Koordinat Kartesius dan Koordinat Kutub serta Kuadran

C. Koordinat Kartesius dan Koordinat Kutub/Polar

Perhatikan ilustrasi berikut

$\begin{array}{|cc|}\hline \text{Kartesius}\to \text{Kutub}& \text{Kutub}\to \text{Kartesius}\\ P(x,y)\to P(r,{\alpha }^0)& P(r,{\alpha }^0)\to P(x,y)\\ \begin{array}{rl}& r=\sqrt{x^2+y^2},\\ & {\displaystyle \tan {\alpha }^0=\frac{y}{x},{\displaystyle {\alpha }^0=\arctan \frac{y}{x}}}\end{array}& \{\begin{array}{c}x=r.\cos {\alpha }^0\\ \\ y=r.\sin {\alpha }^0\end{array}\\ \hline\end{array}$.

D. Perbandingan Trigonometri di Berbagai Kuadran

D. 1 Untuk Sudut 0 sampai dengan 360 derajat.

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

$\begin{array}{ll}(1).& \text{Perbandingan Trigonometri untuk Sudut}({90}^0-\alpha )\\ & \text{a}.\sin \alpha \to \sin ({90}^0-\theta )=\cos \theta \\ & \text{b}.\cos \alpha \to \cos ({90}^0-\theta )=\sin \theta \\ & \text{c}.\tan \alpha \to \tan ({90}^0-\theta )=\cot \theta \\ & \text{d}.\cot \alpha \to \cot ({90}^0-\theta )=\tan \theta \end{array}$.

$\begin{array}{ll}(2).& \text{Perbandingan Trigonometri untuk Sudut}({180}^0-\alpha )\\ & \text{a}.\sin \alpha \to \sin ({180}^0-\theta )=\sin \theta \\ & \text{b}.\cos \alpha \to \cos ({180}^0-\theta )=-\cos \theta \\ & \text{c}.\tan \alpha \to \tan ({180}^0-\theta )=-\tan \theta \\ & \text{d}.\cot \alpha \to \cot ({180}^0-\theta )=-\cot \theta \end{array}$.

$\begin{array}{ll}(3).& \text{Perbandingan Trigonometri untuk Sudut}({180}^0+\alpha )\\ & \text{a}.\sin \alpha \to \sin ({180}^0+\theta )=-\sin \theta \\ & \text{b}.\cos \alpha \to \cos ({180}^0+\theta )=-\cos \theta \\ & \text{c}.\tan \alpha \to \tan ({180}^0+\theta )=\tan \theta \\ & \text{d}.\cot \alpha \to \cot ({180}^0+\theta )=\cot \theta \end{array}$.

$\begin{array}{ll}(4).& \text{Perbandingan Trigonometri untuk Sudut}({360}^0-\alpha )\\ & \text{a}.\sin \alpha \to \sin ({360}^0-\theta )=-\sin \theta \\ & \text{b}.\cos \alpha \to \cos ({360}^0-\theta )=\cos \theta \\ & \text{c}.\tan \alpha \to \tan ({360}^0-\theta )=-\tan \theta \\ & \text{d}.\cot \alpha \to \cot ({360}^0-\theta )=-\cot \theta \end{array}$.

D. 2 Untuk Sudut yang Lain

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

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

$\text{CONTOH SOAL}$.

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

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

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