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