\displaystyle \frac{2\sqrt{5}+5\sqrt{2}+\sqrt{70}}{20}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}}\\ & ={\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\times \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{{(\sqrt{2}+\sqrt{3})}^2-{\left(\sqrt{5}\right)}^2}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{(2+2\sqrt{6}+3)-5}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}}\\ & =\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}\\ & ={\displaystyle \frac{\sqrt{12}+\sqrt{18}+\sqrt{30}}{2\times 6}}\\ & ={\displaystyle \frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12}}\end{array}\end{array}$

Lanjutan Fungsi Eksponen

$\text{C.2 Operasi Bilangan Bentuk Akar}$.

C. 2. 1  Sifat-sifat yang berlaku pada operasi bilangan bentuk akar

$\begin{array}{rl}& \\ 1.& a\sqrt[n]{c}+b\sqrt[n]{c}=(a+b)\sqrt[n]{c}\\ 2.& a\sqrt[n]{c}-b\sqrt[n]{c}=(a-b)\sqrt[n]{c}\\ 3.& \sqrt[n]{a}.\sqrt[n]{b}=\sqrt[n]{ab}\\ 4.& \sqrt[n]{a^n}=a\\ 5.& a\sqrt[n]{c}xb\sqrt[n]{d}=ab\sqrt[n]{cd}\\ 6.& \frac{a\sqrt[n]{c}}{b\sqrt[n]{d}}=\frac{a}{b}.\sqrt[n]{\frac{c}{d}}\\ 7.& \sqrt{(a+b)+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\\ 8.& \sqrt{(a+b)-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}\end{array}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Sederhanakanlah bentuk akar berikut}\\ & \begin{array}{llllll}\text{a}.& \sqrt{8}& \text{f}.& \sqrt[3]{16}& \text{k}.& \sqrt{8x^5},x\ge 0\\ \text{b}.& \sqrt{12}& \text{g}.& \sqrt[3]{32}& \text{l}.& \sqrt{48x^6{y}^{11}},y\ge 0\\ \text{c}.& \sqrt{27}& \text{h}.& \sqrt[3]{54}& \text{m}.& 2\sqrt{8}\times \sqrt{3}\\ \text{d}.& \sqrt{28}& \text{i}.& \sqrt[3]{81}& \text{n}.& 3\sqrt{6}\times 2\sqrt{2}\\ \text{e}.& \sqrt{32}& \text{j}.& \sqrt[3]{625}& \text{o}.& 2\sqrt[3]{6}\times 6\sqrt[3]{9}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \text{a}.& \sqrt{8}=\sqrt{4\times 2}=\sqrt{2^2}\times \sqrt{2}=2\sqrt{2}\\ \text{b}.& \sqrt{12}=\sqrt{4\times 3}=\sqrt{2^2}\times \sqrt{3}=2\sqrt{3}\\ \text{c}.& \sqrt{27}=\sqrt{9\times 3}=\sqrt{3^2}\times \sqrt{3}=3\sqrt{3}\\ \text{d}.& \sqrt{28}=\sqrt{4\times 7}=\sqrt{2^2}\times \sqrt{7}=2\sqrt{7}\\ \text{e}.& \sqrt{32}=\sqrt{16\times 2}=\sqrt{4^2}\times \sqrt{2}=4\sqrt{2}\\ \text{f}.& \sqrt[3]{16}=\sqrt[3]{8\times 2}=\sqrt[3]{2^3}\times \sqrt[3]{2}=2\sqrt[3]{2}\\ \text{g}.& \sqrt[3]{32}=\sqrt[3]{8\times 4}=\sqrt[3]{2^3}\times \sqrt[3]{4}=2\sqrt[3]{4}\\ \text{h}.& \sqrt[3]{54}=\sqrt[3]{27\times 2}=\sqrt[3]{3^3}\times \sqrt[3]{2}=3\sqrt[3]{2}\\ \text{i}.& \sqrt[3]{81}=\sqrt[3]{27\times 3}=\sqrt[3]{3^3}\times \sqrt[3]{3}=3\sqrt[3]{3}\\ \text{j}.& \sqrt[3]{625}=\sqrt[3]{125\times 5}=\sqrt[3]{5^3}\times \sqrt[3]{5}=5\sqrt[3]{5}\\ \text{k}.& \sqrt{8x^5}=\sqrt{4.2.x^4.x^1}=\sqrt{2^2}\times \sqrt{2}\times \sqrt{x^4}\times \sqrt{x}\\ & =2.\sqrt{2}.x^2.\sqrt{x}=2x^2\sqrt{2x},x\ge 0\\ \text{l}.& \sqrt{48x^6{y}^{11}}=\sqrt{16.3.x^6.y^{10}.y^1}\\ & =\sqrt{4^2}\times \sqrt{3}\times \sqrt{x^6}\times \sqrt{y^{10}}\times \sqrt{y}\\ & =4\sqrt{3}.x^3.y^5.\sqrt{y}=4x^3{y}^5\sqrt{3y},\ge 0\\ \text{m}.& 2\sqrt{8}\times \sqrt{3}=2\sqrt{4\times 2}\times \sqrt{3}\\ & =2\sqrt{2^2}\times \sqrt{2}\times \sqrt{3}=2.2.\sqrt{2.3}\\ & =4\sqrt{6}\\ \text{n}.& 3\sqrt{6}\times 2\sqrt{2}=3\sqrt{2\times 3}\times 2\sqrt{2}\\ & =3\times 2\times \sqrt{2^2\times 3}=6\times \sqrt{2^2}\times \sqrt{3}\\ & =6\times 2\times \sqrt{3}=12\sqrt{3}\sqrt{6}\\ \text{o}.& 2\sqrt[3]{6}\times 6\sqrt[3]{9}=2.6.\sqrt[3]{6\times 9}=12\times \sqrt[3]{2.3.3.3}\\ & =12\times \sqrt[3]{{2.3}^3}=12\times \sqrt[3]{2}\times \sqrt[3]{3^3}\\ & =12\times \sqrt[3]{2}\times 3\\ & =36\sqrt[3]{2}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah pangkat rasional dari}\\ & \text{a}.\sqrt{y\sqrt[3]{x^{2}y}}\\ & \text{b}.\sqrt[3]{x^3\sqrt[5]{x^3\sqrt{x^3}}}\\ & \text{c}.\sqrt[3]{x^2\sqrt{x\sqrt[5]{x^2}}}\\ & \text{d}.xyz\sqrt[3]{{\displaystyle \frac{xy}{z^5}}}\sqrt[3]{{\displaystyle \frac{xz}{y^5}}}\sqrt[3]{{\displaystyle \frac{yz}{x^5}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \text{a}.& \sqrt{y\sqrt[3]{x^{2}y}}=\sqrt{y{\left(x^{2}y\right)}^{\frac{1}{3}}}={\left(y{\left(x^{2}y\right)}^{\frac{1}{3}}\right)}^{\frac{1}{2}}\\ & =y^{{.}^{\frac{1}{2}}}.x^{{.}^{\frac{2}{3}.\frac{1}{2}}}.y^{{.}^{\frac{1}{3}.\frac{1}{2}}}=y^{{.}^{\frac{1}{2}+\frac{1}{6}}}{x}^{{.}^{\frac{1}{3}}}=x^{{.}^{\frac{1}{3}}}.y^{{.}^{\frac{4}{6}}}\\ & =x^{{.}^{\frac{1}{3}}}.y^{{.}^{\frac{2}{3}}}\\ \text{b}.& \sqrt[3]{x^3\sqrt[5]{x^3\sqrt{x^3}}}=\sqrt[3]{x^3\sqrt[5]{x^3.x^{{.}^{\frac{3}{2}}}}}=\sqrt[3]{x^3.x^{{.}^{\frac{3}{5}}}{x}^{{.}^{\frac{3}{2.5}}}}\\ & =x^{{.}^{\frac{3}{3}}}.x^{{.}^{\frac{3}{5.3}}}.x^{{.}^{\frac{3}{2.5.3}}}=x^1+x^{{.}^{\frac{1}{5}}}.x^{{.}^{\frac{1}{10}}}\\ & =x^{{.}^{1+\frac{1}{5}+\frac{1}{10}}}=x^{{.}^{\frac{10+2+1}{10}}}=x^{{.}^{\frac{13}{10}}}\\ \text{c}.& \sqrt[3]{x^2\sqrt{x\sqrt[5]{x^2}}}=\sqrt[3]{x^2\sqrt{x.x^{{.}^{\frac{2}{5}}}}}=\sqrt[3]{x^2.x^{{.}^{\frac{1}{2}}}.x^{{.}^{\frac{2}{5.2}}}}\\ & =x^{{.}^{\frac{2}{3}}}.x^{{.}^{\frac{1}{2.3}}}.x^{{.}^{\frac{2}{5.2.3}}}=x^{{.}^{\frac{2}{3}}}.x^{{.}^{\frac{1}{6}}}.x^{{.}^{\frac{1}{15}}}\\ & =x^{{.}^{\frac{20+5+2}{30}}}=x^{{.}^{\frac{27}{30}}}=x^{{.}^{\frac{9}{10}}}\\ \text{d}.& xyz\sqrt[3]{{\displaystyle \frac{xy}{z^5}}}\sqrt[3]{{\displaystyle \frac{xz}{y^5}}}\sqrt[3]{{\displaystyle \frac{yz}{x^5}}}=xyz\sqrt[3]{{\displaystyle \frac{x^2{y}^2{z}^2}{x^5{y}^5{z}^5}}}\\ & =xyz\sqrt[3]{{\displaystyle \frac{1}{x^{(5-2)}{y}^{(5-2)}{z}^{(5-2)}}}}\\ & =xyz\sqrt[3]{{\displaystyle \frac{1}{x^3{y}^3{z}^3}}}=xyz\sqrt[3]{{\displaystyle \frac{1}{(xyz{)}^3}}}\\ & =xyz.{\displaystyle \frac{1}{xyz}={\displaystyle \frac{xyz}{xyz}=1}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}a,b\text{bilangan positif dan}\\ & \sqrt{a^{2}b\sqrt[3]{a{b}^2\sqrt{ab}}}=a^x.b^y,\text{tentukan nilai}x-y\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sqrt{a^{2}b\sqrt[3]{a{b}^2\sqrt{ab}}}=a^x.b^y\\ & \text{perhatikan cara menguraikannya}\\ & \sqrt{a^{2}b\sqrt[3]{a{b}^2\sqrt{ab}}}=\sqrt{a^{2}b\sqrt[3]{a{b}^2.a^{{.}^{\frac{1}{2}}}{b}^{{.}^{\frac{1}{2}}}}}\\ & =\sqrt{a^{2}b\sqrt[3]{a^{{.}^{1+\frac{1}{2}}}{b}^{{.}^{2+\frac{1}{2}}}}}=\sqrt{a^{2}b\sqrt[3]{a^{{.}^{\frac{3}{2}}}{b}^{{.}^{\frac{5}{2}}}}}\\ & =\sqrt{a^{2}b.a^{{.}^{\frac{3}{2.3}}}{b}^{{.}^{\frac{5}{2.3}}}}=\sqrt{a^{2}b.a^{{.}^{\frac{1}{2}}}{b}^{{.}^{\frac{5}{6}}}}\\ & =\sqrt{a^{{.}^{2+\frac{1}{2}}}.b^{{.}^{1+\frac{5}{6}}}}=\sqrt{a^{{.}^{\frac{5}{2}}}{b}^{{.}^{\frac{11}{6}}}}=a^{{.}^{\frac{5}{2.2}}}{b}^{{.}^{\frac{11}{6.2}}}\\ & =a^{{.}^{\frac{5}{4}}}{b}^{{.}^{\frac{11}{12}}}\\ & =a^x{b}^y\\ & \text{maka}x={\displaystyle \frac{5}{4},\text{dan}y=\frac{11}{12}}\\ & x-y={\displaystyle \frac{5}{4}-\frac{11}{12}={\displaystyle \frac{15-11}{12}=\frac{4}{12}={\displaystyle \frac{1}{3}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& (\mathbf{\text{Matematika Dasar UM UGM 2008}})\\ & \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{\sqrt[6]{x^2}\sqrt[3]{x^2\sqrt{x+1}}}{x\sqrt[6]{x+1}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sqrt[6]{x^2}\sqrt[3]{x^2\sqrt{x+1}}}{x\sqrt[6]{x+1}}}\\ & ={\displaystyle \frac{\sqrt[6]{x^2}.\sqrt[3.2]{{(x^2.\sqrt{x+1})}^2}}{\sqrt[6]{x^6}.\sqrt[6]{x+1}}={\displaystyle \frac{\sqrt[6]{x^2}.\sqrt[6]{x^4.(x+1)}}{\sqrt[6]{x^6(x+1)}}}}\\ & ={\displaystyle \frac{\sqrt[6]{x^{(2+4)}(x+1)}}{\sqrt[6]{x^6(x+1)}}={\displaystyle \frac{\sqrt[6]{x^6(x+1)}}{\sqrt[6]{x^6(x+1)}}}}\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai dari}\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ & \text{berikut uraiannya}\\ & \text{Misalkan}\\ & x^2=x^2,\text{maka}{x}^2=1+(x^2-1)\\ & x^2=1+(x-1)(x+1)\\ & x^2=1+(x-1)\sqrt{(x+1{)}^2}\\ & x^2=1+(x-1)\sqrt{1+((x+1{)}^2-1)}\\ & x^2=1+(x-1)\sqrt{1+(x+1-1)(x+1+1)}\\ & x^2=1+(x-1)\sqrt{1+x(x+2)}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{(x+2{)}^2}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+((x+2{)}^2-1)}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+2-1)(x+2+1)}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{(x+3{)}^2}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+((x+3{)}^2-1)}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+3-1)(x+3+1)}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{(x+4{)}^2}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+((x+4{)}^2-1)}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+4-1)(x+4+1)}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)(x+5)}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{...}}}}}\\ & x=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\cdots }}}}}\\ & \text{maka}\\ & \cdots =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ & \text{Jelas tampak bahwa nilai}x\text{yang memenuhi adalah}\\ & x=3\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakanlah bentuk}\\ & \text{a}.\sqrt{3+2\sqrt{2}}\text{d}.\sqrt{21-4\sqrt{5}}\\ & \text{b}.\sqrt{6-\sqrt{32}}\text{e}.\sqrt{6-2\sqrt{8}}\\ & \text{c}.\sqrt{7+4\sqrt{3}}\text{f}.\sqrt{5+\sqrt{24}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Inga}& \text{t bahwa}:\sqrt{a+b\pm 2\sqrt{ab}}=\sqrt{a}\pm \sqrt{b},a\ge b\\ \text{a}.& \sqrt{3+2\sqrt{2}}=\sqrt{2+1+2\sqrt{2.1}}=\sqrt{2}+1\\ \text{b}.& \sqrt{6-\sqrt{32}}=\sqrt{6-\sqrt{4.4.2}}=\sqrt{4+2-2\sqrt{4.2}}\\ & =\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \text{c}.& \sqrt{7+4\sqrt{3}}=\sqrt{4+3+2.2\sqrt{3}}=\sqrt{4+3+2\sqrt{4.3}}\\ & =\sqrt{4}+\sqrt{3}=2+\sqrt{3}\\ \text{d}.& \sqrt{21-4\sqrt{5}}=\sqrt{20+1-2.2\sqrt{5}}=\sqrt{20+1-2\sqrt{4.5}}\\ & =\sqrt{20+1-2\sqrt{20.1}}=\sqrt{20}-1=\sqrt{4.5}-1=2\sqrt{5}-1\\ \text{e}.& \sqrt{6-2\sqrt{8}}=\sqrt{4+2-2\sqrt{4.2}}=\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \text{f}.& \sqrt{5+\sqrt{24}}=\sqrt{3+2+\sqrt{4.3.2}}=\sqrt{3+2+2\sqrt{3.2}}\\ & =\sqrt{3}+\sqrt{2}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Sederhanakan bentuk berikut}\\ & \text{a}.\sqrt{0,3+\sqrt{0,08}}\\ & \text{b}.\sqrt{94+2\sqrt{2013}}\\ & \text{c}.\sqrt{17+4\sqrt{15}}=a\sqrt{3}+b\sqrt{5},\text{tentukan}b-a\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Inga}& \text{t bahwa}:\sqrt{p+q\pm 2\sqrt{pq}}=\sqrt{p}\pm \sqrt{q},p\ge q\\ \text{a}.& \sqrt{0,3+\sqrt{0,08}}=\sqrt{0,3+\sqrt{4.(0,02)}}=\sqrt{0,3+2\sqrt{0,02}}\\ & =\sqrt{0,2+0,1+2\sqrt{(0,2).(0,1)}}=\sqrt{0,2}+\sqrt{0,1}\\ \text{b}.& \sqrt{94+2\sqrt{2013}}=\sqrt{61+33+2\sqrt{61.33}}=\sqrt{61}+\sqrt{33}\\ \text{c}.& \sqrt{17+4\sqrt{15}}=\sqrt{17+2.2\sqrt{15}}=\sqrt{17+2\sqrt{4.15}}\\ & =\sqrt{17+2\sqrt{60}}=\sqrt{12+5+2\sqrt{12.5}}=\sqrt{12}+\sqrt{5}\\ & =\sqrt{4.3}+\sqrt{1.5}=2\sqrt{3}+1\sqrt{5}=a\sqrt{3}+b\sqrt{5}\{\begin{array}{ll}a& =2\\ b& =1\end{array}\\ & \text{maka}b-a=1-2=-1\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Bentuk paling sederhana dari}\\ & \sqrt[4]{49-20\sqrt{6}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sqrt[4]{49-20\sqrt{6}}\\ & =\sqrt{\sqrt{49-2.10\sqrt{6}}}=\sqrt{\sqrt{49-2\sqrt{100.6}}}\\ & =\sqrt{\sqrt{49-2\sqrt{600}}}=\sqrt{\sqrt{25+24-2\sqrt{25.24}}}\\ & =\sqrt{\sqrt{25}-\sqrt{24}}=\sqrt{5-\sqrt{24}}=\sqrt{5-\sqrt{4.6}}\\ & =\sqrt{5-2\sqrt{6}}=\sqrt{3+2-2\sqrt{3.2}}=\sqrt{3}-\sqrt{2}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Bentuk paling sederhana dari}\\ & {\left(\sqrt{52+6\sqrt{43}}\right)}^3-{\left(\sqrt{52-6\sqrt{43}}\right)}^3\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Inga}& \text{tlah bentuk}:(A-B{)}^3=A^3-B^3-3AB(A-B)\\ & \iff A^3-B^3=(A-B{)}^3+3AB(A-B)\\ \bullet & \sqrt{52+6\sqrt{43}}=\sqrt{52+2.3\sqrt{43}}=\sqrt{52+2\sqrt{43.9}}\\ & =\sqrt{43+9+2\sqrt{43.9}}=\sqrt{43}+\sqrt{9}=\sqrt{43}+3\\ \bullet & \sqrt{52-6\sqrt{43}}=\sqrt{52-2.3\sqrt{43}}=\sqrt{52-2\sqrt{43.9}}\\ & =\sqrt{43+9-2\sqrt{43.9}}=\sqrt{43}-\sqrt{9}=\sqrt{43}-3\\ & \text{misalkan}\{\begin{array}{ll}A& =\sqrt{43}+3\\ B& =\sqrt{43}-3\end{array}\\ & A^3-B^3=(A-B{)}^3+3AB(A-B)\\ & ={(\sqrt{43}+3-(\sqrt{43}-3))}^3+3(\sqrt{43}+3)(\sqrt{43}-3)(\sqrt{43}+3-(\sqrt{43}-3))\\ & ={\left(6\right)}^3+3({\sqrt{43}}^2-3^2)\left(6\right)\\ & =216+18(43-9)\\ & =216+18.34\\ & =216+612\\ & =828\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Sederhanakanlah bentuk akar berikut}\\ & \begin{array}{ll}\text{a}.& 3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ \text{b}.& 5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ \text{c}.& 3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ \text{d}.& (2+\sqrt{2})(4-\sqrt{2})\\ \text{e}.& (3\sqrt{2}+\sqrt{3})(\sqrt{2}-2\sqrt{3})\\ \text{f}.& (4\sqrt{3}-3\sqrt{5})(2\sqrt{3}+\sqrt{5})\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \text{a}.& 3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ & =3\sqrt{2}+5\sqrt{4.2}-\sqrt{16.2}\\ & =3\sqrt{2}+5.2\sqrt{2}-4\sqrt{2}\\ & =(3+10-4)\sqrt{2}=9\sqrt{2}\\ \text{b}.& 5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ & =5\sqrt{3}+5\sqrt{9.3}-2\sqrt{25.3}\\ & =5\sqrt{3}+5.3\sqrt{3}-2.5\sqrt{3}\\ & =(5+15-10)\sqrt{3}=10\sqrt{3}\\ \text{c}.& 3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ & =3\sqrt{25.2}-4\sqrt{16.2}-\sqrt{1.2}\\ & =3.5\sqrt{2}-4.4\sqrt{2}-1\sqrt{2}\\ & =(15-16-1)\sqrt{2}=-2\sqrt{2}\\ \text{d}.& (2+\sqrt{2})(4-\sqrt{2})\\ & =2.4-2.\sqrt{2}+4.\sqrt{2}-\sqrt{2.2}\\ & =8+(4-2)\sqrt{2}-2\\ & =6+2\sqrt{2}\\ \text{e}.& (3\sqrt{2}+\sqrt{3})(\sqrt{2}-2\sqrt{3})\\ & =3\sqrt{2.2}-3.2.\sqrt{2.3}+\sqrt{3.2}-2\sqrt{3.3}\\ & =3.2-6\sqrt{6}+1\sqrt{6}-3.2\\ & =6-6+(1-6)\sqrt{6}=-5\sqrt{6}\\ \text{f}.& (4\sqrt{3}-3\sqrt{5})(2\sqrt{3}+\sqrt{5})\\ & =4.2.\sqrt{3.3}+4\sqrt{3.5}-3.2.\sqrt{5.3}-3\sqrt{5.5}\\ & =8.3+4\sqrt{15}-6\sqrt{15}-3.5\\ & =24-15+(4-6)\sqrt{15}=9-2\sqrt{15}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X 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 X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.

Lanjutan 3 Materi Geometri Ruang (Dimensi Tiga)

B. 3 Kedudukan Garis Terhadap Garis dalam Ruang

Pada bangun sebuah kubus di mana bangun ruang ini dibatasi oleh tiga pasang bidang persegi. Setiap daerah persegi membatasi kubus yang disebut sebagai sisi kubus. Setiap dua sisi yang tidak sejajar  akan saling berpotongan pada sebuah garis yang disebut rusuk, yaitu AB, BC, AE, dan lain-lainnya. Perhatikan ilustrasi kubus ABCD.EFGH berikut

Sehingga sebuah kubus memiliki 12 rusuk yang sama panjang. Jika Anda perhatikan susunan dan struktur dari rusuk-rusk kubus kubus di atas, maka 12 di atas dapat dibagi menjadi 3 kelompok, yaitu:

• kelompok pertama : AB, DC, HG, dan EF
• kelompok kedua : AD, BC, FG, dan EH
• kelompok ketiga : AE, BF, CG, dan DH

Selanjutnya Anda juga dapat menemukan atau melihat bahwa setiap tiga rusuk bertemu di suatu titik yang selanjutnya disebut titik sudut kubus. Sebuah kubus memiliki 8 titik sudut. Terdapat pasangan-pasangan titik sudut yang tidak terletak pada sebuah bidang sisi, yaitu titik A dengan G, B dengan H, C dengan E, serta D dengan F, pasangan titik yang demikian disebut dengan pasangan titik yang berhadapan.

Ruas garis yang menghubungkan dua buah titik yang berhadapan disebut diagonal ruang kubus. Karena terdapat empat pasang titik berhadapan, maka terdapat 4 buah diagonal ruang, yaitu: AB, BH, CE, dan DF. Selain itu juga karena sisi kubus ada 6 buah dan masing-masing memiliki dua diagonal, sehingga terdapat 12 diagonal sisi, yaitu: AC, BD, AF, BE, EG, FH, AH, DE, BG, dan CF.

Anda juga dapat menemukan pasangan rusuk yang sejajar tetapi tidak terletak pada sebuah bidang sisi, misalnya AB dengan HG, dan lain-lainnya. Dari sana Anda akan menemukan 6 pasang rusuk yang yang berhadapan. Bidang yang melalui dua rusuk yang berhadapan disebut bidang diagonal. Sehingga dalam sebuah kubus terdapat 6 buah bidang diagonal.

Perhatikan letak rusuk AB dan DH, kedua rusuk itu tidak terletak pada sebuah bidang, maka dikatakan AB dan DH dua rusuk saling bersilangan demikian juga garis yang lain dengan kondisi semisal. Selain itu hubungan dua garis adalah saling sejajar dan saling tegak lurus.

B. 4 Kedudukan Garis terhadap Bidang yang Sejajar

Jarak anatar suatu garis  $g$  dan bidang  $\alpha$ yang saling sejajar adalah jarak antara sebarang titik A pada  $g$  dengan bidang  $\alpha$. Jika proyeksi titik A pada garis $g$ ke bidang $\alpha$ yang saling sejajar adalah A', maka AA' adalah jarak antara garis  $g$  dan bidang  $\alpha$ yang saling sejajar.

Berikut ilustrasinya

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 9.& \text{Diketahui kubus ABCD.EFGH dengan rusuk 8 cm}\\ & \text{Tentukanlah jarak titik A ke bidang}\\ & \text{a}\text{BCGF}\\ & \text{b}\text{BCHE}\\ \\ & \text{Jawab}:\\ & \text{Perhatikan ilustrasi berikut ini}\end{array}$

$.\begin{array}{rl}\text{a}.\text{titik}& A\text{ke}\text{BCGF}=\text{A}\text{ke}\text{B}=\text{AB}=8cm\\ \text{b}.\text{titik}& \text{A}\text{ke}\text{BCHE}=A\text{ke tengah-tengan}\text{BE}\\ & ={\displaystyle \frac{1}{2}\text{diagonal sisi kubus}={\displaystyle \frac{1}{2}\left(8\sqrt{2}\right)}}\\ & =4\sqrt{2}cm\end{array}$.

$\begin{array}{l}\\ 10.& \text{Pada kubus ABCD.EFGH dengan panjang rusuk}\\ & 6cm,\text{titik S dan R berturut turut adalah pusat}\\ & \text{bidang EFGH dan ABCD. Tentukanlah jarak}\\ & \text{antara garis RF dan DS}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi kubus ABCD.EFGH berikut}\end{array}$.

$.\begin{array}{rl}& \text{Diketahui bahwa}:\\ & BD=6\sqrt{2}cm(\text{diagonal sisi kubus})\\ & RD={\displaystyle \frac{1}{2}BD={\displaystyle \frac{1}{2}\left(6\sqrt{2}\right)=3\sqrt{2}cm}}\\ & DS={\displaystyle \frac{1}{2}(sisi)\sqrt{6}={\displaystyle \frac{1}{2}(6)\sqrt{6}=3\sqrt{6}cm}}\\ & \text{Langkah selanjutnya}\\ & \begin{array}{rl}\left[DSR\right]& =\left[DSR\right]\\ {\displaystyle \frac{1}{2}\times DS\times RM}& ={\displaystyle \frac{1}{2}\times DR\times RS}\\ RM& ={\displaystyle \frac{{\displaystyle \frac{1}{2}\times DR\times RS}}{{\displaystyle \frac{1}{2}\times DS}}}\\ & ={\displaystyle \frac{DR\times RS}{DS}}\\ & ={\displaystyle \frac{3\sqrt{2}\times 6}{3\sqrt{6}}}\\ & ={\displaystyle \frac{18\sqrt{2}}{3\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}}\\ & ={\displaystyle \frac{18\sqrt{12}}{18}=\sqrt{12}=\sqrt{4.3}}\\ & =2\sqrt{3}cm\end{array}\\ \\ \text{Jadi},& \text{jarak garis RF ke DS}=2\sqrt{3}m\end{array}$

B. 5 Kedudukan antara Dua Bidang yang Sejajar

Jarak antara dua bidang $\alpha$ dan $\beta$ yang saling sejajar adalah sama dengan jarak antara sebarang titik A pada bidang  $\alpha$  dan A' pada bidang  $\beta$  dengan A' adalah proyeksi titik A pada bidang  $\beta$.

Berikut ilustrasinya

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 11.& \text{Diketahui kubus ABCD.EFGH dengan rusuk 8 cm}\\ & \text{Tentukanlah jarak bidang ABCD ke bidang}\\ & \text{a}\text{BCGF}\\ & \text{b}\text{EFGH}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.\text{bidang}& \text{ABCD}\text{ke}\text{BCGF}=\text{tidak terdefinisi}\\ \text{alasan}& :\mathbf{\text{karena saling tegak lurus}}\\ \text{b}.\text{bidang}& \text{ABCD}\text{ke}\text{EFGH}=A\text{ke }\text{E}=8cm\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Pada kubus ABCD.EFGH dengan panjang rusuk}\\ & 8cm,\text{titik P, Q, R, dan S berturut turut berada}\\ & \text{di pertengahan rusuk BC, CG, DH, dan AD.}\\ & \text{Tentukanlah jarak antara bidang ABGH dan}\\ & \text{bidang PQRS}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi kubus ABCD.EFGH berikut}\end{array}$.

$.\begin{array}{rl}& \text{Diketahui bahwa}:\\ & BP={\displaystyle \frac{1}{2}BC={\displaystyle \frac{1}{2}8=4cm(\text{rusuk kubus})}}\\ & \sin \mathrm{\angle }PB{P}^{\prime }={\displaystyle \frac{P{P}^{\prime }}{BP}\iff P{P}^{\prime }=BP\times \sin \mathrm{\angle }PB{P}^{\prime }}\\ & =4\times \sin {45}^{\circ }=4\times {\displaystyle \frac{1}{2}\sqrt{2}=2\sqrt{2}cm}\\ & DS={\displaystyle \frac{1}{2}(sisi)\sqrt{6}={\displaystyle \frac{1}{2}(6)\sqrt{6}=3\sqrt{6}cm}}\\ & \text{Jadi, jarak bidang ABGH ke PQRS adalah}2\sqrt{2}cm\end{array}$.

$\begin{array}{l}\\ 13.& \text{Pada sebuah kubus ABCD.EFGH, tunjukkan}\\ & \text{bahwa bidang AFH sejajar dengan bidang BDG}\\ \\ & \text{Bukti}:\\ & \text{Perhatikan ilustrasi kubus ABCD.EFGH berikut}\end{array}$

$.\begin{array}{rl}& \text{Untuk menunjukkan bidang AFH dan BDG}\\ & \text{itu sejajar, maka harus ditunjukkan bahwa}\\ & \text{kedua bidang itu masing-masing memuat}\\ & \text{dua garis berpotongan yang sepasang-sepasang}.\\ & \text{Bidang ADHE}//\text{bidang BCGF},\text{sedang bidang}\\ & \text{ABGH memotong kedua bidang yang sejajar itu}\\ & \text{berupa garis AH dan BG, maka AH}//\text{BG}.\\ & \text{Demikian pula bidang ABCD dan EFGH yang}\\ & \text{dipotong oleh bidang BDHF, masing-masing}\\ & \text{berupa garis BD dan FH, maka BD}//\text{FH}\\ & \text{Berdasarkan fakta di atas, yaitu}:\\ & \text{garis AH}//\text{BG dan BD}//\text{FH, maka sudah}\\ & \text{cukup menunjukkan bahwa}\text{bidang AFH}\\ & \text{bidang BDG sejajar}◼\end{array}$.

DAFTAR PUSTAKA

1. Rasiman. 2000. Diktat Geometri. Semarang: IKIP Semarang
2. Tampomas, H. 1999. Seribu Pena Matematika SMU Kelas 3. Jakarta: ERLANGGA.

Lanjutan 2 Materi Geometri Ruang (Dimensi Tiga)

 $\text{B. 3. Kedudukan Titik terhadap Bidang}$.

Perhatikan kubus ABCD.EFGH berikut

Pandang titik A terhadap bidang EFGH. Tampak bahwa titik A terletak tidak pada bidang EFGH termasuk juga titik-titik yang lain yang tidak terletak pada bidang EFGH tersebut yaitu: titik B, C, dan D. Walaupun demikian pada kubus ABCD.EFGH tersebut terdapat beberapa titik yang terletak pada bidang EFGH, yaitu: titik E, F, G, G, dan P. Selanjutnya hubungan kedudukan suatu titik terhadap bidang dapat kita tuliskan sebagaimana dalam tabel berikut:

$\begin{array}{|cll|}\hline \text{No}& \text{Kedudukan}& \text{Keterangan}\\ 1.& \text{pada bidang}& \text{titik terletak pada bidang}\\ 2.& \text{di luar bidang}& \text{titik berada di luar bidang}\\ \hline\end{array}$

Sebagai tambahan penjelasan perhatikan pula gambar limas D.ABC berikut

Pada limas D.ABC di atas terlihat jelas bagwa titik D terletak di luar bidan ABC, tetapi titik A atau titik B ataupun titik C, semuanya terletak pada bidang ABC pada bangun limas D.ABC di atas.

Selanjutnya dalam penentuan jarak antar titik dengan suatu bidang adalah panjang ruas garis secara tegak lurus yang menghungkan titik tersebut dengan bidang yang dimaksud.

Sebagai ilustrasi adalah gambar berikut

Pada ilsutrasi gambar di atas jarak titik A ke bidang V adalah sepanjang ruas garis AB yang mana ruas garis AB tegak lurus dengan bidang V.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 7.& \text{Diketahui kubus ABCD.EFGH dengan rusuk 8 cm}.\\ & \text{Tentukanlah jarak titik C ke bidang BDG}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah ilustrasi}\end{array}$.

Jika gambarnya dipartisi lagi di bagian segitiga GCG' maka akan tampak seperti ilustrasi berikut

$.\begin{array}{rl}\text{Jelas}& \text{bahwa}\\ & \{\begin{array}{ll}GC& =8cm(\text{dari soal})\\ BD& =AC=8\sqrt{2}cm(\text{diagonal sisi kubus})\\ C{G}^{\prime }& ={\displaystyle \frac{1}{2}AC=4\sqrt{2}cm}\end{array}\\ \text{maka}& \text{dengan rumus Pythagoras dapat}\\ \text{panja}& \text{ng}G{G}^{\prime },\text{yaitu}:\\ \left(G{G}^{\prime }\right)& {}^2={\left(G^{\prime }C\right)}^2+C{G}^2\\ G{G}^{\prime }& =\sqrt{{\left(G^{\prime }C\right)}^2+C{G}^2}=\sqrt{{\left(4\sqrt{2}\right)}^2+8^2}\\ & =\sqrt{32+64}=\sqrt{96}=\sqrt{16.6}=4\sqrt{6}cm\\ \text{Perha}& \text{tikan}△GC{G}^{\prime }\\ & \begin{array}{rl}& \text{Dengan perbandingan luas}△GC{G}^{\prime }=△GC{G}^{\prime }\\ & {\displaystyle \frac{1}{2}\times C{C}^{\prime }\times G{G}^{\prime }={\displaystyle \frac{1}{2}\times C{G}^{\prime }\times CG}}\\ & {\displaystyle \frac{1}{2}\times C{C}^{\prime }\times \left(4\sqrt{6}\right)={\displaystyle \frac{1}{2}\times \left(4\sqrt{2}\right)\times 8}}\\ & C{C}^{\prime }={\displaystyle \frac{{\displaystyle \frac{1}{2}\times \left(4\sqrt{2}\right)\times 8}}{{\displaystyle \frac{1}{2}\times \left(4\sqrt{6}\right)}}}\\ & ={\displaystyle \frac{8}{\sqrt{3}}}\\ & ={\displaystyle \frac{8}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}}\\ & ={\displaystyle \frac{8}{3}\sqrt{3}cm}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Diketahui rusuk kubus ABCD.EFGH adalah 6 cm}\\ & \text{Tentukanlah jarak titik E ke bidang BDG}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah ilustrasi berikut ini}\end{array}$.

$.\begin{array}{rl}\text{Jelas}& \text{bahwa}\\ & \{\begin{array}{ll}AB& =BC=CG=Q{Q}^{\prime }=6cm(\text{dari soal})\\ AC& =EG=6\sqrt{2}cm(\text{diagonal sisi kubus})\\ EQ& =QG={\displaystyle \frac{1}{2}(\mathbf{\text{sisi}})\sqrt{6}=3\sqrt{6}cm}\end{array}\\ \text{Perh}& \text{atikan}△EQG\\ & \begin{array}{rl}& \text{Dengan perbandingan luas}△EQG=△EQG\\ & {\displaystyle \frac{1}{2}\times QG\times E{E}^{\prime }={\displaystyle \frac{1}{2}\times Q{Q}^{\prime }\times EG}}\\ & {\displaystyle \frac{1}{2}\times \left(3\sqrt{6}\right)\times E{E}^{\prime }={\displaystyle \frac{1}{2}\times 6\times 6\sqrt{2}}}\\ & E{E}^{\prime }={\displaystyle \frac{{\displaystyle \frac{1}{2}\times \left(6\sqrt{2}\right)\times 6}}{{\displaystyle \frac{1}{2}\times \left(3\sqrt{6}\right)}}}\\ & =4\sqrt{3}cm\end{array}\end{array}$

Lanjutan Materi Geometri Ruang (Dimensi Tiga)

 $\text{B. 2. Kedudukan Titik terhadap Garis}$.

$\begin{array}{|cll|}\hline \text{No}& \text{Kedudukan}& \text{Keterangan}\\ 1.& \text{pada garis}& \text{titik berimpit pada garis}\\ 2.& \text{di luar garis}& \text{titik berada di luar garis}\\ \hline\end{array}$.

Pada contoh kubus ABCD.EFGH di atas adalah :

• terletak pada garis : Titik A terletak pada tiga garis yaitu ruas garis AB, AD, dan AE
• terletak di luar garis : Titik A di luar rus garis BC, CD, BF, CG, DH, EF, ED, FG, dan GH.

Secara definisi jarak antara suatu titik dengan garis adalah panjang ruas garis yang ditarik dari titik tersebut ke garis tersebut secara tegak lurus.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 4.& \text{Diketahui kubus ABCD.EFGH dengan rusuk}.\\ & \text{10 cm. Tentukanlah jarak titik F ke garis AC}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah gambar kubus berikut}\end{array}$.

$.\text{perhatikan pula gambar kedua berikut}$

$.\begin{array}{rl}\text{Tampak}& \text{bahwa jakak titik F ke garis AC}\\ \text{adalah s}& \text{ama dengan jarak titik F ke P}\\ \text{dengan}& \text{panjang rusuk 10}cm,\text{yaitu}:\\ \text{Alternat}& \text{if 1}\\ \text{Dengan}& \text{memandang segitiga BFP kita}\\ \text{gunakan}& \text{rumus Pythagoras, yaitu}:\\ P{F}^2& =P{B}^2+B{F}^2\\ PF& =\sqrt{P{B}^2+B{F}^2}\\ & =\sqrt{{\left(5\sqrt{2}\right)}^2+{10}^2}\\ & \text{ingat}\text{PB setengah diagonal sisi}\\ & =\sqrt{50+100}=\sqrt{150}=\sqrt{25.6}\\ & =5\sqrt{6}cm\\ \text{Alternat}& \text{if 2}\\ \text{Gunaka}& \text{n rumus luas segitiga, yaitu}:\\ \text{Luas}△& \text{PBF}=\text{Luas}△\text{PBF, atau}\\ \left[PBF\right]& =\left[PBF\right]\\ & \text{karena}△\text{PBF segitiga sama sisi}\\ & \text{maka AC=CF=FA=}10\sqrt{2}\text{dan ketiga}\\ & \text{sudutnya masing-masing}{60}^{\circ }\\ {\displaystyle \frac{1}{2}\times AC}& \times FP={\displaystyle \frac{1}{2}\times AF\times FC\times \sin \mathrm{\angle }AFC}\\ {\displaystyle \frac{1}{2}\times AC}& \times FP={\displaystyle \frac{1}{2}\times AC\times AC\times \sin \mathrm{\angle }AFC}\\ FP& =AC\times \sin \mathrm{\angle }AFC=10\sqrt{2}\times \sin {60}^{\circ }\\ & =10\sqrt{2}\times \left({\displaystyle \frac{1}{2}\sqrt{3}}\right)=5\sqrt{6}cm\\ \text{Alternat}& \text{if 3}\\ \text{Perhatik}& \text{an gambar berikut}\end{array}$.

$.\begin{array}{rl}\text{Tampak}& \text{bahwa jakak titik F ke garis AC}\\ \text{seperti}& \text{jarak titik B ke Q}={\displaystyle \frac{1}{2}a\sqrt{6}}\\ \text{maka}\text{P}& \text{F}=BQ={\displaystyle \frac{1}{2}a\sqrt{6}}\\ PF& ={\displaystyle \frac{1}{2}(\text{sisi})\sqrt{6}}\\ & ={\displaystyle \frac{1}{2}(10)\sqrt{6}}\\ & =5\sqrt{6}cm\end{array}$.

$\begin{array}{l}\\ 5.& \text{Diketahui limas beraturan T.ABCD. Panjang}\\ & \text{rusuk alasnya 8 cm dan panjang rusuk tegaknya}\\ & \text{12 cm. Tentukanlah jarak B ke TD}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah gambar limas beraturan T.ABCD}\\ & \text{berikut}\end{array}$.

$.\text{perhatikan pula gambar kedua berikut}$

$.\begin{array}{rl}\text{Gunaka}& \text{n rumus luas segitiga, yaitu}:\\ \text{Luas}△& \text{TBD}=\text{Luas}△\text{TBD atau}\\ \left[TBD\right]& =\left[TBD\right]\\ & \text{karena}△\text{TBD segitiga sama kaki}\\ & \text{maka TB=TD=}12cm\text{dan BD}\\ & \text{adalah diagonal sisi alas}=8\sqrt{2}cm\\ & \text{dengan menghitung tinggi}△TBD\\ & \text{dengan alas TD dengan tinggi ditarik dari}\\ & \text{B ke arak rusuk TD, maka akan ketemu}\\ & \text{jarak titik B ke garis TD}.\\ & \text{Berikut perhitungannya}\\ {\displaystyle \frac{1}{2}\times TD}& \times Tinggi={\displaystyle \frac{1}{2}\times TP\times DB}\\ {\displaystyle \frac{1}{2}\times TD}& \times Tinggi={\displaystyle \frac{1}{2}\times \left(\sqrt{T{B}^2-P{B}^2}\right)\times DB}\\ {\displaystyle \frac{1}{2}\times 12}& \times Tinggi={\displaystyle \frac{1}{2}\times \left(\sqrt{{12}^2-{\left(4\sqrt{2}\right)}^2}\right)\times 8\sqrt{2}}\\ & \text{ingat bahwa}PB={\displaystyle \frac{1}{2}BD}\\ Tinggi& ={\displaystyle \frac{{\displaystyle \frac{1}{2}\times \left(\sqrt{{12}^2-{\left(4\sqrt{2}\right)}^2}\right)\times 8\sqrt{2}}}{6}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}\times \sqrt{144-32}\times 8\sqrt{2}}}{6}}\\ & ={\displaystyle \frac{\sqrt{112}\times 4\sqrt{2}}{6}}\\ & ={\displaystyle \frac{\sqrt{16\times 7}\times 4\sqrt{2}}{6}}\\ & ={\displaystyle \frac{4\sqrt{7}\times 2\sqrt{2}}{3}}\\ & ={\displaystyle \frac{8}{3}\sqrt{7.2}}\\ & ={\displaystyle \frac{8}{3}\sqrt{14}cm}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Diketahui limas beraturan T.ABCD. Panjang}\\ & \text{rusuk alasnya}5\sqrt{2}cm\text{dan panjang rusuk}\\ & \text{tegaknya 13 cm. Tentukanlah jarak A ke TC}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah gambar limas beraturan T.ABCD}\\ & \text{berikut}\end{array}$.

$.\begin{array}{rl}\text{Gunaka}& \text{n rumus luas segitiga, yaitu}:\\ \left[TAC\right]& =\left[TAC\right]\\ \text{dengan}& \text{proses pengerjaaan sama semisal}\\ \text{no.5 di}& \text{ atas, maka}\\ & \begin{array}{rl}{\displaystyle \frac{1}{2}\times TC}& \times Tinggi={\displaystyle \frac{1}{2}\times T{T}^{\prime }\times AC}\\ {\displaystyle \frac{1}{2}\times TC}& \times Tinggi={\displaystyle \frac{1}{2}\times \left(\sqrt{T{A}^2-A{T}^{\prime 2}}\right)\times AC}\\ {\displaystyle \frac{1}{2}\times 13}& \times Tinggi={\displaystyle \frac{1}{2}\times \left(\sqrt{{13}^2-{\left({\displaystyle \frac{10}{2}}\right)}^2}\right)\times 10}\\ 13& \times Tinggi=12\times 10\\ & Tinggi={\displaystyle \frac{120}{13}}\end{array}\end{array}$