Pembuktian Rumus Penjumlahan Trigonometri dan Contoh Soal

Posted on
5/5 (1)

Pada tulisan ini akan dijelaskan mengenai pembuktian rumus penjumlahan dan pengurangan fungsi trigonometri, sebelumnya sudah dijelaskan pula mengenai pembuktian rumus perkalian sinus dan cosinus, jika belum membacanya sebaikanya Kamu baca dulu tulisan tersebut karena ada kaitannya.

Berikut adalah rumus penjumlahan dan pengurangan trigonometri yang wajib Kamu pahami dan hafalkan:

Sekali lagi, agar dapat membuktikan rumus penjumlahan dan pengurangan fungsi trigonometri Kamu harus paham dulu rumus perkalian sinus dan cosinus, karena pembuktiannya dimulai dari rumus perkalian. Simaklah pembahasan dibawah ini!

Misalkan \(\alpha + \beta = A\) dan \(\alpha – \beta = B\)

Sekarang jumlahkan kedua persamaan tersebut!

\(\alpha + \beta = A\)

\(\alpha – \beta = B\)

Didapatkan hasilnya

\(2 \alpha = A + B\)

\(\alpha = \frac{A + B}{2}\)

Sekarang kurangkan kedua persamaan tersebut!

\(\alpha + \beta = A\)

\(\alpha – \beta = B\)

Didapatkan hasilnya

\(2 \beta = A – B\)

\(\beta = \frac{A – B}{2}\)

Sekarang gunakanlah rumus perkalian fungsi sinus dan cosinus yang sudah dibahas sebelumnya.

\(2 \sin \alpha \cos \beta = \sin \left( \alpha + \beta \right) + \sin \left( \alpha – \beta \right)\)
\(\sin \left( \alpha + \beta \right) + \sin \left( \alpha – \beta \right) = 2 \sin \alpha \cos \beta\)
\(\sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A – B}{2} \right)\) terbukti
\(2 \cos \alpha \sin \beta = \sin \left( \alpha + \beta \right) – \sin \left( \alpha – \beta \right)\)
\(\sin \left( \alpha + \beta \right) – \sin \left( \alpha – \beta \right) = 2 \cos \alpha \sin \beta\)
\(\sin A – \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A – B}{2} \right)\) terbukti
\(2 \cos \alpha \cos \beta = \cos \left( \alpha + \beta \right) + \cos \left( \alpha – \beta \right)\)
\(\cos \left( \alpha + \beta \right) + \cos \left( \alpha – \beta \right) = 2 \cos \alpha \cos \beta\)
\(\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A – B}{2} \right)\) terbukti
\(– 2 \sin \alpha \sin \beta = \cos \left( \alpha + \beta \right) – \cos \left( \alpha – \beta \right)\)
\(\cos \left( \alpha + \beta \right) – \cos \left( \alpha – \beta \right) = – 2 \sin \alpha \sin \beta\)
\(\cos A – \cos B = – 2 \sin \left( \frac{A + B}{2} \right) \sin \left( \frac{A – B}{2} \right)\) terbukti

Contoh

Nomor 1
sederhanakanlah bentuk-bentuk berikut:
a. \(\sin 3x – \sin x\)
b. \(\cos 5 \alpha + \cos 3 \alpha\)
c. \(\cos (x+h) – \cos x\)
d. \(\sin \left( \alpha + \beta \right) + \sin \left( \alpha – \beta \right)\)

Jawab:

Nomor 1a
\(\sin 3x – \sin x\)
Misalkan \(3x = A\) dan \(x = B\)
\(\sin A – \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A – B}{2} \right)\)
\(\sin 3x – \sin x = 2 \cos \left( \frac{3x + x}{2} \right) \sin \left( \frac{3x – x}{2} \right)\)
\(\sin 3x – \sin x = 2 \cos \left( \frac{4x}{2} \right) \sin \left( \frac{2x}{2} \right)\)
\(\sin 3x – \sin x = 2 \cos 2x \sin x\)
Nomor 1b
\(\cos 5 \alpha + \cos 3 \alpha\)
Misalkan \(5 \alpha = A\) dan \(3 \alpha = B\)
\(\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A – B}{2} \right)\)
\(\cos 5 \alpha + \cos 3 \alpha = 2 \cos \left( \frac{5 \alpha + 3 \alpha}{2} \right) \cos \left( \frac{5 \alpha – 3 \alpha}{2} \right)\)
\(\cos 5 \alpha + \cos 3 \alpha = 2 \cos \left( \frac{8 \alpha}{2} \right) \cos \left( \frac{2 \alpha}{2} \right)\)
\(\cos 5 \alpha + \cos 3 \alpha = 2 \cos 4 \alpha \cos \alpha\)
Nomor 1c
\(\cos (x+h) – \cos x\)
Misalkan \((x+h) = A\) dan \(x = B\)
\(\cos A – \cos B = – 2 \sin \left( \frac{A + B}{2} \right) \sin \left( \frac{A – B}{2} \right)\)
\(\cos (x+h) – \cos x = – 2 \sin \left( \frac{(x+h) + x}{2} \right) \sin \left( \frac{(x+h) – x}{2} \right)\)
\(\cos (x+h) – \cos x = – 2 \sin \left( \frac{2x+h}{2} \right) \sin \left( \frac{h}{2} \right)\)
Nomor 1d
\(\sin \left( \alpha + \beta \right) + \sin \left( \alpha – \beta \right)\)
Misalkan \(\left( \alpha + \beta \right) = A\) dan \(\left( \alpha – \beta \right) = B\)
\(\sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A – B}{2} \right)\)
\(\sin \left( \alpha + \beta \right) + \sin \left( \alpha – \beta \right)\)
\(= 2 \sin \left( \frac{\left( \alpha + \beta \right) + \left( \alpha – \beta \right)}{2} \right) \cos \left( \frac{\left( \alpha + \beta \right) – \left( \alpha – \beta \right)}{2} \right)\)
\(= 2 \sin \left( \frac{2 \alpha}{2} \right) \cos \left( \frac{2 \beta}{2} \right)\)
\(= 2 \sin \alpha \cos \beta\)

Nomor 2
Berapakah nilai dari:
a. \(\cos 75^{\circ} – \cos 15^{\circ}\)
b. \(\frac{\cos 15^{\circ} – \cos 75^{\circ}}{\sin 15^{\circ} – \sin 75^{\circ}}\)

Jawab:

Nomor 2a
\(\cos 75^{\circ} – \cos 15^{\circ}\)
Misalkan \(75^{\circ} = A\) dan \(15^{\circ} = B\)
\(\cos A – \cos B = – 2 \sin \left( \frac{A + B}{2} \right) \sin \left( \frac{A – B}{2} \right)\)
\(\cos 75^{\circ} – \cos 15^{\circ} = – 2 \sin \left( \frac{75^{\circ} + 15^{\circ}}{2} \right) \sin \left( \frac{75^{\circ} – 15^{\circ}}{2} \right)\)
\(\cos 75^{\circ} – \cos 15^{\circ} = – 2 \sin \left( \frac{90^{\circ}}{2} \right) \sin \left( \frac{60^{\circ}}{2} \right)\)
\(\cos 75^{\circ} – \cos 15^{\circ} = – 2 \sin 45^{\circ} \sin 30^{\circ}\)
\(\cos 75^{\circ} – \cos 15^{\circ} = – 2 \times \frac{1}{2} \sqrt{2} \times \frac{1}{2}\)
\(\cos 75^{\circ} – \cos 15^{\circ} = – \frac{1}{2} \sqrt{2}\)
Nomor 2b
\(\frac{\cos 15^{\circ} – \cos 75^{\circ}}{\sin 15^{\circ} – \sin 75^{\circ}}\)
\(= \frac{- 2 \sin \left( \frac{15^{\circ} + 75^{\circ}}{2} \right) \sin \left( \frac{15^{\circ} – 75^{\circ}}{2} \right)}{2 \cos \left( \frac{15^{\circ} + 75^{\circ}}{2} \right) \sin \left( \frac{15^{\circ} – 75^{\circ}}{2} \right)}\)
\(= – \frac{\sin \left( \frac{15^{\circ} + 75^{\circ}}{2} \right)}{\cos \left( \frac{15^{\circ} + 75^{\circ}}{2} \right)}\)
\(= – \frac{\sin \left( \frac{90^{\circ}}{2} \right)}{\cos \left( \frac{90^{\circ}}{2} \right)}\)
\(= – \frac{\sin 45^{\circ}}{\cos 45^{\circ}}\)
\(= – \tan 45^{\circ}\)
\(= – 1\)

Nomor 3
Buktikan identitas berikut:
a. \(\frac{\sin 7 \alpha – \sin 5 \alpha}{\cos 7 \alpha + \cos 5 \alpha} = \tan \alpha\)
b. \(\frac{\sin \beta + \sin 3 \beta}{\cos \beta + \cos 3 \beta} = \tan 2 \beta\)

Jawab:

Nomor 3a
\(\frac{\sin 7 \alpha – \sin 5 \alpha}{\cos 7 \alpha – \cos 5 \alpha}\)
\(= \frac{2 \cos \left( \frac{7 \alpha + 5 \alpha}{2} \right) \sin \left( \frac{7 \alpha – 5 \alpha}{2} \right)}{2 \cos \left( \frac{7 \alpha + 5 \alpha}{2} \right) \cos \left( \frac{7 \alpha – 5 \alpha}{2} \right)}\)
\(= \frac{\sin \left( \frac{7 \alpha – 5 \alpha}{2} \right)}{\cos \left( \frac{7 \alpha – 5 \alpha}{2} \right)}\)
\(= \frac{\sin \left( \frac{2 \alpha}{2} \right)}{\cos \left( \frac{2 \alpha}{2} \right)}\)
\(= \frac{\sin \alpha}{\cos \alpha}\)
\(= \tan \alpha\)
Nomor 3b
\(\frac{\sin \beta + \sin 3 \beta}{\cos \beta + \cos 3 \beta}\)
\(= \frac{2 \sin \left( \frac{\beta + 3 \beta}{2} \right) \cos \left( \frac{\beta – 3 \beta}{2} \right)}{2 \cos \left( \frac{\beta + 3 \beta}{2} \right) \cos \left( \frac{\beta – 3 \beta}{2} \right)}\)
\(= \frac{\sin \left( \frac{\beta + 3 \beta}{2} \right)}{\cos \left( \frac{\beta + 3 \beta}{2} \right)}\)
\(= \frac{\sin \left( \frac{4 \beta}{2} \right)}{\cos \left( \frac{4 \beta}{2} \right)}\)
\(= \frac{\sin 2 \beta}{\cos 2 \beta}\)
\(= \tan 2 \beta\)

Itulah penjelasan lengkap mengenai pembuktian rumus penjumlahan trigonometri yang dibahas oleh Edumatik, semoga Kamu paham dengan apa yang dijelaskan.

Jika Kamu suka dengan artikel ini jangan lupa untuk memberi bintang dan share artikel ini yaa. See you, bye

Beri nilai tulisan ini!

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