Dengan Segitiga Sama Sisi
Menggunakan keistimewaan segitiga sama sisi yang semua sudutnya 60 ° 60\degree 60° .
Misalkan kita mempunyai segitiga sama sisi dengan panjang sisi 1. Kemudian kita buat garis tinggi
dari c c c sehingga segitiga terbelah menjadi 2 segitiga siku-siku. Karena sudut
∠ C A D = 60 ° \angle CAD=60\degree ∠ C A D = 60° maka sudut ∠ A C D = 90 ° − 60 ° = 30 ° \angle ACD=90\degree - 60\degree = 30 \degree ∠ A C D = 90° − 60° = 30° . Karena sudut
∠ C B D = 60 ° \angle CBD=60\degree ∠ CB D = 60° maka sudut ∠ B C D = 90 ° − 60 ° = 30 ° \angle BCD=90\degree - 60\degree = 30 \degree ∠ BC D = 90° − 60° = 30° .
Karena kedua segitiga siku-siku mempunyai sudut internal sama, berbagi satu garis sebagai sisi dan
salah satu sisi mempunyai panjang sama maka kedua segitiga adalah kongruen. Jadi sisi didepan sudut
30 ° 30\degree 30° pada segitiga ACD (yaitu AD) dan BCD (yaitu BD) sama panjang (A D = B D AD = BD A D = B D ). Maka berlaku:
A B = A D + B D = 2 A D A D = 1 2 A B = 1 2 × 1 = 1 2 \begin{aligned}
AB &= AD + BD \\
&= 2AD \\
AD &= \frac{1}{2}AB \\
&= \frac{1}{2} \times 1 \\
&= \frac{1}{2}
\end{aligned}
A B A D = A D + B D = 2 A D = 2 1 A B = 2 1 × 1 = 2 1
Kita sekarang bisa mencari panjang C D CD C D
A D 2 + C D 2 = A C 2 C D 2 = A C 2 − A D 2 = ( A C − A D ) ( A C + A D ) = ( 1 − 1 2 ) ( 1 + 1 2 ) = 1 2 × 3 4 = 3 4 C D = 3 4 = 3 2 \begin{aligned}
AD^2 + CD^2 &= AC^2 \\
CD^2 &= AC^2 - AD^2 \\
&= (AC - AD)(AC + AD) \\
&= (1 - \frac{1}{2})(1 + \frac{1}{2}) \\
&= \frac{1}{2}\times \frac{3}{4} \\
&= \frac{3}{4} \\
CD &= \sqrt{\frac{3}{4}} \\
&= \frac{\sqrt{3}}{2}
\end{aligned}
A D 2 + C D 2 C D 2 C D = A C 2 = A C 2 − A D 2 = ( A C − A D ) ( A C + A D ) = ( 1 − 2 1 ) ( 1 + 2 1 ) = 2 1 × 4 3 = 4 3 = 4 3 = 2 3
Jadi kita bisa mengetahui nilai trigonometri:
sin ( ∠ A C D ) = sin ( 30 ° ) = A D A C = ( 1 2 ) 1 = 1 2 cos ( ∠ A C D ) = cos ( 30 ° ) = C D A C = ( 3 2 ) 1 = 3 2 tan ( ∠ A C D ) = tan ( 30 ° ) = A D C D = ( 1 2 ) ( 3 2 ) = 1 2 × 2 3 = 1 3 × 3 3 = 3 3 sin ( ∠ C A D ) = sin ( 60 ° ) = C D A C = 3 2 cos ( ∠ C A D ) = cos ( 60 ° ) = A D A C = 1 2 tan ( ∠ C A D ) = tan ( 60 ° ) = C D A D = ( 3 2 ) ( 1 2 ) = 3 2 × 2 1 = 3 \begin{aligned}
\sin(\angle ACD) &= \sin(30 \degree) \\
&= \frac{AD}{AC} \\
&= \frac{(\frac{1}{2})}{1} \\
&= \frac{1}{2} \\
\cos(\angle ACD) &= \cos(30 \degree) \\
&= \frac{CD}{AC} \\
&= \frac{(\frac{\sqrt{3}}{2})}{1} \\
&= \frac{\sqrt{3}}{2} \\
\tan(\angle ACD) &= \tan(30 \degree) \\
&= \frac{AD}{CD} \\
&= \frac{(\frac{1}{2})}{\frac{(\sqrt{3}}{2})} \\
&= \frac{1}{2}\times \frac{2}{\sqrt{3}} \\
&= \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\
&= \frac{\sqrt{3}}{3} \\
\sin(\angle CAD) &= \sin(60 \degree) \\
&= \frac{CD}{AC} \\
&= \frac{\sqrt{3}}{2} \\
\cos(\angle CAD) &= \cos(60 \degree) \\
&= \frac{AD}{AC} \\
&= \frac{1}{2} \\
\tan(\angle CAD) &= \tan(60 \degree) \\
&= \frac{CD}{AD} \\
&= \frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})} \\
&= \frac{\sqrt{3}}{2}\times \frac{2}{1} \\
&= \sqrt{3} \\
\end{aligned}
sin ( ∠ A C D ) cos ( ∠ A C D ) tan ( ∠ A C D ) sin ( ∠ C A D ) cos ( ∠ C A D ) tan ( ∠ C A D ) = sin ( 30° ) = A C A D = 1 ( 2 1 ) = 2 1 = cos ( 30° ) = A C C D = 1 ( 2 3 ) = 2 3 = tan ( 30° ) = C D A D = 2 ( 3 ) ( 2 1 ) = 2 1 × 3 2 = 3 1 × 3 3 = 3 3 = sin ( 60° ) = A C C D = 2 3 = cos ( 60° ) = A C A D = 2 1 = tan ( 60° ) = A D C D = ( 2 1 ) ( 2 3 ) = 2 3 × 1 2 = 3
Kesimpulan
Nilai fungsi trigonometri sudut 30 ° 30\degree 30° dan 60 ° 60\degree 60°
sin ( 30 ° ) = 1 2 cos ( 30 ° ) = 3 2 tan ( 30 ° ) = 3 3 sin ( 60 ° ) = 3 2 cos ( 60 ° ) = 1 2 tan ( 60 ° ) = 3 \begin{align*}
\sin(30 \degree) &= \dfrac{1}{2} \\
\cos(30 \degree) &= \dfrac{\sqrt{3}}{2} \\
\tan(30 \degree) &= \dfrac{\sqrt{3}}{3} \\
\sin(60 \degree) &= \dfrac{\sqrt{3}}{2} \\
\cos(60 \degree) &= \dfrac{1}{2} \\
\tan(60 \degree) &= \sqrt{3}
\end{align*}
sin ( 30° ) cos ( 30° ) tan ( 30° ) sin ( 60° ) cos ( 60° ) tan ( 60° ) = 2 1 = 2 3 = 3 3 = 2 3 = 2 1 = 3