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Rumus Perkalian Sinus dan Kosinus

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Rumus-Rumus untuk $2\sin a\cos b$ dan $2\cos a\sin b$

Rumus $2\sin a\cos b$

Perhatikan kembali rumus untuk $\sin \left( {a \pm b} \right)$. Jika rumus $\sin \left( {a + b} \right)$ dan $\sin \left( {a - b} \right)$ dijumlahkan maka diperoleh:

$\begin{array}{l} \sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\\ \underline {\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b} \left( + \right)\\ \sin \left( {a + b} \right) + \sin \left( {a - b} \right) = 2\sin a\cos b \end{array}$

Jadi, diperoleh:

\[\boxed{2\sin a\cos b = \sin \left( {a + b} \right) + \sin \left( {a - b} \right)}\]

Rumus $2\cos a\sin b$

Perhatikan kembali rumus untuk $\sin \left( {a \pm b} \right)$. Jika rumus $\sin \left( {a + b} \right)$ dan $\sin \left( {a - b} \right)$ dikurangkan maka diperoleh:

$\begin{array}{l} \sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\\ \underline {\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b} \left( - \right)\\ \sin \left( {a + b} \right) + \sin \left( {a - b} \right) = 2\cos a\sin b \end{array}$

Jadi, diperoleh:

\[\boxed{2\cos a\sin b = \sin \left( {a + b} \right) - \sin \left( {a - b} \right)}\]

Contoh 1

Nyatakan bentuk-bentuk berikut ini sebagai jumlah atau selisih sinus

  1. $2\sin 2a\cos b$
  2. $2\sin {40^ \circ }\cos {9^ \circ }$
  3. $2\sin x\cos y$
  4. $4\cos 2a\sin 3b$
  5. $5\sin {72^ \circ }\cos {26^ \circ }$

Jawab

  1. $2\sin 2a\cos b = \sin \left( {2a + b} \right) + \sin \left( {2a - b} \right)$
  2. $\begin{array}{l} 2\sin {40^ \circ }\cos {9^ \circ } &= \sin \left( {{{40}^ \circ } + {9^ \circ }} \right) + \sin \left( {{{40}^ \circ } - {9^ \circ }} \right)\\ &= \sin \left( {{{49}^ \circ }} \right) + \sin \left( {{{31}^ \circ }} \right)\\ &= \sin {49^ \circ } + \sin {31^ \circ } \end{array}$
  3. $2\sin x\cos y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right)$
  4. $\begin{array}{l} 4\cos 2a\sin 3b &= 2\left( {2\cos 2a\sin 3b} \right)\\ &= 2\left[ {\sin \left( {2a + 3b} \right) - \sin \left( {2a - 3b} \right)} \right]\\ &= 2\sin \left( {2a + 3b} \right) - 2\sin \left( {2a - 3b} \right) \end{array}$
  5. $\begin{array}{l} 5\sin {72^\circ }\cos {26^\circ } &= \frac{5}{2}\left( {2\sin {{72}^\circ }\cos {{26}^\circ }} \right)\\ &= \frac{5}{2}\left[ {\sin \left( {{{72}^\circ } + {{26}^\circ }} \right) + \sin \left( {{{72}^\circ } - {{26}^\circ }} \right)} \right]\\ &= \frac{5}{2}\left[ {\sin \left( {{{98}^\circ }} \right) + \sin \left( {{{46}^\circ }} \right)} \right]\\ &= \frac{5}{2}\sin {98^\circ } - \frac{5}{2}\sin {46^\circ } \end{array}$

Contoh 2

Tanpa menggunakan tabel trigonometri atau kalkulator, hitunglah nilai eksak dari:

  1. $2\sin 37{\frac{1}{2}^ \circ }\cos 7{\frac{1}{2}^ \circ }$
  2. $2\cos 37{\frac{1}{2}^ \circ }\sin 7{\frac{1}{2}^ \circ }$

Jawab

  1. $\begin{array}{l} 2\sin 37{\frac{1}{2}^ \circ }\cos 7{\frac{1}{2}^ \circ } &= \sin \left( {37{{\frac{1}{2}}^ \circ } + 7{{\frac{1}{2}}^ \circ }} \right) + \sin \left( {37{{\frac{1}{2}}^ \circ } - 7{{\frac{1}{2}}^ \circ }} \right)\\ &= \sin {45^ \circ } + \sin {30^ \circ }\\ &= \frac{1}{2}\sqrt 2 + \frac{1}{2}\\ &= \frac{1}{2}\left( {\sqrt 2 + 1} \right) \end{array}$
  2. $\begin{array}{l} 2\cos 37{\frac{1}{2}^ \circ }\sin 7{\frac{1}{2}^ \circ } &= \sin \left( {37{{\frac{1}{2}}^ \circ } + 7{{\frac{1}{2}}^ \circ }} \right) - \sin \left( {37{{\frac{1}{2}}^ \circ } - 7{{\frac{1}{2}}^ \circ }} \right)\\ &= \sin {45^ \circ } - \sin {30^ \circ }\\ &= \frac{1}{2}\sqrt 2 - \frac{1}{2}\\ &= \frac{1}{2}\left( {\sqrt 2 - 1} \right) \end{array}$

Rumus-Rumus untuk $2\sin a\sin b$ dan $2\cos a\cos b$

Rumus untuk $2\sin a\sin b$

Perhatikan kembali rumus untuk $\cos \left( {a \pm b} \right)$. Jika rumus $\cos \left( {a + b} \right)$ dan $\cos \left( {a - b} \right)$ dijumlahkan maka diperoleh:

$\begin{array}{l} \cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\\ \underline {\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b} \left( + \right)\\ \cos \left( {a + b} \right) + \cos \left( {a - b} \right) = 2\cos a\cos b \end{array}$

Jadi, diperoleh:

\[\boxed{2\cos a\cos b = \cos \left( {a + b} \right) + \cos \left( {a - b} \right)}\]

Rumus untuk $2\cos a\cos b$

Perhatikan kembali rumus untuk $\cos \left( {a \pm b} \right)$. Jika rumus $\cos \left( {a + b} \right)$ dan $\cos \left( {a - b} \right)$ dikurangkan maka diperoleh:

$\begin{array}{l} \cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\\ \underline {\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b} \left( - \right)\\ \cos \left( {a + b} \right) - \cos \left( {a - b} \right) = -2\sin a\sin b \end{array}$

Jadi, diperoleh:

\[\boxed{2\sin a\sin b = \cos \left( {a - b} \right) - \cos \left( {a + b} \right)}\]

Contoh 3

Nyatakan bentuk-bentuk berikut ini sebagai jumlah atau selisih sinus

  1. $2\cos 2a\cos b$
  2. $2\cos {40^ \circ }\cos {9^ \circ }$
  3. $2\cos x\cos y$
  4. $4\sin 2a\sin 3b$
  5. $5\sin {72^ \circ }\sin {26^ \circ }$

Jawab

  1. $2\cos 2a\cos b = \cos \left( {2a + b} \right) + \cos \left( {2a - b} \right)$
  2. $\begin{array}{l} 2\cos {40^ \circ }\cos {9^ \circ } &= \cos \left( {{{40}^ \circ } + {9^ \circ }} \right) + \cos \left( {{{40}^ \circ } - {9^ \circ }} \right)\\ &= \cos \left( {{{49}^ \circ }} \right) + \cos \left( {{{31}^ \circ }} \right)\\ &= \cos {49^ \circ } + \cos {31^ \circ } \end{array}$
  3. $2\cos x\cos y = \cos \left( {x + y} \right) + \cos \left( {x - y} \right)$
  4. $\begin{array}{l} 4\sin 2a\sin 3b &= 2\left( {2\sin 2a\sin 3b} \right)\\ &= 2\left[ {\cos \left( {2a - 3b} \right) - \cos \left( {2a + 3b} \right)} \right]\\ &= 2\cos \left( {2a - 3b} \right) - 2\cos \left( {2a + 3b} \right) \end{array}$
  5. $\begin{array}{l} 5\sin {72^\circ }\sin {26^\circ } &= \frac{5}{2}\left( {2\sin {{72}^\circ }\sin {{26}^\circ }} \right)\\ &= \frac{5}{2}\left[ {\cos \left( {{{72}^\circ } - {{26}^\circ }} \right) - \cos \left( {{{72}^\circ } + {{26}^\circ }} \right)} \right]\\ &= \frac{5}{2}\left[ {\cos \left( {{{46}^\circ }} \right) - \cos \left( {{{98}^\circ }} \right)} \right]\\ &= \frac{5}{2}\cos {46^\circ } - \frac{5}{2}\cos {98^\circ } \end{array}$

Contoh 4

Tanpa menggunakan tabel trigonometri atau kalkulator, hitunglah nilai eksak dari:

  1. $2\cos 37{\frac{1}{2}^ \circ }\cos 7{\frac{1}{2}^ \circ }$
  2. $2\sin 37{\frac{1}{2}^ \circ }\sin 7{\frac{1}{2}^ \circ }$

Jawab

  1. $\begin{array}{l} 2\cos 37{\frac{1}{2}^ \circ }\cos 7{\frac{1}{2}^ \circ } &= \cos \left( {37{{\frac{1}{2}}^ \circ } + 7{{\frac{1}{2}}^ \circ }} \right) + \cos \left( {37{{\frac{1}{2}}^ \circ } - 7{{\frac{1}{2}}^ \circ }} \right)\\ &= \cos {45^ \circ } + \cos {30^ \circ }\\ &= \frac{1}{2}\sqrt 2 + \frac{1}{2}\sqrt 3\\ &= \frac{1}{2}\left( {\sqrt 2 + \sqrt 3} \right) \end{array}$
  2. $\begin{array}{l} 2\sin 37{\frac{1}{2}^ \circ }\sin 7{\frac{1}{2}^ \circ } &= \cos \left( {37{{\frac{1}{2}}^ \circ } - 7{{\frac{1}{2}}^ \circ }} \right) - \cos \left( {37{{\frac{1}{2}}^ \circ } + 7{{\frac{1}{2}}^ \circ }} \right)\\ &= \cos {30^ \circ } - \cos {45^ \circ }\\ &= \frac{1}{2}\sqrt 3 - \frac{1}{2}\sqrt 2\\ &= \frac{1}{2}\left( {\sqrt 3 - \sqrt 2} \right) \end{array}$
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nurhamim86
nurhamim86 A Mathematics Teacher who also likes the IT world.

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