Hubungan Perbandingan Trigonometri dan Identitas Trigonometri
Materi lengkap disertai contoh soal dan latihan
1. Perbandingan Trigonometri Dasar
Definisi pada Segitiga Siku-Siku
Pada segitiga siku-siku dengan sudut \(\alpha\), sisi depan (d), sisi samping (s), dan sisi miring (m):
\(\sin \alpha = \dfrac{\text{depan}}{\text{miring}} = \dfrac{d}{m}\)
\(\csc \alpha = \dfrac{1}{\sin \alpha} = \dfrac{m}{d}\)
\(\cos \alpha = \dfrac{\text{samping}}{\text{miring}} = \dfrac{s}{m}\)
\(\sec \alpha = \dfrac{1}{\cos \alpha} = \dfrac{m}{s}\)
\(\tan \alpha = \dfrac{\text{depan}}{\text{samping}} = \dfrac{d}{s}\)
\(\cot \alpha = \dfrac{1}{\tan \alpha} = \dfrac{s}{d}\)
Tabel Nilai Trigonometri Sudut Istimewa
| Sudut | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | \(\frac{1}{2}\) | \(\frac{\sqrt{2}}{2}\) | \(\frac{\sqrt{3}}{2}\) | 1 |
| cos | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{\sqrt{2}}{2}\) | \(\frac{1}{2}\) | 0 |
| tan | 0 | \(\frac{\sqrt{3}}{3}\) | 1 | \(\sqrt{3}\) | ∞ |
Contoh Soal – Perbandingan Trigonometri Dasar
🟢 Mudah
1. Diketahui segitiga siku-siku dengan sisi depan = 3 dan sisi miring = 5. Tentukan sin α!
Pembahasan:
\(\sin \alpha = \dfrac{d}{m} = \dfrac{3}{5}\)
2. Jika cos α = 4/5, tentukan sec α!
Pembahasan:
\(\sec \alpha = \dfrac{1}{\cos \alpha} = \dfrac{1}{\frac{4}{5}} = \dfrac{5}{4}\)
3. Tentukan nilai tan 45°!
Pembahasan:
Dari tabel sudut istimewa: \(\tan 45° = 1\)
4. Diketahui sin α = 5/13. Tentukan sisi samping jika sisi miring = 13!
Pembahasan:
d = 5, m = 13. Dengan Pythagoras: \(s = \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = \sqrt{144} = 12\)
5. Tentukan nilai sin 30° + cos 60°!
Pembahasan:
\(\sin 30° + \cos 60° = \dfrac{1}{2} + \dfrac{1}{2} = 1\)
🟡 Sedang
1. Jika tan α = 3/4 dan α di kuadran I, tentukan sin α dan cos α!
Pembahasan:
tan α = d/s = 3/4, maka d = 3, s = 4.
\(m = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\)
\(\sin \alpha = \dfrac{3}{5}, \quad \cos \alpha = \dfrac{4}{5}\)
2. Tentukan nilai \(2\sin 60° \cdot \cos 30°\)!
Pembahasan:
\(2 \cdot \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{3}}{2} = 2 \cdot \dfrac{3}{4} = \dfrac{3}{2}\)
3. Jika sin α = 7/25, tentukan tan α! (α di kuadran I)
Pembahasan:
d = 7, m = 25. \(s = \sqrt{625 – 49} = \sqrt{576} = 24\)
\(\tan \alpha = \dfrac{7}{24}\)
4. Tentukan nilai \(\dfrac{\sin 45°}{\cos 45°} + \tan 45°\)!
Pembahasan:
\(\dfrac{\sin 45°}{\cos 45°} = \tan 45° = 1\)
Jadi: \(1 + 1 = 2\)
5. Jika cot α = 5/12, tentukan csc α!
Pembahasan:
cot α = s/d = 5/12, maka s = 5, d = 12.
\(m = \sqrt{25 + 144} = \sqrt{169} = 13\)
\(\csc \alpha = \dfrac{m}{d} = \dfrac{13}{12}\)
🔴 Sulit
1. Jika \(\sin \alpha = \dfrac{3}{5}\) dan α di kuadran II, tentukan nilai \(\cos \alpha + \tan \alpha\)!
Pembahasan:
Di kuadran II: sin positif, cos negatif, tan negatif.
\(\cos \alpha = -\dfrac{4}{5}\) (karena kuadran II)
\(\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}\)
\(\cos \alpha + \tan \alpha = -\dfrac{4}{5} + \left(-\dfrac{3}{4}\right) = -\dfrac{16}{20} – \dfrac{15}{20} = -\dfrac{31}{20}\)
2. Buktikan bahwa \(\dfrac{\sin 60° – \sin 30°}{\cos 30° – \cos 60°} = 1\)!
Pembahasan:
\(\dfrac{\frac{\sqrt{3}}{2} – \frac{1}{2}}{\frac{\sqrt{3}}{2} – \frac{1}{2}} = \dfrac{\frac{\sqrt{3}-1}{2}}{\frac{\sqrt{3}-1}{2}} = 1\) ✓
3. Jika \(\sec \alpha – \tan \alpha = \dfrac{1}{3}\), tentukan nilai \(\sec \alpha + \tan \alpha\)!
Pembahasan:
Gunakan identitas: \(\sec^2\alpha – \tan^2\alpha = 1\)
\((\sec\alpha – \tan\alpha)(\sec\alpha + \tan\alpha) = 1\)
\(\dfrac{1}{3} \cdot (\sec\alpha + \tan\alpha) = 1\)
\(\sec\alpha + \tan\alpha = 3\)
4. Tentukan semua perbandingan trigonometri dari α jika \(\tan \alpha = -\dfrac{5}{12}\) dan \(90° < \alpha < 180°\)!
Pembahasan:
Kuadran II: sin(+), cos(−), tan(−). d = 5, s = 12, \(m = 13\)
\(\sin\alpha = \dfrac{5}{13},\; \cos\alpha = -\dfrac{12}{13},\; \tan\alpha = -\dfrac{5}{12}\)
\(\csc\alpha = \dfrac{13}{5},\; \sec\alpha = -\dfrac{13}{12},\; \cot\alpha = -\dfrac{12}{5}\)
5. Sederhanakan: \(\dfrac{\sin^2 30° + \cos^2 60°}{\tan 45° \cdot \sin 90°}\)!
Pembahasan:
\(\dfrac{(\frac{1}{2})^2 + (\frac{1}{2})^2}{1 \cdot 1} = \dfrac{\frac{1}{4} + \frac{1}{4}}{1} = \dfrac{1}{2}\)
Latihan Soal – Perbandingan Trigonometri Dasar
🟢 Mudah
- Tentukan cos α jika sisi samping = 8 dan sisi miring = 17!
- Tentukan nilai sin 60°!
- Jika tan α = 1, tentukan sudut α (0° ≤ α ≤ 90°)!
- Tentukan cot 30°!
- Jika sin α = 12/13, tentukan cos α (kuadran I)!
🟡 Sedang
- Jika cos α = 5/13, tentukan semua perbandingan trigonometri di kuadran I!
- Hitung \(\sin^2 45° + \cos^2 45°\)!
- Jika sec α = 2, tentukan cos α dan tan α!
- Tentukan nilai \(\tan 60° – \tan 30°\)!
- Jika csc α = 5/3, tentukan sin α dan cos α!
🔴 Sulit
- Jika \(\sin\alpha = -\dfrac{4}{5}\) dan α di kuadran III, tentukan \(\cos\alpha + \cot\alpha\)!
- Sederhanakan \(\dfrac{\sin 30° \cdot \tan 60°}{\cos 30° \cdot \cot 60°}\)!
- Jika \(\cos\alpha = -\dfrac{7}{25}\) dan \(180° < \alpha < 270°\), tentukan semua perbandingan trigonometri!
- Buktikan \(\sin^2 60° – \sin^2 30° = \sin 90° \cdot \sin 30°\)!
- Jika \(\tan\alpha + \cot\alpha = \dfrac{25}{12}\), tentukan \(\sin\alpha \cdot \cos\alpha\)!
2. Hubungan Antar Perbandingan Trigonometri
Hubungan Kebalikan (Reciprocal)
\(\csc \alpha = \dfrac{1}{\sin \alpha}\)
\(\sec \alpha = \dfrac{1}{\cos \alpha}\)
\(\cot \alpha = \dfrac{1}{\tan \alpha}\)
Hubungan Perbandingan (Quotient)
\(\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha}\)
\(\cot \alpha = \dfrac{\cos \alpha}{\sin \alpha}\)
Hubungan Kuadrat (Pythagorean)
\(\sin^2\alpha + \cos^2\alpha = 1\)
\(1 + \tan^2\alpha = \sec^2\alpha\)
\(1 + \cot^2\alpha = \csc^2\alpha\)
Tanda Perbandingan Trigonometri di Setiap Kuadran
Contoh Soal – Hubungan Antar Perbandingan
🟢 Mudah
1. Jika sin α = 3/5, tentukan cos α menggunakan identitas Pythagoras!
Pembahasan:
\(\sin^2\alpha + \cos^2\alpha = 1\)
\(\cos^2\alpha = 1 – \left(\dfrac{3}{5}\right)^2 = 1 – \dfrac{9}{25} = \dfrac{16}{25}\)
\(\cos\alpha = \dfrac{4}{5}\) (kuadran I)
2. Tunjukkan bahwa \(\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}\) untuk α = 30°!
Pembahasan:
\(\dfrac{\sin 30°}{\cos 30°} = \dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} = \tan 30°\) ✓
3. Jika cos α = 5/13, tentukan tan α! (kuadran I)
Pembahasan:
\(\sin\alpha = \sqrt{1 – \frac{25}{169}} = \sqrt{\frac{144}{169}} = \dfrac{12}{13}\)
\(\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{12/13}{5/13} = \dfrac{12}{5}\)
4. Tentukan nilai \(\sin^2 60° + \cos^2 60°\)!
Pembahasan:
\(\left(\dfrac{\sqrt{3}}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 = \dfrac{3}{4} + \dfrac{1}{4} = 1\) ✓ (sesuai identitas)
5. Jika tan α = 2, tentukan sec²α!
Pembahasan:
\(\sec^2\alpha = 1 + \tan^2\alpha = 1 + 4 = 5\)
🟡 Sedang
1. Jika \(\sin\alpha = \dfrac{8}{17}\) dan α di kuadran II, tentukan tan α!
Pembahasan:
\(\cos^2\alpha = 1 – \dfrac{64}{289} = \dfrac{225}{289}\), di kuadran II cos negatif:
\(\cos\alpha = -\dfrac{15}{17}\)
\(\tan\alpha = \dfrac{8/17}{-15/17} = -\dfrac{8}{15}\)
2. Buktikan bahwa \(1 + \cot^2\alpha = \csc^2\alpha\) dari identitas dasar!
Pembahasan:
Dari \(\sin^2\alpha + \cos^2\alpha = 1\), bagi kedua ruas dengan \(\sin^2\alpha\):
\(1 + \dfrac{\cos^2\alpha}{\sin^2\alpha} = \dfrac{1}{\sin^2\alpha}\)
\(1 + \cot^2\alpha = \csc^2\alpha\) ✓
3. Jika \(\tan\alpha = -\dfrac{3}{4}\) dan α di kuadran IV, tentukan sin α + cos α!
Pembahasan:
Kuadran IV: sin(−), cos(+). d=3, s=4, m=5
\(\sin\alpha = -\dfrac{3}{5},\; \cos\alpha = \dfrac{4}{5}\)
\(\sin\alpha + \cos\alpha = -\dfrac{3}{5} + \dfrac{4}{5} = \dfrac{1}{5}\)
4. Sederhanakan \(\dfrac{1 – \sin^2\alpha}{\cos\alpha}\)!
Pembahasan:
\(\dfrac{1 – \sin^2\alpha}{\cos\alpha} = \dfrac{\cos^2\alpha}{\cos\alpha} = \cos\alpha\)
5. Jika cot α = 7/24, tentukan sin α dan cos α!
Pembahasan:
cot = s/d = 7/24. \(m = \sqrt{49+576} = \sqrt{625} = 25\)
\(\sin\alpha = \dfrac{24}{25},\; \cos\alpha = \dfrac{7}{25}\)
🔴 Sulit
1. Buktikan \(\dfrac{\tan\alpha}{1 – \cot\alpha} + \dfrac{\cot\alpha}{1 – \tan\alpha} = 1 + \sec\alpha\csc\alpha\)!
Pembahasan:
Substitusi \(\tan\alpha = \dfrac{s}{c}\), \(\cot\alpha = \dfrac{c}{s}\) dimana s = sin α, c = cos α:
\(\dfrac{\frac{s}{c}}{1-\frac{c}{s}} + \dfrac{\frac{c}{s}}{1-\frac{s}{c}} = \dfrac{\frac{s}{c}}{\frac{s-c}{s}} + \dfrac{\frac{c}{s}}{\frac{c-s}{c}}\)
\(= \dfrac{s^2}{c(s-c)} + \dfrac{c^2}{s(c-s)} = \dfrac{s^2}{c(s-c)} – \dfrac{c^2}{s(s-c)}\)
\(= \dfrac{s^3 – c^3}{sc(s-c)} = \dfrac{(s-c)(s^2+sc+c^2)}{sc(s-c)} = \dfrac{1+sc}{sc} = \dfrac{1}{sc} + 1\)
\(= \csc\alpha\sec\alpha + 1\) ✓
2. Jika \(\sin\alpha + \cos\alpha = \dfrac{7}{5}\), tentukan \(\sin\alpha \cdot \cos\alpha\)!
Pembahasan:
Kuadratkan: \((\sin\alpha + \cos\alpha)^2 = \dfrac{49}{25}\)
\(\sin^2\alpha + 2\sin\alpha\cos\alpha + \cos^2\alpha = \dfrac{49}{25}\)
\(1 + 2\sin\alpha\cos\alpha = \dfrac{49}{25}\)
\(\sin\alpha\cos\alpha = \dfrac{1}{2}\left(\dfrac{49}{25} – 1\right) = \dfrac{24}{50} = \dfrac{12}{25}\)
3. Sederhanakan \(\dfrac{\sec\alpha – \cos\alpha}{\sin\alpha}\)!
Pembahasan:
\(\dfrac{\frac{1}{\cos\alpha} – \cos\alpha}{\sin\alpha} = \dfrac{\frac{1-\cos^2\alpha}{\cos\alpha}}{\sin\alpha} = \dfrac{\sin^2\alpha}{\cos\alpha \cdot \sin\alpha} = \dfrac{\sin\alpha}{\cos\alpha} = \tan\alpha\)
4. Jika \(\tan\alpha – \sin\alpha = m\) dan \(\tan\alpha + \sin\alpha = n\), buktikan \(m^2 – n^2 = -4\sqrt{mn}\) salah, dan tentukan hubungan yang benar!
Pembahasan:
\(n^2 – m^2 = (n+m)(n-m) = 2\tan\alpha \cdot 2\sin\alpha = 4\tan\alpha\sin\alpha\)
\(mn = (\tan\alpha)^2 – \sin^2\alpha = \dfrac{\sin^2\alpha}{\cos^2\alpha} – \sin^2\alpha = \sin^2\alpha\left(\dfrac{1-\cos^2\alpha}{\cos^2\alpha}\right) = \dfrac{\sin^4\alpha}{\cos^2\alpha}\)
\(4\sqrt{mn} = 4\cdot\dfrac{\sin^2\alpha}{\cos\alpha} = 4\tan\alpha\sin\alpha\)
Jadi hubungan yang benar: \(n^2 – m^2 = 4\sqrt{mn}\) ✓
5. Sederhanakan \(\dfrac{1+\sin\alpha}{\cos\alpha} + \dfrac{\cos\alpha}{1+\sin\alpha}\)!
Pembahasan:
\(= \dfrac{(1+\sin\alpha)^2 + \cos^2\alpha}{\cos\alpha(1+\sin\alpha)}\)
\(= \dfrac{1+2\sin\alpha+\sin^2\alpha+\cos^2\alpha}{\cos\alpha(1+\sin\alpha)}\)
\(= \dfrac{2+2\sin\alpha}{\cos\alpha(1+\sin\alpha)} = \dfrac{2(1+\sin\alpha)}{\cos\alpha(1+\sin\alpha)} = \dfrac{2}{\cos\alpha} = 2\sec\alpha\)
Latihan Soal – Hubungan Antar Perbandingan
🟢 Mudah
- Jika cos α = 7/25, tentukan sin α! (kuadran I)
- Tentukan sec²α jika tan α = 3!
- Jika sin α = 1/2, tentukan csc α!
- Tentukan cot α jika tan α = √3!
- Verifikasi bahwa sin²30° + cos²30° = 1!
🟡 Sedang
- Jika sin α = −5/13 dan α di kuadran III, tentukan cos α dan tan α!
- Sederhanakan \(\dfrac{\sin^2\alpha}{\cos^2\alpha} + 1\)!
- Jika sec α = 5/3, tentukan tan α dan cot α!
- Buktikan \(1 + \tan^2\alpha = \sec^2\alpha\) dari identitas dasar!
- Jika cos α = 3/5 dan α di kuadran IV, tentukan csc α!
🔴 Sulit
- Jika \(\sin\alpha – \cos\alpha = \dfrac{1}{3}\), tentukan \(\sin\alpha\cos\alpha\)!
- Sederhanakan \(\dfrac{1-\cos\alpha}{\sin\alpha} + \dfrac{\sin\alpha}{1-\cos\alpha}\)!
- Buktikan \((\sin\alpha + \cos\alpha)^2 + (\sin\alpha – \cos\alpha)^2 = 2\)!
- Jika \(\tan\alpha + \cot\alpha = 3\), tentukan \(\tan^2\alpha + \cot^2\alpha\)!
- Sederhanakan \(\dfrac{\csc\alpha – \sin\alpha}{\cos\alpha}\)!
3. Identitas Trigonometri
Identitas Dasar
| No | Identitas | Keterangan |
|---|---|---|
| 1 | \(\sin^2\alpha + \cos^2\alpha = 1\) | Identitas Pythagoras |
| 2 | \(\tan^2\alpha + 1 = \sec^2\alpha\) | Turunan dari (1) ÷ cos²α |
| 3 | \(1 + \cot^2\alpha = \csc^2\alpha\) | Turunan dari (1) ÷ sin²α |
| 4 | \(\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}\) | Hubungan pembagian |
| 5 | \(\cot\alpha = \dfrac{\cos\alpha}{\sin\alpha}\) | Hubungan pembagian |
Strategi Membuktikan Identitas Trigonometri
- Pilih satu ruas (biasanya yang lebih kompleks) dan ubah menjadi ruas lainnya.
- Ubah semua ke sin dan cos jika kesulitan.
- Gunakan identitas Pythagoras untuk menyederhanakan.
- Faktorisasi jika memungkinkan.
- Kalikan dengan konjugat jika terdapat bentuk (1 ± sin) atau (1 ± cos).
Contoh Soal – Identitas Trigonometri
🟢 Mudah
1. Buktikan \(\sin^2\alpha = 1 – \cos^2\alpha\)!
Pembahasan:
Dari identitas: \(\sin^2\alpha + \cos^2\alpha = 1\)
Pindahkan: \(\sin^2\alpha = 1 – \cos^2\alpha\) ✓
2. Sederhanakan \(\sin\alpha \cdot \csc\alpha\)!
Pembahasan:
\(\sin\alpha \cdot \csc\alpha = \sin\alpha \cdot \dfrac{1}{\sin\alpha} = 1\)
3. Sederhanakan \(\cos\alpha \cdot \tan\alpha\)!
Pembahasan:
\(\cos\alpha \cdot \tan\alpha = \cos\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} = \sin\alpha\)
4. Buktikan \(\tan\alpha \cdot \cot\alpha = 1\)!
Pembahasan:
\(\tan\alpha \cdot \cot\alpha = \dfrac{\sin\alpha}{\cos\alpha} \cdot \dfrac{\cos\alpha}{\sin\alpha} = 1\) ✓
5. Sederhanakan \(\sec^2\alpha – \tan^2\alpha\)!
Pembahasan:
Dari identitas \(1 + \tan^2\alpha = \sec^2\alpha\):
\(\sec^2\alpha – \tan^2\alpha = 1\)
🟡 Sedang
1. Buktikan \(\dfrac{\sin\alpha}{1+\cos\alpha} = \dfrac{1-\cos\alpha}{\sin\alpha}\)!
Pembahasan:
Cross multiply: \(\sin\alpha \cdot \sin\alpha = (1-\cos\alpha)(1+\cos\alpha)\)
\(\sin^2\alpha = 1 – \cos^2\alpha\) ✓ (identitas Pythagoras)
2. Sederhanakan \(\dfrac{1}{\sin^2\alpha} – \dfrac{1}{\tan^2\alpha}\)!
Pembahasan:
\(= \csc^2\alpha – \cot^2\alpha\)
Dari identitas \(1 + \cot^2\alpha = \csc^2\alpha\):
\(\csc^2\alpha – \cot^2\alpha = 1\)
3. Buktikan \(\cos^4\alpha – \sin^4\alpha = \cos^2\alpha – \sin^2\alpha\)!
Pembahasan:
Ruas kiri = selisih kuadrat:
\((\cos^2\alpha + \sin^2\alpha)(\cos^2\alpha – \sin^2\alpha) = 1 \cdot (\cos^2\alpha – \sin^2\alpha)\)
\(= \cos^2\alpha – \sin^2\alpha\) = ruas kanan ✓
4. Sederhanakan \(\dfrac{\tan\alpha + \cot\alpha}{\csc\alpha}\)!
Pembahasan:
\(= \dfrac{\frac{\sin\alpha}{\cos\alpha} + \frac{\cos\alpha}{\sin\alpha}}{\frac{1}{\sin\alpha}} = \dfrac{\frac{\sin^2\alpha + \cos^2\alpha}{\sin\alpha\cos\alpha}}{\frac{1}{\sin\alpha}}\)
\(= \dfrac{\frac{1}{\sin\alpha\cos\alpha}}{\frac{1}{\sin\alpha}} = \dfrac{1}{\cos\alpha} = \sec\alpha\)
5. Buktikan \(\dfrac{1-\sin\alpha}{\cos\alpha} = \dfrac{\cos\alpha}{1+\sin\alpha}\)!
Pembahasan:
Cross multiply: \((1-\sin\alpha)(1+\sin\alpha) = \cos^2\alpha\)
\(1 – \sin^2\alpha = \cos^2\alpha\) ✓
🔴 Sulit
1. Buktikan \(\dfrac{\sin\alpha}{1-\cos\alpha} – \dfrac{\sin\alpha}{1+\cos\alpha} = 2\cot\alpha\)!
Pembahasan:
\(= \dfrac{\sin\alpha(1+\cos\alpha) – \sin\alpha(1-\cos\alpha)}{(1-\cos\alpha)(1+\cos\alpha)}\)
\(= \dfrac{\sin\alpha + \sin\alpha\cos\alpha – \sin\alpha + \sin\alpha\cos\alpha}{1-\cos^2\alpha}\)
\(= \dfrac{2\sin\alpha\cos\alpha}{\sin^2\alpha} = \dfrac{2\cos\alpha}{\sin\alpha} = 2\cot\alpha\) ✓
2. Buktikan \((\sec\alpha + \tan\alpha)(\sec\alpha – \tan\alpha) = 1\)!
Pembahasan:
\(= \sec^2\alpha – \tan^2\alpha\)
Dari identitas: \(\sec^2\alpha = 1 + \tan^2\alpha\)
\(\sec^2\alpha – \tan^2\alpha = 1\) ✓
3. Buktikan \(\dfrac{\tan^2\alpha – \sin^2\alpha}{\tan^2\alpha \cdot \sin^2\alpha} = 1\)!
Pembahasan:
Ruas kiri \(= \dfrac{1}{\sin^2\alpha} – \dfrac{1}{\tan^2\alpha} = \csc^2\alpha – \cot^2\alpha\)
Tunggu, kita cek ulang:
\(\dfrac{\tan^2\alpha – \sin^2\alpha}{\tan^2\alpha \cdot \sin^2\alpha} = \dfrac{\frac{\sin^2\alpha}{\cos^2\alpha} – \sin^2\alpha}{\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\sin^2\alpha}\)
\(= \dfrac{\sin^2\alpha\left(\frac{1}{\cos^2\alpha}-1\right)}{\frac{\sin^4\alpha}{\cos^2\alpha}} = \dfrac{\sin^2\alpha \cdot \frac{1-\cos^2\alpha}{\cos^2\alpha}}{\frac{\sin^4\alpha}{\cos^2\alpha}} = \dfrac{\sin^2\alpha \cdot \frac{\sin^2\alpha}{\cos^2\alpha}}{\frac{\sin^4\alpha}{\cos^2\alpha}} = \dfrac{\frac{\sin^4\alpha}{\cos^2\alpha}}{\frac{\sin^4\alpha}{\cos^2\alpha}} = 1\) ✓
4. Sederhanakan \(\dfrac{1}{1-\sin\alpha} + \dfrac{1}{1+\sin\alpha}\)!
Pembahasan:
\(= \dfrac{(1+\sin\alpha) + (1-\sin\alpha)}{(1-\sin\alpha)(1+\sin\alpha)} = \dfrac{2}{1-\sin^2\alpha} = \dfrac{2}{\cos^2\alpha} = 2\sec^2\alpha\)
5. Buktikan \(\dfrac{\sin^3\alpha + \cos^3\alpha}{\sin\alpha + \cos\alpha} = 1 – \sin\alpha\cos\alpha\)!
Pembahasan:
Gunakan rumus \(a^3+b^3 = (a+b)(a^2-ab+b^2)\):
\(\dfrac{(\sin\alpha+\cos\alpha)(\sin^2\alpha – \sin\alpha\cos\alpha + \cos^2\alpha)}{\sin\alpha+\cos\alpha}\)
\(= \sin^2\alpha – \sin\alpha\cos\alpha + \cos^2\alpha = 1 – \sin\alpha\cos\alpha\) ✓
Latihan Soal – Identitas Trigonometri
🟢 Mudah
- Sederhanakan \(\sin\alpha \cdot \sec\alpha\)!
- Buktikan \(\csc^2\alpha – \cot^2\alpha = 1\)!
- Sederhanakan \(\cos^2\alpha(1 + \tan^2\alpha)\)!
- Buktikan \(\sin\alpha \cdot \cot\alpha = \cos\alpha\)!
- Sederhanakan \(\dfrac{\sin^2\alpha + \cos^2\alpha}{\cos\alpha}\)!
🟡 Sedang
- Buktikan \(\dfrac{\cos\alpha}{1-\sin\alpha} = \sec\alpha + \tan\alpha\)!
- Sederhanakan \(\dfrac{\sin^2\alpha – 1}{\sin\alpha – \csc\alpha}\)!
- Buktikan \(\sin^4\alpha – \cos^4\alpha = 2\sin^2\alpha – 1\)!
- Sederhanakan \((\csc\alpha – \cot\alpha)(\csc\alpha + \cot\alpha)\)!
- Buktikan \(\dfrac{\tan\alpha}{\sec\alpha} = \sin\alpha\)!
🔴 Sulit
- Buktikan \(\dfrac{1+\tan^2\alpha}{1+\cot^2\alpha} = \tan^2\alpha\)!
- Buktikan \(\dfrac{\cos\alpha}{1+\sin\alpha} + \dfrac{1+\sin\alpha}{\cos\alpha} = 2\sec\alpha\)!
- Buktikan \((\sin\alpha + \cos\alpha)^2 + (\sin\alpha – \cos\alpha)^2 = 2\)!
- Sederhanakan \(\dfrac{\sec^4\alpha – \tan^4\alpha}{\sec^2\alpha + \tan^2\alpha}\)!
- Buktikan \(\dfrac{\cot^2\alpha – 1}{\csc^2\alpha} = 1 – 2\sin^2\alpha\)!