Jika besar sudut ß adalah 18⁰, maka nilai dari: sin

Berikut ini adalah pertanyaan dari Tannng pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Jika besar sudut ß adalah 18⁰, maka nilai dari: sin ß = ..... cos ß = .... tan ß = ...

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jika besar sudut β adalah 18⁰, maka nilai dari:
$\begin{aligned}&\bullet&\sin\beta&=\boxed{\,\bf\frac{\sqrt{5}-1}{4}\,}\\&\bullet&\cos\beta&=\boxed{\bf\,\frac{\sqrt{10+2\sqrt{5}}}{4}\,}\\&\bullet&\tan\beta&=\boxed{\bf\,\sqrt{\frac{5-2\sqrt{5}}{5}}\,}\end{aligned}$
 

Pembahasan

Menentukan nilai sin β

β = 18°. Maka 5β = 90°, atau 3β = 90° – 2β.

$\begin{aligned}&\sin3\beta=\sin(90^{\circ}-2\beta)\\&{\Rightarrow\ }\sin(2\beta+\beta)=\cos2\beta\\&{\Rightarrow\ }\sin2\beta\cos\beta+\cos2\beta\sin\beta=\cos2\beta\\&{\Rightarrow\ }\sin2\beta\cos\beta=\cos2\beta(1-\sin\beta)\\&{\Rightarrow\ }2\sin\beta\cos^2\beta=(1-2\sin^2\beta)(1-\sin\beta)\\&{\Rightarrow\ }2\sin\beta(1-\sin^2\beta)=(1-2\sin^2\beta)(1-\sin\beta)\\&{\Rightarrow\ }2\sin\beta\cancel{(1-\sin\beta)}(1+\sin\beta)=(1-2\sin^2\beta)\cancel{(1-\sin\beta)}\end{aligned}$
$\begin{aligned}&{\Rightarrow\ }2\sin\beta(1+\sin\beta)=1-2\sin^2\beta\\&{\Rightarrow\ }2\sin\beta+2\sin^2\beta+2\sin^2\beta=1\\&{\Rightarrow\ }4\sin^2\beta+2\sin\beta-1=0\\&\quad\textsf{Gunakan Rumus ABC:}\\&{\Rightarrow\ }\sin\beta=\frac{-2\pm\sqrt{2^2-4\cdot4\cdot(-1)}}{2\cdot4}\\&{\Rightarrow\ }\sin\beta=\frac{-2\pm\sqrt{4+16}}{8}=\frac{-2\pm\sqrt{20}}{8}\\&{\Rightarrow\ }\sin\beta=\frac{-2\pm2\sqrt{5}}{8}\\&{\Rightarrow\ }\sin\beta=\frac{-1\pm\sqrt{5}}{4}\\\end{aligned}$
$\begin{aligned}&\quad\textsf{$\beta\in$ Kuadran I}\implies\sin\beta \ge 0\\&{\Rightarrow\ }\sin\beta=\frac{-1+\sqrt{5}}{4}\\&\therefore\ \sin\beta=\boxed{\,\bf\frac{\sqrt{5}-1}{4}\,}\\\end{aligned}$
$\blacksquare$

Menentukan nilai cos β

$\begin{aligned}\cos\beta&=\sqrt{1-\sin^2\beta}\\&=\sqrt{1-\left(\frac{\sqrt{5}-1}{4}\right)^2}\\&=\sqrt{1-\frac{5+1-2\sqrt{5}}{16}}\\&=\sqrt{1-\frac{6-2\sqrt{5}}{16}}\\&=\sqrt{\frac{16-6+2\sqrt{5}}{16}}\\&=\sqrt{\frac{10+2\sqrt{5}}{16}}\\\therefore\ \cos\beta&=\boxed{\bf\,\frac{\sqrt{10+2\sqrt{5}}}{4}\,}\end{aligned}$
$\blacksquare$

Menentukan nilai tan β

$\begin{aligned}\tan\beta&=\frac{\sin\beta}{\cos\beta}=\sqrt{\frac{\sin^2\beta}{\cos^2\beta}}\\&=\sqrt{\frac{\left(\frac{\sqrt{5}-1}{4}\right)^2}{\left(\frac{10+2\sqrt{5}}{16}\right)}}\\&=\sqrt{\frac{\cancel{\frac{1}{16}}\left(5+1-2\sqrt{5}\right)}{\cancel{\frac{1}{16}}\left(10+2\sqrt{5}\right)}}\\&=\sqrt{\frac{6-2\sqrt{5}}{10+2\sqrt{5}}\times\frac{10-2\sqrt{5}}{10-2\sqrt{5}}}\\&=\sqrt{\frac{60-20\sqrt{5}-12\sqrt{5}+20}{100-20}}\end{aligned}$
$\begin{aligned}\tan\beta&=\sqrt{\frac{80-32\sqrt{5}}{80}}=\sqrt{\frac{\cancel{16}\left(5-2\sqrt{5}\right)}{\cancel{16}\cdot5}}\end{aligned}$
$\begin{aligned}\therefore\ \tan\beta&=\boxed{\bf\,\sqrt{\frac{5-2\sqrt{5}}{5}}\,}\end{aligned}$
$\blacksquare$