Rumus Jumlah dan Selisih pada Sinus dan Kosinus

Rumus $\sin A + \sin B$

Perhatikan kembali rumus $2\sin a\cos b$, yaitu:

$2\sin a\cos b = \sin \left( {a + b} \right) + \sin \left( {a - b} \right)$

Jika kita misalkan $a + b = A$ dan $a - b = B$ maka diperoleh.

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

Dengan elimininasi substitusi $a + b = A$ dan $a - b = B$, diperoleh nilai $a$ dan $b$ sebagai berikut:

$\begin{array}{*{20}{c}} \begin{array}{l} a + b = A\\ \underline {a - b = B} \left( + \right)\\ 2a = A + B\\ a = \frac{1}{2}\left( {A + B} \right) \end{array}&\begin{array}{l} a + b = A\\ \underline {a - b = B} \left( - \right)\\ 2b = A - B\\ b = \frac{1}{2}\left( {A - B} \right) \end{array} \end{array}$

Dengan substitusi nilai $a = \frac{1}{2}\left( {A + B} \right)$ dan $b = \frac{1}{2}\left( {A - B} \right)$ ke persamaan sebelumnya diperoleh:

$\begin{array}{l} \sin A + \sin B &= 2\sin a\cos b\\ &= 2\sin \frac{1}{2}\left( {A + B} \right)\cos \frac{1}{2}\left( {A - B} \right) \end{array}$

Jadi, diperoleh:

\[\boxed{\sin A + \sin B = 2\sin \frac{1}{2}\left( {A + B} \right)\cos \frac{1}{2}\left( {A - B} \right)}\]

Rumus $\sin A - \sin B$

Perhatikan kembali rumus $2\cos a\sin b$, yaitu:

$2\cos a\sin b = \sin \left( {a + b} \right) - \sin \left( {a - b} \right)$

Jika kita misalkan $a + b = A$ dan $a - b = B$ maka diperoleh.

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

Dengan elimininasi substitusi $a + b = A$ dan $a - b = B$, diperoleh nilai $a$ dan $b$ sebagai berikut:

$\begin{array}{*{20}{c}} \begin{array}{l} a + b = A\\ \underline {a - b = B} \left( + \right)\\ 2a = A + B\\ a = \frac{1}{2}\left( {A + B} \right) \end{array}&\begin{array}{l} a + b = A\\ \underline {a - b = B} \left( - \right)\\ 2b = A - B\\ b = \frac{1}{2}\left( {A - B} \right) \end{array} \end{array}$

Dengan substitusi nilai $a = \frac{1}{2}\left( {A + B} \right)$ dan $b = \frac{1}{2}\left( {A - B} \right)$ ke persamaan sebelumnya diperoleh:

$\begin{array}{l} \sin A - \sin B &= 2\cos a\sin b\\ &= 2\cos \frac{1}{2}\left( {A + B} \right)\sin \frac{1}{2}\left( {A - B} \right) \end{array}$

Jadi, diperoleh:

\[\boxed{\sin A - \sin B = 2\cos \frac{1}{2}\left( {A + B} \right)\sin \frac{1}{2}\left( {A - B} \right)}\]

Rumus $\cos A + \cos B$

Perhatikan kembali rumus $2\cos a\cos b$, yaitu:

$2\cos a\cos b = \cos \left( {a + b} \right) + \cos \left( {a - b} \right)$

Jika kita misalkan $a + b = A$ dan $a - b = B$ maka diperoleh.

$\begin{array}{l} 2\cos a\cos b &= \cos \left( {a + b} \right) + \cos \left( {a - b} \right)\\ \Leftrightarrow \cos \left( {a + b} \right) + \cos \left( {a - b} \right) &= 2\cos a\cos b\\ \Leftrightarrow \cos A + \cos B &= 2\cos a\cos b \end{array}$

Dengan elimininasi substitusi $a + b = A$ dan $a - b = B$, diperoleh nilai $a$ dan $b$ sebagai berikut:

$\begin{array}{*{20}{c}} \begin{array}{l} a + b = A\\ \underline {a - b = B} \left( + \right)\\ 2a = A + B\\ a = \frac{1}{2}\left( {A + B} \right) \end{array}&\begin{array}{l} a + b = A\\ \underline {a - b = B} \left( - \right)\\ 2b = A - B\\ b = \frac{1}{2}\left( {A - B} \right) \end{array} \end{array}$

Dengan substitusi nilai $a = \frac{1}{2}\left( {A + B} \right)$ dan $b = \frac{1}{2}\left( {A - B} \right)$ ke persamaan sebelumnya diperoleh:

$\begin{array}{l} \cos A + \cos B &= 2\cos a\cos b\\ &= 2\cos \frac{1}{2}\left( {A + B} \right)\cos \frac{1}{2}\left( {A - B} \right) \end{array}$

Jadi, diperoleh:

\[\boxed{\cos A + \cos B = 2\cos \frac{1}{2}\left( {A + B} \right)\cos \frac{1}{2}\left( {A - B} \right)}\]

Rumus $\cos A - \cos B$

Perhatikan kembali rumus $2\sin a\sin b$, yaitu:

$2\sin a\sin b = \cos \left( {a - b} \right) - \cos \left( {a - b} \right)$

Jika kita misalkan $a + b = A$ dan $a - b = B$ maka diperoleh.

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

Dengan elimininasi substitusi $a + b = A$ dan $a - b = B$, diperoleh nilai $a$ dan $b$ sebagai berikut:

$\begin{array}{*{20}{c}} \begin{array}{l} a + b = A\\ \underline {a - b = B} \left( + \right)\\ 2a = A + B\\ a = \frac{1}{2}\left( {A + B} \right) \end{array}&\begin{array}{l} a + b = A\\ \underline {a - b = B} \left( - \right)\\ 2b = A - B\\ b = \frac{1}{2}\left( {A - B} \right) \end{array} \end{array}$

Dengan substitusi nilai $a = \frac{1}{2}\left( {A + B} \right)$ dan $b = \frac{1}{2}\left( {A - B} \right)$ ke persamaan sebelumnya diperoleh:

$\begin{array}{l} \sin B - \sin A &= 2\sin a\sin b\\ &= 2\cos \frac{1}{2}\left( {A + B} \right)\sin \frac{1}{2}\left( {A + B} \right)\\ \sin A - \sin B &= - 2\cos \frac{1}{2}\left( {A + B} \right)\sin \frac{1}{2}\left( {A + B} \right) \end{array}$

Jadi, diperoleh:

\[\boxed{\cos A - \cos B = - 2\sin \frac{1}{2}\left( {A + B} \right)\sin \frac{1}{2}\left( {A - B} \right)}\]

Contoh 1

Hitunglah nilai eksak dari:

1. $\sin {75^ \circ } + \sin {15^ \circ }$
2. $\cos {75^ \circ } - \cos {15^ \circ }$

Jawab

$\begin{array}{l} \sin {75^ \circ } + \sin {15^ \circ } &= 2\sin \frac{1}{2}\left( {{{75}^ \circ } + {{15}^ \circ }} \right)\cos \frac{1}{2}\left( {{{75}^ \circ } - {{15}^ \circ }} \right)\\ &= 2\sin \frac{1}{2}\left( {{{90}^ \circ }} \right)\cos \frac{1}{2}\left( {{{60}^ \circ }} \right)\\ &= 2\sin \left( {{{45}^ \circ }} \right)\cos \left( {{{30}^ \circ }} \right)\\ &= 2.\frac{1}{2}\sqrt 2 .\frac{1}{2}\sqrt 3 \\ &= \frac{1}{2}\sqrt 6 \end{array}$
$\begin{array}{l} \cos {75^ \circ } - \cos {15^ \circ } &= - 2\sin \frac{1}{2}\left( {{{75}^ \circ } + {{15}^ \circ }} \right)\sin \frac{1}{2}\left( {{{75}^ \circ } - {{15}^ \circ }} \right)\\ &= - 2\sin \frac{1}{2}\left( {{{90}^ \circ }} \right)\sin \frac{1}{2}\left( {{{60}^ \circ }} \right)\\ &= - 2\sin \left( {{{45}^ \circ }} \right)\sin \left( {{{30}^ \circ }} \right)\\ &= - 2.\frac{1}{2}\sqrt 2 .\frac{1}{2}\\ &= - \frac{1}{2}\sqrt 2 \end{array}$

Contoh 2

Buktikan bahwa $\frac{{\cos 2x - \cos 4x}}{{\sin 2x\sin 3x}} = \sec x$

Jawab

$\begin{array}{l} \frac{{\cos 2x - \cos 4x}}{{\sin 2x\sin 3x}} &= \frac{{ - 2\sin \frac{1}{2}\left( {2x + 4x} \right)\sin \frac{1}{2}\left( {2x - 4x} \right)}}{{\sin 2x\sin 3x}}\\ &= \frac{{ - 2\sin 3x\sin \left( { - x} \right)}}{{\sin 2x\sin 3x}}\\ &= \frac{{2\sin 3x\sin x}}{{\sin 2x\sin 3x}}\\ &= \frac{{\bcancel{2}\bcancel{{\sin 3x}}\bcancel{{\sin x}}}}{{\bcancel{2}\bcancel{{\sin x}}\cos x\bcancel{{\sin 3x}}}}\\ &= \frac{1}{{\cos x}}\\ &= \sec x \end{array}$

Contoh 3

Jika diketahui $A + B + C = {180^ \circ }$, buktikan bahwa:

$\sin A + \sin B + \sin C = 4\cos \frac{1}{2}A\cos \frac{1}{2}B\cos \frac{1}{2}C$

Jawab

$\begin{array}{l} A + B + C = {180^ \circ }\\ \sin A + \sin B + \sin C &= \sin A + \sin B + \sin \left( {{{180}^ \circ } - \left( {A + B} \right)} \right)\\ &= \sin A + \sin B + \sin {180^ \circ }\cos \left( {A + B} \right) - \cos {180^ \circ }\sin \left( {A + B} \right)\\ &= \sin A + \sin B + 0.\cos \left( {A + B} \right) - \left( { - 1} \right).\sin \left( {A + B} \right)\\ &= \sin A + \sin B + \sin \left( {A + B} \right)\\ &= \sin A + \sin B + \sin A\cos B + \cos A\sin B\\ &= \sin A + \sin A\cos B + \sin B + \cos A\sin B\\ &= \sin A\left( {1 + \cos B} \right) + \sin B\left( {1 + \cos A} \right)\\ &= \left( {2\sin \frac{1}{2}A\cos \frac{1}{2}A} \right)\left( {2{{\cos }^2}\frac{1}{2}B} \right) + \left( {2\sin \frac{1}{2}B\cos \frac{1}{2}B} \right)\left( {2{{\cos }^2}\frac{1}{2}A} \right)\\ &= 4\sin \frac{1}{2}A\cos \frac{1}{2}A{\cos ^2}\frac{1}{2}B + 4\sin \frac{1}{2}B\cos \frac{1}{2}B{\cos ^2}\frac{1}{2}A\\ &= 4\cos \frac{1}{2}A\cos \frac{1}{2}B\left( {\sin \frac{1}{2}A\cos \frac{1}{2}B + \cos \frac{1}{2}A\sin \frac{1}{2}B} \right)\\ &= 4\cos \frac{1}{2}A\cos \frac{1}{2}B\sin \left( {\frac{1}{2}A + \frac{1}{2}B} \right)\\ &= 4\cos \frac{1}{2}A\cos \frac{1}{2}B\sin \frac{1}{2}\left( {A + B} \right)\\ &= 4\cos \frac{1}{2}A\cos \frac{1}{2}B\sin \frac{1}{2}\left( {{{180}^ \circ } - C} \right)\\ &= 4\cos \frac{1}{2}A\cos \frac{1}{2}B\sin \left( {{{90}^ \circ } - \frac{1}{2}C} \right)\\ &= 4\cos \frac{1}{2}A\cos \frac{1}{2}B\sin \frac{1}{2}C\left( {terbukti} \right) \end{array}$