Penjumlahan dan pengurangan fungsi trigonometri merupakan salah satu materi matematika yang cukup menarik untuk dibahas. Kalau kebetulan kamu ingin belajar tentang materi ini lebih dalam, simak penjelasan lengkapnya berikut. Kami juga telah menyediakan soal latihan yang bisa dikerjakan untuk mengasah kemampuanmu.
Di sini, kamu akan belajar tentang Penjumlahan & Pengurangan Fungsi Trigonometri melalui video yang dibawakan oleh Bapak Anton Wardaya. Kamu akan diajak untuk memahami materi hingga metode menyelesaikan soal.
Selain itu, kamu juga akan mendapatkan latihan soal interaktif dalam 3 tingkat kesulitan (mudah, sedang, sukar). Dengan begitu, kamu bisa langsung mempraktikkan materi yang telah dijelaskan.
Sekarang, kamu bisa mulai belajar dengan 2 video dan 3 set latihan soal yang ada di halaman ini. Apabila materi ini berguna, bagikan ke teman atau rekan kamu supaya mereka juga mendapatkan manfaatnya.
Kamu dapat download modul & contoh soal serta kumpulan latihan soal lengkap dalam bentuk pdf pada list dibawah ini:
Pembuktian Penjumlahan & Pengurangan Fungsi Trigonometri
Contoh Soal Penjumlahan & Pengurangan Fungsi Trigonometri
Latihan Soal Penjumlahan & Pengurangan Fungsi Trigonometri (Mudah)
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Pertanyaan ke 1 dari 5
1. Pertanyaan
$\sin15^{\circ}+\sin105^{\circ}=…$
Betul$\sin15^{\circ}+\sin105^{\circ}$$=2\times\sin\left(\frac{15^{\circ}+105^{\circ}}{2}\right)$$\times\cos\left(\frac{15^{\circ}-105^{\circ}}{2}\right)$
$=2\sin60^{\circ}\cos\left(-45^{\circ}\right)$
$=2\times\frac{1}{2}\sqrt{3}\times\frac{\sqrt{2}}{2}$
$=\frac{1}{2}\sqrt{6}$
Salah$\sin15^{\circ}+\sin105^{\circ}$$=2\times\sin\left(\frac{15^{\circ}+105^{\circ}}{2}\right)$$\times\cos\left(\frac{15^{\circ}-105^{\circ}}{2}\right)$
$=2\sin60^{\circ}\cos\left(-45^{\circ}\right)$
$=2\times\frac{1}{2}\sqrt{3}\times\frac{\sqrt{2}}{2}$
$=\frac{1}{2}\sqrt{6}$
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Pertanyaan ke 2 dari 5
2. Pertanyaan
$\cos\frac{5\pi}{12}+\cos\frac{7\pi}{12}=…$
Betul$\cos\frac{5\pi}{12}+\cos\frac{7\pi}{12}$$=2\times\cos\frac{1}{2}\left(\frac{5\pi}{12}+\frac{7\pi}{12}\right)$$\times\cos\frac{1}{2}\left(\frac{5\pi}{12}-\frac{7\pi}{12}\right)$
$=2\times\cos\frac{\pi}{2}\cos\left(-\frac{\pi}{12}\right)$
$=0$
Salah$\cos\frac{5\pi}{12}+\cos\frac{7\pi}{12}$$=2\times\cos\frac{1}{2}\left(\frac{5\pi}{12}+\frac{7\pi}{12}\right)$$\times\cos\frac{1}{2}\left(\frac{5\pi}{12}-\frac{7\pi}{12}\right)$
$=2\times\cos\frac{\pi}{2}\cos\left(-\frac{\pi}{12}\right)$
$=0$
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Pertanyaan ke 3 dari 5
3. Pertanyaan
Jika $\cos10^{\circ}=a,$ $\cos40^{\circ}+\cos20^{\circ}=…$
Betul$\cos40^{\circ}+\cos20^{\circ}$$=2\times\cos\left(\frac{40^{\circ}+20^{\circ}}{2}\right)$$\times\cos\left(\frac{40^{\circ}-20^{\circ}}{2}\right)$
$=2\cos30^{\circ}\cos10^{\circ}$
$=\sqrt{3}a$
Salah$\cos40^{\circ}+\cos20^{\circ}$$=2\times\cos\left(\frac{40^{\circ}+20^{\circ}}{2}\right)$$\times\cos\left(\frac{40^{\circ}-20^{\circ}}{2}\right)$
$=2\cos30^{\circ}\cos10^{\circ}$
$=\sqrt{3}a$
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Pertanyaan ke 4 dari 5
4. Pertanyaan
$\cos\frac{5\pi}{12}-\cos\frac{\pi}{12}=…$
Betul$\cos\frac{5\pi}{12}-\cos\frac{\pi}{12}=-2\sin\frac{1}{2}\left(\frac{5\pi}{12}+\frac{\pi}{12}\right)$$\sin\frac{1}{2}\left(\frac{5\pi}{12}-\frac{\pi}{12}\right)$
$=-2\sin\frac{\pi}{4}\sin\frac{\pi}{6}$
$=-\frac{\sqrt{2}}{2}$
$=-\frac{1}{2}\sqrt{2}$
Salah$\cos\frac{5\pi}{12}-\cos\frac{\pi}{12}=-2\sin\frac{1}{2}\left(\frac{5\pi}{12}+\frac{\pi}{12}\right)$$\sin\frac{1}{2}\left(\frac{5\pi}{12}-\frac{\pi}{12}\right)$
$=-2\sin\frac{\pi}{4}\sin\frac{\pi}{6}$
$=-\frac{\sqrt{2}}{2}$
$=-\frac{1}{2}\sqrt{2}$
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Pertanyaan ke 5 dari 5
5. Pertanyaan
$\sin536^{\circ}-\sin4^{\circ}=…$
Betul$\sin536^{\circ}-\sin4^{\circ}=2\times\cos\left(\frac{536^{\circ}+4^{\circ}}{2}\right)$$\times\sin\left(\frac{536^{\circ}-4^{\circ}}{2}\right)$
$=2\cos270^{\circ}\sin266^{\circ}$
$=0$
Salah$\sin536^{\circ}-\sin4^{\circ}=2\times\cos\left(\frac{536^{\circ}+4^{\circ}}{2}\right)$$\times\sin\left(\frac{536^{\circ}-4^{\circ}}{2}\right)$
$=2\cos270^{\circ}\sin266^{\circ}$
$=0$
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Pertanyaan ke 1 dari 5
1. Pertanyaan
$\frac{\sin\left(\frac{\pi}{3}+\alpha\right)+\sin\left(\frac{\pi}{3}-\alpha\right)}{\cos\left(\frac{\pi}{2}+\alpha\right)-\cos\left(\frac{\pi}{2}-\alpha\right)}=…$ Untuk $\alpha=-\frac{\pi}{6}$
Betul$\begin{aligned}\frac{\sin\left(\frac{\pi}{3}+\alpha\right)+\sin\left(\frac{\pi}{3}-\alpha\right)}{\cos\left(\frac{\pi}{2}+\alpha\right)-\cos\left(\frac{\pi}{2}-\alpha\right)} & =\frac{2\sin\frac{\pi}{3}\cos\alpha}{-2\sin\frac{\pi}{2}\sin\alpha}\\
& =-\frac{\frac{1}{2}\sqrt{3}}{1}\cot\left(-\frac{\pi}{6}\right)\\
& =\frac{1}{2}\sqrt{3}-\sqrt{3}\\
& =\frac{3}{2}
\end{aligned}
$Salah$\begin{aligned}\frac{\sin\left(\frac{\pi}{3}+\alpha\right)+\sin\left(\frac{\pi}{3}-\alpha\right)}{\cos\left(\frac{\pi}{2}+\alpha\right)-\cos\left(\frac{\pi}{2}-\alpha\right)} & =\frac{2\sin\frac{\pi}{3}\cos\alpha}{-2\sin\frac{\pi}{2}\sin\alpha}\\
& =-\frac{\frac{1}{2}\sqrt{3}}{1}\cot\left(-\frac{\pi}{6}\right)\\
& =\frac{1}{2}\sqrt{3}-\sqrt{3}\\
& =\frac{3}{2}
\end{aligned}
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Pertanyaan ke 2 dari 5
2. Pertanyaan
$\frac{\cos\left(\frac{\pi}{3}+x\right)-\cos\left(\frac{\pi}{3}-x\right)}{\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}-x\right)}=…$
Betul$\begin{aligned}\frac{\cos\left(\frac{\pi}{3}+x\right)-\cos\left(\frac{\pi}{3}-x\right)}{\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}-x\right)} & =\frac{-2\sin\frac{\pi}{3}\sin x}{+2\cos\frac{\pi}{3}\sin x}\\
& =-\frac{\frac{1}{2}}{\frac{1}{2}\sqrt{3}}\\
& =-\frac{1}{3}\sqrt{3}
\end{aligned}
$Salah$\begin{aligned}\frac{\cos\left(\frac{\pi}{3}+x\right)-\cos\left(\frac{\pi}{3}-x\right)}{\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}-x\right)} & =\frac{-2\sin\frac{\pi}{3}\sin x}{+2\cos\frac{\pi}{3}\sin x}\\
& =-\frac{\frac{1}{2}}{\frac{1}{2}\sqrt{3}}\\
& =-\frac{1}{3}\sqrt{3}
\end{aligned}
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Pertanyaan ke 3 dari 5
3. Pertanyaan
Jika $p=\cos20^{\circ}-\cos40^{\circ}-\cos80^{\circ}$ dan $q=(\cos465^{\circ}+\cos165^{\circ})$$(\cos75^{\circ}+\sin105^{\circ}),$ nilai $3p-2p=…$
Betul$P=\cos20^{\circ}-\left(\cos40^{\circ}+\cos80^{\circ}\right)$
$=\cos20^{\circ}-2\cos60^{\circ}\cos20^{\circ}$
$=0$
$Q=\left(\cos465^{\circ}+\cos75^{\circ}\right)$$\left(\cos75^{\circ}+\sin105^{\circ}\right)$
$=2\cos270^{\circ}\cos105$$\left(\cos75^{\circ}+\sin105^{\circ}\right)$
$=0$
Maka $3p-2q=0$
Salah$P=\cos20^{\circ}-\left(\cos40^{\circ}+\cos80^{\circ}\right)$
$=\cos20^{\circ}-2\cos60^{\circ}\cos20^{\circ}$
$=0$
$Q=\left(\cos465^{\circ}+\cos75^{\circ}\right)$$\left(\cos75^{\circ}+\sin105^{\circ}\right)$
$=2\cos270^{\circ}\cos105$$\left(\cos75^{\circ}+\sin105^{\circ}\right)$
$=0$
Maka $3p-2q=0$
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Pertanyaan ke 4 dari 5
4. Pertanyaan
Jika $A=\sin22,5^{\circ}+\cos67,5^{\circ}$ dan $B=\sin67,5^{\circ}+\cos22,5^{\circ},$ nilai $\frac{A+B}{\cos22,5^{\circ}}=…$
Betul$A=\sin22,5^{\circ}+\cos22,5^{\circ}$$+\sin67,5^{\circ}+\cos67,5^{\circ}$
$=2\sin45^{\circ}\sin22,5^{\circ}$$+2\cos45^{\circ}\cos22,5^{\circ}$
$=2\sqrt{2}\cos22,5^{\circ}$
$\Rightarrow\frac{A+B}{\cos22,5^{\circ}}=2\sqrt{2}$
Salah$A=\sin22,5^{\circ}+\cos22,5^{\circ}$$+\sin67,5^{\circ}+\cos67,5^{\circ}$
$=2\sin45^{\circ}\sin22,5^{\circ}$$+2\cos45^{\circ}\cos22,5^{\circ}$
$=2\sqrt{2}\cos22,5^{\circ}$
$\Rightarrow\frac{A+B}{\cos22,5^{\circ}}=2\sqrt{2}$
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Pertanyaan ke 5 dari 5
5. Pertanyaan
Jika $\alpha-\beta=\frac{\pi}{7}$ dan $\alpha+\beta=2\pi,$ berapakah $\cos\alpha-\cos\beta$ ?
Betul$\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$
$=-2\sin\pi\sin\frac{\pi}{14}$
$=0$
Salah$\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$
$=-2\sin\pi\sin\frac{\pi}{14}$
$=0$
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Pertanyaan ke 1 dari 5
1. Pertanyaan
Nilai dari $\cos1^{\circ}+\cos3^{\circ}$$+…+\cos181^{\circ}=…$
BetulKarena $\sin\left(x+1\right)^{\circ}-\sin\left(x-1\right)^{\circ}$$=2\cos x^{\circ}\sin1^{\circ},$ maka $\cos x^{\circ}=\frac{\sin\left(x+1\right)^{\circ}-\sin\left(x-1\right)^{\circ}}{2\sin1^{\circ}}$
Jadi
$\cos1^{\circ}+\cos3+…+\cos181^{\circ}$$=\frac{\left(\sin2^{\circ}-\sin0^{\circ}\right)+\left(\sin4^{\circ}-\sin2^{\circ}\right)+…+\left(\sin182^{\circ}-\sin180^{\circ}\right)}{2\sin1^{\circ}}$
$=\frac{\sin182^{\circ}-\sin0^{\circ}}{2\sin1^{\circ}}$$=\frac{-\sin22^{\circ}}{2\sin1^{\circ}}$
$=-\cos1^{\circ}$
SalahKarena $\sin\left(x+1\right)^{\circ}-\sin\left(x-1\right)^{\circ}$$=2\cos x^{\circ}\sin1^{\circ},$ maka $\cos x^{\circ}=\frac{\sin\left(x+1\right)^{\circ}-\sin\left(x-1\right)^{\circ}}{2\sin1^{\circ}}$
Jadi
$\cos1^{\circ}+\cos3+…+\cos181^{\circ}$$=\frac{\left(\sin2^{\circ}-\sin0^{\circ}\right)+\left(\sin4^{\circ}-\sin2^{\circ}\right)+…+\left(\sin182^{\circ}-\sin180^{\circ}\right)}{2\sin1^{\circ}}$
$=\frac{\sin182^{\circ}-\sin0^{\circ}}{2\sin1^{\circ}}$$=\frac{-\sin22^{\circ}}{2\sin1^{\circ}}$
$=-\cos1^{\circ}$
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Pertanyaan ke 2 dari 5
2. Pertanyaan
Nilai dari $\cos20^{\circ}+\cos70^{\circ}+\frac{1}{2}\sqrt{3}+\cos50^{\circ}$ adalah…
BetulKarena
$\cos\alpha+\cos3\alpha+\cos5\alpha+\cos7\alpha$$=2\cos4\alpha\cos\alpha+2\cos4\alpha\cos3\alpha$
$=2\cos4\alpha\left(\cos3\alpha+\cos\alpha\right)$
$=4\cos4\alpha\cos2\alpha+\cos\alpha$
$=4\cdot\frac{\sin8\alpha}{2\sin4\alpha}\cdot\frac{\sin4\alpha}{2\sin\alpha}\cdot\frac{\sin2\alpha}{2\sin\alpha}$
$=\frac{1}{2}\frac{\sin8\alpha}{\sin\alpha}$
Untuk $\alpha=20^{\circ},$ jawaban soal $=\frac{1}{2}\times\frac{\sin160^{\circ}}{\sin20^{\circ}}$$=\frac{1}{2}$
SalahKarena
$\cos\alpha+\cos3\alpha+\cos5\alpha+\cos7\alpha$$=2\cos4\alpha\cos\alpha+2\cos4\alpha\cos3\alpha$
$=2\cos4\alpha\left(\cos3\alpha+\cos\alpha\right)$
$=4\cos4\alpha\cos2\alpha+\cos\alpha$
$=4\cdot\frac{\sin8\alpha}{2\sin4\alpha}\cdot\frac{\sin4\alpha}{2\sin\alpha}\cdot\frac{\sin2\alpha}{2\sin\alpha}$
$=\frac{1}{2}\frac{\sin8\alpha}{\sin\alpha}$
Untuk $\alpha=20^{\circ},$ jawaban soal $=\frac{1}{2}\times\frac{\sin160^{\circ}}{\sin20^{\circ}}$$=\frac{1}{2}$
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Pertanyaan ke 3 dari 5
3. Pertanyaan
Jika $\alpha,\beta$ di kuadran $1$ , $\sin\alpha-\sin\beta=\frac{3}{5}$ dan $\cos\alpha+\cos\beta=\frac{4}{5},$ maka $\sin(\frac{\alpha+\beta}{2})=…$
Betul$\frac{3}{4}=\frac{\sin\alpha-\sin\beta}{\cos\alpha+\cos\beta}$
$=\frac{2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)}{2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)}$
$=\tan\left(\frac{\alpha-\beta}{2}\right)$
$\Rightarrow\sin\left(\frac{\alpha-\beta}{2}\right)=\frac{3}{5}$
Jadi,
$\frac{3}{5}=\sin\alpha-\sin\beta$
$=2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$
$=\frac{6}{5}\cos\left(\frac{\alpha+\beta}{2}\right)$
$\Rightarrow\cos\left(\frac{\alpha+\beta}{2}\right)=\frac{1}{2}$
$\Rightarrow\cos\left(\alpha+\beta\right)=2\left(\frac{1}{2}\right)^{2}-1=-\frac{1}{2}$
$\Rightarrow\sin\left(\frac{\alpha+\beta}{2}\right)$$=\frac{1}{2}\sqrt{3}$
Salah$\frac{3}{4}=\frac{\sin\alpha-\sin\beta}{\cos\alpha+\cos\beta}$
$=\frac{2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)}{2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)}$
$=\tan\left(\frac{\alpha-\beta}{2}\right)$
$\Rightarrow\sin\left(\frac{\alpha-\beta}{2}\right)=\frac{3}{5}$
Jadi,
$\frac{3}{5}=\sin\alpha-\sin\beta$
$=2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$
$=\frac{6}{5}\cos\left(\frac{\alpha+\beta}{2}\right)$
$\Rightarrow\cos\left(\frac{\alpha+\beta}{2}\right)=\frac{1}{2}$
$\Rightarrow\cos\left(\alpha+\beta\right)=2\left(\frac{1}{2}\right)^{2}-1=-\frac{1}{2}$
$\Rightarrow\sin\left(\frac{\alpha+\beta}{2}\right)$$=\frac{1}{2}\sqrt{3}$
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Pertanyaan ke 4 dari 5
4. Pertanyaan
Nilai dari $\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}=…$
BetulMisal $x=\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$$=\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$,
karena $\sin\left(x+\frac{2\pi}{7}\right)-\sin x=2\cos\left(x+\frac{\pi}{7}\right)\sin\frac{\pi}{7},$
maka $\cos\left(x+\frac{\pi}{7}\right)=\frac{\sin\left(x+\frac{2\pi}{7}\right)-\sin x}{2\sin\frac{\pi}{7}}$
untuk $x=0,\frac{2\pi}{7},\,\frac{4\pi}{7}$
$x=\frac{\sin\frac{2\pi}{7}-\sin0}{2\sin\frac{\pi}{7}}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2\sin\frac{\pi}{7}}$$+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2\sin\frac{\pi}{7}}$$=\frac{\sin\frac{\pi}{7}}{2\sin\frac{\pi}{7}}=\frac{1}{2}$
SalahMisal $x=\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$$=\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$,
karena $\sin\left(x+\frac{2\pi}{7}\right)-\sin x=2\cos\left(x+\frac{\pi}{7}\right)\sin\frac{\pi}{7},$
maka $\cos\left(x+\frac{\pi}{7}\right)=\frac{\sin\left(x+\frac{2\pi}{7}\right)-\sin x}{2\sin\frac{\pi}{7}}$
untuk $x=0,\frac{2\pi}{7},\,\frac{4\pi}{7}$
$x=\frac{\sin\frac{2\pi}{7}-\sin0}{2\sin\frac{\pi}{7}}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2\sin\frac{\pi}{7}}$$+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2\sin\frac{\pi}{7}}$$=\frac{\sin\frac{\pi}{7}}{2\sin\frac{\pi}{7}}=\frac{1}{2}$
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Pertanyaan ke 5 dari 5
5. Pertanyaan
Nilai $k,\,0 < k < 90,$ sehingga $\frac{\cos3^{\circ}-\sin6^{\circ}-\cos9^{\circ}}{\sin9^{\circ}-\cos6^{\circ}-\sin3^{\circ}}=\cot k^{\circ}$ adalah …
Betul$\frac{\cos3^{\circ}-\sin6^{\circ}-\cos9^{\circ}}{\sin9^{\circ}-\cos6^{\circ}-\sin3^{\circ}}=\frac{-2\sin6^{\circ}\sin\left(-3^{\circ}\right)-\sin6{}^{\circ}}{2\cos6^{\circ}\sin3^{\circ}-\cos6^{\circ}}$
$=\frac{\sin6^{\circ}\left(2\sin3^{\circ}-1\right)}{\cos6^{\circ}\left(2\sin3^{\circ}-1\right)}$
$=\tan6^{\circ}$
$=\cot84^{\circ}$
$\Rightarrow k=84$
Salah$\frac{\cos3^{\circ}-\sin6^{\circ}-\cos9^{\circ}}{\sin9^{\circ}-\cos6^{\circ}-\sin3^{\circ}}=\frac{-2\sin6^{\circ}\sin\left(-3^{\circ}\right)-\sin6{}^{\circ}}{2\cos6^{\circ}\sin3^{\circ}-\cos6^{\circ}}$
$=\frac{\sin6^{\circ}\left(2\sin3^{\circ}-1\right)}{\cos6^{\circ}\left(2\sin3^{\circ}-1\right)}$
$=\tan6^{\circ}$
$=\cot84^{\circ}$
$\Rightarrow k=84$