Tugas 2 Kalkulus: Trigonometri (Evaluasi Sudut, Identitas, dan Penyelesaian Segitiga)

1. Hitung Sin, Cos, Tan di bawah ini

a. \(\sin 120^\circ\) \[ \sin 120^\circ = \sin (180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

b. \(\cos 150^\circ\) \[ \cos 150^\circ = \cos (180^\circ - 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \]

c. \(\tan 210^\circ\) \[ \tan 210^\circ = \tan (180^\circ + 30^\circ) = \tan 30^\circ = \frac{\sqrt{3}}{3} \]

d. \(\cos 225^\circ\) \[ \cos 225^\circ = \cos (180^\circ + 45^\circ) = -\cos 45^\circ = -\frac{\sqrt{2}}{2} \]

e. \(\sin 315^\circ\) \[ \sin 315^\circ = \sin (360^\circ - 45^\circ) = -\sin 45^\circ = -\frac{\sqrt{2}}{2} \]

f. \(\tan 300^\circ\) \[ \tan 300^\circ = \tan (360^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3} \]

2. Hitunglah

a. \(\cos (210^\circ - 150^\circ) = ?\) \[ \cos (210^\circ - 150^\circ) = \cos 60^\circ = \frac{1}{2} \]

b. \(\sin (300^\circ - 210^\circ) = ?\) \[ \sin (300^\circ - 210^\circ) = \sin 90^\circ = 1 \] Atau menggunakan rumus: \[ \sin(300^\circ - 210^\circ) = \sin 300^\circ \cos 210^\circ - \cos 300^\circ \sin 210^\circ \] \[ = \left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right) \] \[ = \frac{3}{4} + \frac{1}{4} = 1 \]

c. \(\tan (60^\circ + 120^\circ) = ?\) \[ \tan (60^\circ + 120^\circ) = \tan 180^\circ = 0 \]

d. \(\sin 15^\circ\) \[ \sin 15^\circ = \sin (45^\circ - 30^\circ) \] \[ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \] \[ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) \] \[ = \frac{\sqrt{6} - \sqrt{2}}{4} \]

e. \(\cos 105^\circ\) \[ \cos 105^\circ = \cos (60^\circ + 45^\circ) \] \[ = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ \] \[ = \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) \] \[ = \frac{\sqrt{2} - \sqrt{6}}{4} \]

f. \(\tan 75^\circ\) \[ \tan 75^\circ = \tan (45^\circ + 30^\circ) \] \[ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \] \[ = \frac{1 + \frac{\sqrt{3}}{3}}{1 - (1)\left(\frac{\sqrt{3}}{3}\right)} = \frac{\frac{3+\sqrt{3}}{3}}{\frac{3-\sqrt{3}}{3}} \] \[ = \frac{3+\sqrt{3}}{3-\sqrt{3}} \cdot \frac{3+\sqrt{3}}{3+\sqrt{3}} \] \[ = \frac{9 + 6\sqrt{3} + 3}{9 - 3} \] \[ = \frac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3} \]

3. Hitung

a. \(\cos^2 60^\circ - \sin^2 60^\circ\) \[ \cos^2 60^\circ - \sin^2 60^\circ = \cos (2 \cdot 60^\circ) = \cos 120^\circ = -\frac{1}{2} \]

b. \(\tan 120^\circ\) \[ \tan 120^\circ = \tan (180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3} \]

c. \(2 \cos 120^\circ \cos 30^\circ\) \[ 2 \cos 120^\circ \cos 30^\circ = 2 \left(-\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{3}}{2} \]

d. \(2 \sin 75^\circ \sin 15^\circ\) \[ 2 \sin 75^\circ \sin 15^\circ = \cos(75^\circ - 15^\circ) - \cos(75^\circ + 15^\circ) \] \[ = \cos 60^\circ - \cos 90^\circ \] \[ = \frac{1}{2} - 0 = \frac{1}{2} \]

e. \(\cos 120^\circ - \cos 30^\circ\) \[ \cos 120^\circ - \cos 30^\circ = -\frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{-1 - \sqrt{3}}{2} \]

4. Soal Segitiga

a. Bila \(a=30\), \(b=50\), \(c=60\). Hitung Luas \(\Delta\). \[ s = \frac{a+b+c}{2} = \frac{30+50+60}{2} = \frac{140}{2} = 70 \] \[ \text{Luas} = \sqrt{s(s-a)(s-b)(s-c)} \] \[ = \sqrt{70(70-30)(70-50)(70-60)} \] \[ = \sqrt{70 \cdot 40 \cdot 20 \cdot 10} = \sqrt{560.000} \] \[ = \sqrt{56 \cdot 10.000} = 100\sqrt{56} = 200\sqrt{14} \]

b. Bila \(\alpha=45^\circ\), \(a=40\), \(\beta=60^\circ\). Hitung: \(\gamma, b, c\). \[ \gamma = 180^\circ - (\alpha + \beta) = 180^\circ - (45^\circ + 60^\circ) = 180^\circ - 105^\circ = 75^\circ \] \[ \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} \implies \frac{40}{\sin 45^\circ} = \frac{b}{\sin 60^\circ} \] \[ b = \frac{40 \cdot \sin 60^\circ}{\sin 45^\circ} = \frac{40 \cdot \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = 40 \cdot \frac{\sqrt{3}}{\sqrt{2}} = 20\sqrt{6} \] \[ \frac{a}{\sin \alpha} = \frac{c}{\sin \gamma} \implies \frac{40}{\sin 45^\circ} = \frac{c}{\sin 75^\circ} \] \[ c = \frac{40 \cdot \sin 75^\circ}{\sin 45^\circ} = \frac{40 \cdot \left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)}{\frac{\sqrt{2}}{2}} \] \[ = 10(\sqrt{6}+\sqrt{2}) \cdot \frac{2}{\sqrt{2}} = 20 \cdot \frac{\sqrt{6}+\sqrt{2}}{\sqrt{2}} = 20(\sqrt{3} + 1) \]

c. Bila \(\alpha=45^\circ\), \(b=70\), \(c=80\). Hitung: \(a, \beta, \gamma\). \[ a^2 = b^2 + c^2 - 2bc \cos \alpha \] \[ a^2 = 70^2 + 80^2 - 2(70)(80) \cos 45^\circ \] \[ a^2 = 4900 + 6400 - 11200\left(\frac{\sqrt{2}}{2}\right) \] \[ a^2 = 11300 - 5600\sqrt{2} \] \[ a = \sqrt{11300 - 5600\sqrt{2}} \approx 58.14 \] Untuk mencari \(\beta\) dan \(\gamma\) menggunakan Aturan Sinus: \[ \frac{\sin \beta}{b} = \frac{\sin \alpha}{a} \] \[ \sin \beta = \frac{70 \cdot \sin 45^\circ}{\sqrt{11300 - 5600\sqrt{2}}} \approx 0.8513 \] \[ \beta \approx \arcsin(0.8513) \approx 58.36^\circ \] \[ \gamma = 180^\circ - (45^\circ + 58.36^\circ) = 180^\circ - 103.36^\circ = 76.64^\circ \]

d. Bila \(b=55\), \(c=70\), \(\alpha=30^\circ\). Hitung Luas \(\Delta\). \[ \text{Luas} = \frac{1}{2} \cdot b \cdot c \cdot \sin \alpha \] \[ = \frac{1}{2} \cdot 55 \cdot 70 \cdot \sin 30^\circ \] \[ = \frac{1}{2} \cdot 3850 \cdot \frac{1}{2} \] \[ = \frac{3850}{4} = 962.5 \]