*# SINUS SUDUT GANDA #*
\(\sin 2\alpha^{\circ} = 2 \cdot \sin \alpha^{\circ} \cdot \cos \alpha^{\circ}\)
Pembuktian:
\(\sin \left ( \alpha ^{\circ} + \beta ^{\circ}\right ) = \sin \alpha ^{\circ} \cdot \cos \beta ^{\circ} + \cos \alpha ^{\circ} \cdot \sin \beta ^{\circ}\)
\(\sin 2\alpha ^{\circ} = \sin \left ( \alpha ^{\circ} + \alpha ^{\circ} \right )\)
\(\sin 2\alpha ^{\circ} = \sin \alpha ^{\circ} \cdot \cos \alpha ^{\circ} + \cos \alpha ^{\circ} \cdot \sin \alpha ^{\circ}\)
\(\sin 2\alpha ^{\circ} = 2 \cdot \sin \alpha ^{\circ} \cdot \cos \alpha ^{\circ}\)
Contoh Soal:
1. Diketahui \(\sin A = -\frac{5}{13}\) dimana \(A\) di kuadran III. Dengan menggunakan rumus sudut ganda, hitunglah \(\sin 2A\).
Penyelesaian:
\(r^{2} = x^{2}+y^{2}\rightarrow x^{2} = r^{2}-y^{2}\)
\(x^{2} = 13^{2}-\left ( -5 \right )^{2}\)
\(x^{2} = 168-25\)
\(x^{2} = 144\)
\(x = 12\), karena di kuadran III
\(\cos A = -\frac{x}{r}\)
\(\cos A = -\frac{12}{13}\)
\(\sin 2A = 2\cdot \sin A\cdot \cos A\)
\(\sin 2A = 2\left ( -\frac{5}{13} \right )\left ( -\frac{12}{13} \right )\)
\(\sin 2A = \frac{120}{169}\)
2. Diketahui \(\cos A = -\frac{4}{5}\) dan \(\sin B = \frac{5}{13}\). Sudut \(A\) dan \(B\) tumpul. Hitunglah \(\sin \left ( A + B \right )\).
Penyelesaian:
\(\cos A = -\frac{4}{5}\), maka \(\sin A = \frac{3}{5}\), kuadran II.
\(\sin B = \frac{5}{13}\), maka \(\cos B = -\frac{12}{13}\), kuadran II.
\(\sin \left ( A+B \right ) = \sin A\cdot \cos B + \cos A\cdot \sin B\)
\(= \frac{3}{5}\cdot \left ( -\frac{12}{13} \right ) + \left ( -\frac{4}{5} \right )\cdot \frac{5}{13}\)
\(= -\frac{36}{65} - \frac{20}{65} = \frac{56}{65}\)
*# COSINUS SUDUT GANDA #*
\( \cos 2A = \cos^{2} A - \sin^{2} A\)
\(= 1 - \sin^{2} A\)
\(= 2 \cos^{2} A\)
Pembuktian rumus \( \cos (\alpha + \beta )\)
\( \cos (\alpha + \beta ) = \cos \alpha . \cos \beta - \sin \alpha . \sin \beta \)
\( \cos (\alpha + \alpha) = \cos \alpha . \cos \alpha - \sin \alpha . \sin \alpha \)
\( \cos 2 \alpha = \cos^{2} \alpha - \sin^{2} \alpha \) \( \mathbf{(Terbukti)}\)
Pembuktian \( \cos 2 \alpha = \cos^{2} \alpha - \sin^{2} \alpha \)
\( \cos 2 \alpha = \cos^{2} \alpha - \sin^{2} \alpha \)
\(= \cos^{2} \alpha - (1 - \cos^{2} \alpha ) \)
\(= \cos^{2} \alpha - 1 + \cos^{2} \alpha \)
\(= 2 \cos^{2} - 1 \) \( \mathbf{(Terbukti)}\)
\( \mathbf{atau} \)
\( \cos 2 \alpha = \cos^{2} \alpha - \sin^{2} \alpha \)
\(= (1 - \sin^{2} \alpha ) - \sin^{2} \alpha \)
\(= 1 - \sin^{2} \alpha - \sin^{2} \alpha \)
\(= 1 - 2 \sin^{2} \alpha \) \( \mathbf{(Terbukti)}\)
Pembuktian \( \sin^{2} + \cos^{2} = 1 \) :
Dengan menggunakan Phytagoras
\( \sin^{2} + \cos^{2} = 1 \)
\( b^{2} = a^{2} + c^{2} \)
\( \left ( \frac{a}{\sin \alpha } \right )^{2} = (b . \sin\alpha )^{2} + (b . \cos \alpha )^{2}\)
\(\frac{a^{2}}{\sin ^{2} \alpha } = b^{2} . \sin^{2}\alpha + b^{2} . \cos^{2}\alpha \)
\( \frac{a^{2}}{b^{2} . \sin \alpha } = \sin^{2} \alpha + cos^{2} \alpha \)
\( \frac{a^{2}}{a^{2}} = \sin^{2} + \cos^{2} \)
\( 1 = \sin^{2} + \cos^{2} \) \( \mathbf{(Terbukti)}\)
CONTOH SOAL
1. Diketahui \( \cos A = -\frac{24}{25}\), dimana A di kuadran III. Dengan menggunakan rumus sudut ganda,
Hitunglah nilai \( \cos 2A = \).....
\(\mathbf{Jawaban :}\)</span>
\( \cos 2A = 2 \cos^{2} \alpha - 1\)
\(= 2 \left ( -\frac{24}{25} \right )^{2} - 1\)
\(= 2 \left ( -\frac{276}{625} \right ) - 1\)
\(= \frac{1.152}{625} - 1\)
\(= \frac{527}{625}\)
Jadi nilai \( \cos 2A = \frac{527}{625}\)
2. Carilah nilai dari \(6-12 \sin^{2}\frac{\pi}{12}=...\)
\(\mathbf{Jawaban :}\)
\(\pi = 180^{\circ}\)
\(6 - 12 \sin^{2} \frac{\pi }{12} = 6(1-2 \sin^{2} \frac{\pi}{12})\)
\(= 6 \cos \left ( 2 . \frac{\pi}{12} \right )\)
\(= 6 \cos \frac{\pi}{6}\)
\(= 6 \cos \frac{180^{\circ}}{6}\)
\(= 6 \cos 30^{\circ}\)
\(= 6 . \frac{1}{2}\sqrt{3}\)
\(= 3\sqrt{3}\)
*# TANGEN SUDUT GANDA #*
\(\tan 2 \alpha^{\circ} = \frac{2 tan \alpha ^{\circ}}{1- \tan ^{2} \alpha ^{\circ}}\)
Pembuktian :
\(\tan \left ( \alpha ^{\circ} + \beta^{\circ} \right ) = \frac{\tan \alpha ^{\circ}}{\tan \beta ^{\circ}}\)
Misal :
\(\tan 2\alpha ^{\circ}\)= \(\tan \left ( \alpha ^{\circ} + \alpha ^{\circ} \right )\)
\(\tan \left ( \alpha ^{\circ} + \alpha ^{\circ} \right )\) = \(\frac{\tan \alpha ^{\circ} + \tan \alpha ^{\circ}}{1 - \tan ^{\circ}\tan \alpha ^{\circ}}\)
\(\tan 2 \alpha^{\circ}\)= \(\frac{ 2 tan \alpha^{\circ}}{1 - \tan ^{2}\alpha ^{\circ}}\)
Contoh Soal!
1. Sederhanakan bentuk berikut.
\(\frac{2 \tan 3 \alpha }{1 - tan^{2} 3\alpha }\)
Jawab :
Misalkan \(3\alpha = \beta\) , Sehingga :
\(=\frac{2 \tan \beta }{1 - \tan ^{2}\beta }\)
\(=\tan 2 \beta\)
\(=\tan 2 \left ( 3\alpha \right )\)
\(=\tan 6\alpha\)
2. Jika \(\tan x = \frac{1}{2}\) dan \(\tan y = \frac{1}{3}\)
Hitunglah :
\(\tan 2x\)
\(\tan 2x = \frac{2 \tan x}{1 - \tan ^{2} x }\)
\(\tan 2x = \frac{2.\frac{1}{2}}{1-\left ( \frac{1}{2} \right )^{2}}\)
\(\tan 2x = \frac{1}{1-\frac{1}{4}}\)
\(\tan 2x = \frac{1}{\frac{3}{4}}\)
\(\tan 2x = \frac{4}{3}\)
