Rumus Perkalian Trigonometri untuk Sinus dan Cosinus
Misalkan diketahui dua sudut yaitu A dan B, berikut rumus perkalian antara sinus dan cosinus pada sudut A dan B :
$ \begin{align} \sin A \cos B & = \frac{1}{2}[ \sin (A+B) + \sin (A- B) ] \\ \cos A \sin B & = \frac{1}{2}[ \sin (A+B) - \sin (A- B) ] \\ \cos A \cos B & = \frac{1}{2}[ \cos (A+B) + \cos (A- B) ] \\ \sin A \sin B & = - \frac{1}{2}[ \cos (A+B) - \cos (A- B) ] \end{align} $
$ \begin{align} \sin A \cos B & = \frac{1}{2}[ \sin (A+B) + \sin (A- B) ] \\ \cos A \sin B & = \frac{1}{2}[ \sin (A+B) - \sin (A- B) ] \\ \cos A \cos B & = \frac{1}{2}[ \cos (A+B) + \cos (A- B) ] \\ \sin A \sin B & = - \frac{1}{2}[ \cos (A+B) - \cos (A- B) ] \end{align} $
*). Kita menggunakan rumus jumlah dan selisih sudut, yaitu :
$ \begin{align} \sin (A + B) & = \sin A \cos B + \cos A \sin B \\ \sin (A - B) & = \sin A \cos B - \cos A \sin B \\ \cos (A+B) & = \cos A \cos B - \sin A \sin B \\ \cos (A-B) & = \cos A \cos B + \sin A \sin B \\ \end{align} $
$\clubsuit $ Pembuktian Rumus : $ \sin A \cos B = \frac{1}{2}[ \sin (A+B) + \sin (A- B) ] $
$ \begin{array}{cc} \sin (A + B) = \sin A \cos B + \cos A \sin B & \\ \sin (A - B) = \sin A \cos B - \cos A \sin B & + \\ \hline \sin (A + B) + \sin (A - B ) = 2 \sin A \cos B & \end{array} $
Sehingg terbukti : $ \sin A \cos B = \frac{1}{2}[ \sin (A + B) + \sin (A - B ) ] $
$\clubsuit $ Pembuktian Rumus : $ \cos A \sin B = \frac{1}{2}[ \sin (A+B) - \sin (A- B) ] $
$ \begin{array}{cc} \sin (A + B) = \sin A \cos B + \cos A \sin B & \\ \sin (A - B) = \sin A \cos B - \cos A \sin B & - \\ \hline \sin (A + B) - \sin (A - B ) = 2 \cos A \sin B & \end{array} $
Sehingg terbukti : $ \cos A \sin B = \frac{1}{2}[ \sin (A+B) - \sin (A- B) ] $
$\clubsuit $ Pembuktian Rumus : $ \cos A \cos B = \frac{1}{2}[ \cos (A+B) + \cos (A- B) ] $
$ \begin{array}{cc} \cos (A+B) = \cos A \cos B - \sin A \sin B & \\ \cos (A-B) = \cos A \cos B + \sin A \sin B & + \\ \hline \cos (A + B) + \cos (A - B ) = 2 \cos A \cos B & \end{array} $
Sehingg terbukti : $ \cos A \cos B = \frac{1}{2}[ \cos (A+B) + \cos (A- B) ] $
$\clubsuit $ Pembuktian Rumus : $ \sin A \sin B = -\frac{1}{2}[ \cos (A+B) - \cos (A- B) ] $
$ \begin{array}{cc} \cos (A+B) = \cos A \cos B - \sin A \sin B & \\ \cos (A-B) = \cos A \cos B + \sin A \sin B & - \\ \hline \cos (A + B) - \cos (A - B ) = -2 \sin A \sin B & \end{array} $
Sehingg terbukti : $ \sin A \sin B = -\frac{1}{2}[ \cos (A+B) - \cos (A- B) ] $
Contoh :
1). Tentukan nilai dari trigonometri berikut :
a). $ \sin 75^\circ \cos 15^\circ $
b). $ \cos 67\frac{1}{2}^\circ \sin 22\frac{1}{2}^\circ $
c). $ \cos 105^\circ \cos 15^\circ $
d). $ \sin 127\frac{1}{2}^\circ \sin 97\frac{1}{2}^\circ $
Penyelesaian :
a). Gunakan rumus : $ \sin A \cos B = \frac{1}{2}[ \sin (A+B) + \sin (A- B) ] $
dengan besar sudut $ A = 75^\circ \, $ dan $ B = 15^\circ $
$ \begin{align} \sin A \cos B & = \frac{1}{2}[ \sin (A+B) + \sin (A- B) ] \\ \sin 75^\circ \cos 15^\circ & = \frac{1}{2}[ \sin (75^\circ +15^\circ ) + \sin (75^\circ - 15^\circ ) ] \\ & = \frac{1}{2}[ \sin (90^\circ ) + \sin (60^\circ ) ] \\ & = \frac{1}{2}[ 1 + \frac{1}{2}\sqrt{3} ] \\ & = \frac{1}{4}( 2 + \sqrt{3} ) \end{align} $
Jadi, nilai $ \sin 75^\circ \cos 15^\circ = \frac{1}{4}( 2 + \sqrt{3} ) $
b). Gunakan rumus : $ \cos A \sin B = \frac{1}{2}[ \sin (A+B) - \sin (A- B) ] $
dengan besar sudut $ A = 67\frac{1}{2}^\circ \, $ dan $ B = 22\frac{1}{2}^\circ $
$ \begin{align} \cos A \sin B & = \frac{1}{2}[ \sin (A+B) - \sin (A- B) ] \\ \cos 67\frac{1}{2}^\circ \sin 22\frac{1}{2}^\circ & = \frac{1}{2}[ \sin ( 67\frac{1}{2}^\circ + 22\frac{1}{2}^\circ ) - \sin (67\frac{1}{2}^\circ - 22\frac{1}{2}^\circ) ] \\ & = \frac{1}{2}[ \sin ( 90^\circ ) - \sin (45^\circ) ] \\ & = \frac{1}{2}[ 1 - \frac{1}{2} \sqrt{2} ] \\ & = \frac{1}{4}( 2 - \sqrt{2} ) \end{align} $
Jadi, nilai $ \cos 67\frac{1}{2}^\circ \sin 22\frac{1}{2}^\circ = \frac{1}{4}( 2 - \sqrt{2} ) $
c). Gunakan rumus : $ \cos A \cos B = \frac{1}{2}[ \cos (A+B) + \cos (A- B) ] $
dengan besar sudut $ A = 105^\circ \, $ dan $ B = 15^\circ $
$ \begin{align} \cos A \cos B & = \frac{1}{2}[ \cos (A+B) + \cos (A- B) ] \\ \cos 105^\circ \cos 15^\circ & = \frac{1}{2}[ \cos (105^\circ + 15^\circ ) + \cos (105^\circ - 15^\circ ) ] \\ & = \frac{1}{2}[ \cos (120^\circ ) + \cos (90^\circ ) ] \\ & = \frac{1}{2}[ - \cos (60^\circ ) + 0 ] \\ & = \frac{1}{2}[ - \frac{1}{2} + 0 ] \\ & = - \frac{1}{4} \end{align} $
Jadi, nilai $ \cos 105^\circ \cos 15^\circ = - \frac{1}{4} $
d). Gunakan rumus : $ \sin A \sin B = -\frac{1}{2}[ \cos (A+B) - \cos (A- B) ] $
dengan besar sudut $ A = 127\frac{1}{2}^\circ \, $ dan $ B = 97\frac{1}{2}^\circ $
$ \begin{align} \sin A \sin B & = -\frac{1}{2}[ \cos (A+B) - \cos (A- B) ] \\ \sin 127\frac{1}{2}^\circ \sin 97\frac{1}{2}^\circ & = -\frac{1}{2}[ \cos (127\frac{1}{2}^\circ + 97\frac{1}{2}^\circ) - \cos (127\frac{1}{2}^\circ - 97\frac{1}{2}^\circ ) ] \\ & = -\frac{1}{2}[ \cos (225^\circ) - \cos (30^\circ ) ] \\ & = -\frac{1}{2}[ \cos (180^\circ + 45^\circ) - \cos (30^\circ ) ] \\ & = -\frac{1}{2}[ -\cos (45^\circ) - \cos (30^\circ ) ] \\ & = -\frac{1}{2}[ -\frac{1}{2}\sqrt{2} - \frac{1}{2}\sqrt{3} ] \\ & = \frac{1}{4} ( \sqrt{2} + \sqrt{3} ) \end{align} $
Jadi, nilai $ \sin 127\frac{1}{2}^\circ \sin 97\frac{1}{2}^\circ = \frac{1}{4} ( \sqrt{2} + \sqrt{3} ) $
Rumus Trigonometri Penjumlahan dan Pengurangan
Misalkan diketahui dua sudut P dan Q, berlaku rumus penjumlahan dan pengurangannya :
$ \begin{align} \sin P + \sin Q & = 2 \sin \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) \\ \sin P - \sin Q & = 2 \cos \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) \\ \cos P + \cos Q & = 2 \cos \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) \\ \cos P - \cos Q & = -2 \sin \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) \\ \tan P + \tan Q & = \frac{2\sin(P+Q)}{\cos (P+Q) + \cos (P-Q) } \\ \tan P - \tan Q & = \frac{2\sin(P-Q)}{\cos (P+Q) + \cos (P-Q) } \end{align} $
$ \begin{align} \sin P + \sin Q & = 2 \sin \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) \\ \sin P - \sin Q & = 2 \cos \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) \\ \cos P + \cos Q & = 2 \cos \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) \\ \cos P - \cos Q & = -2 \sin \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) \\ \tan P + \tan Q & = \frac{2\sin(P+Q)}{\cos (P+Q) + \cos (P-Q) } \\ \tan P - \tan Q & = \frac{2\sin(P-Q)}{\cos (P+Q) + \cos (P-Q) } \end{align} $
*). Kita menggunakan rumus perkalian trigonometri sebelumnya.
*). Misalkan $ A + B = P \, $ dan $ A - B = Q $ , maka dengan eliminasi kedua persamaan kita peroleh : $ A = \frac{1}{2}(P+Q) \, $ dan $ A = \frac{1}{2}(P-Q) $
*). Substitusi bentuk permisalan di atas ke persamaan yang digunakan.
$\spadesuit $ Pembuktian Rumus : $ \sin P + \sin Q = 2 \sin \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) $
$ \begin{align} \sin A \cos B & = \frac{1}{2}[ \sin (A+B) + \sin (A- B) ] \\ \sin \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) & = \frac{1}{2}[ \sin P + \sin Q ] \\ 2\sin \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) & = \sin P + \sin Q \end{align} $
Sehingga tebukti rumus $ \sin P + \sin Q = 2 \sin \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) $
$\spadesuit $ Pembuktian Rumus : $ \sin P - \sin Q = 2 \cos \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) $
$ \begin{align} \cos A \sin B & = \frac{1}{2}[ \sin (A+B) - \sin (A- B) ] \\ \cos \frac{1}{2}(P+Q) \sin \frac{1}{2}(P - Q) & = \frac{1}{2}[ \sin P - \sin Q ] \\ 2 \cos \frac{1}{2}(P+Q) \sin \frac{1}{2}(P - Q) & = \sin P - \sin Q \end{align} $
Sehingga tebukti rumus $ \sin P - \sin Q = 2 \cos \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) $
$\spadesuit $ Pembuktian Rumus : $ \cos P + \cos Q = 2 \cos \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) $
$ \begin{align} \cos A \cos B & = \frac{1}{2}[ \cos (A+B) + \cos (A- B) ] \\ \cos \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) & = \frac{1}{2}[ \cos P + \cos Q ] \\ 2\cos \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) & = \cos P + \cos Q \end{align} $
Sehingga tebukti rumus $ \cos P + \cos Q = 2 \cos \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) $
$\spadesuit $ Pembuktian Rumus : $ \cos P - \cos Q = -2 \sin \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) $
$ \begin{align} \sin A \sin B & = -\frac{1}{2}[ \cos (A+B) - \cos (A- B) ] \\ \sin \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) & = -\frac{1}{2}[ \cos P - \cos Q ] \\ -2\sin \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) & = \cos P - \cos Q \end{align} $
Sehingga tebukti rumus $ \cos P - \cos Q = -2 \sin \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) $
$\spadesuit $ Pembuktian Rumus : $ \tan P + \tan Q = \frac{2\sin(P+Q)}{\cos (P+Q) + \cos (P-Q) } $
*). Gunakan rumus :
$ \sin (P+Q) = \sin P\cos Q + \cos P \sin Q \, $ dan $ 2 \cos P \cos Q = \cos (P+Q) + \cos (P-Q) $
$ \begin{align} \tan P + \tan Q & = \frac{\sin P}{\cos P} + \frac{\sin Q}{\cos Q} \\ & = \frac{\sin P\cos Q}{\cos P \cos Q} + \frac{\cos P \sin Q }{\cos P \cos Q} \\ & = \frac{\sin P\cos Q + \cos P \sin Q }{\cos P \cos Q} \\ & = \frac{\sin (P+Q) }{\cos P \cos Q} \\ & = \frac{2\sin (P+Q) }{2\cos P \cos Q} \\ & = \frac{2\sin (P+Q) }{\cos (P+Q) + \cos (P-Q)} \end{align} $
Sehingga tebukti rumus $ \tan P + \tan Q = \frac{2\sin(P+Q)}{\cos (P+Q) + \cos (P-Q) } $
$\spadesuit $ Pembuktian Rumus : $ \tan P - \tan Q = \frac{2\sin(P-Q)}{\cos (P+Q) + \cos (P-Q) } $
*). Gunakan rumus :
$ \sin (P-Q) = \sin P\cos Q - \cos P \sin Q \, $ dan $ 2 \cos P \cos Q = \cos (P+Q) + \cos (P-Q) $
$ \begin{align} \tan P - \tan Q & = \frac{\sin P}{\cos P} - \frac{\sin Q}{\cos Q} \\ & = \frac{\sin P\cos Q}{\cos P \cos Q} - \frac{\cos P \sin Q }{\cos P \cos Q} \\ & = \frac{\sin P\cos Q - \cos P \sin Q }{\cos P \cos Q} \\ & = \frac{\sin (P-Q) }{\cos P \cos Q} \\ & = \frac{2\sin (P-Q) }{2\cos P \cos Q} \\ & = \frac{2\sin (P-Q) }{\cos (P+Q) + \cos (P-Q)} \end{align} $
Sehingga tebukti rumus $ \tan P + \tan Q = \frac{2\sin(P-Q)}{\cos (P+Q) + \cos (P-Q) } $
Contoh :
2). Tentukan nilai dari :
a). $ \sin 105^\circ + \sin 15 ^\circ $
b). $ \sin 105^\circ - \sin 15 ^\circ $
c). $ \cos 105^\circ + \cos 15 ^\circ $
d). $ \tan 105^\circ + \tan 15 ^\circ $
Penyelesaian :
a). Nilai $ \sin 105^\circ + \sin 15 ^\circ $
$\begin{align} \sin P + \sin Q & = 2 \sin \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) \\ \sin 105^\circ + \sin 15 ^\circ & = 2 \sin \frac{1}{2}(105^\circ+ 15 ^\circ) \cos \frac{1}{2}(105^\circ-15 ^\circ) \\ & = 2 \sin (60 ^\circ) \cos (45 ^\circ) \\ & = 2 .\frac{1}{2}\sqrt{3} . \frac{1}{2}\sqrt{2} \\ & = \frac{1}{2}\sqrt{6} \end{align} $
Jadi, nilai $ \sin 105^\circ + \sin 15 ^\circ = \frac{1}{2}\sqrt{6} $
b). Nilai $ \sin 105^\circ - \sin 15 ^\circ $
$\begin{align} \sin P - \sin Q & = 2 \cos \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) \\ \sin 105^\circ - \sin 15 ^\circ & = 2 \cos \frac{1}{2}(105^\circ+ 15 ^\circ) \sin \frac{1}{2}(105^\circ-15 ^\circ) \\ & = 2 \cos (60 ^\circ) \sin (45 ^\circ) \\ & = 2 .\frac{1}{2} . \frac{1}{2}\sqrt{2} \\ & = \frac{1}{2}\sqrt{2} \end{align} $
Jadi, nilai $ \sin 105^\circ - \sin 15 ^\circ = \frac{1}{2}\sqrt{2} $
c). Nilai $ \cos 105^\circ + \cos 15 ^\circ $
$\begin{align} \cos P + \cos Q & = 2 \cos \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) \\ \cos 105^\circ + \cos 15 ^\circ & = 2 \cos \frac{1}{2}(105^\circ+ 15 ^\circ) \cos \frac{1}{2}(105^\circ-15 ^\circ) \\ & = 2 \cos (60 ^\circ) \cos (45 ^\circ) \\ & = 2 .\frac{1}{2} . \frac{1}{2}\sqrt{2} \\ & = \frac{1}{2}\sqrt{2} \end{align} $
Jadi, nilai $ \cos 105^\circ + \cos 15 ^\circ = \frac{1}{2}\sqrt{2} $
d). Nilai $ \tan 105^\circ + \tan 15 ^\circ $
$\begin{align} \tan P + \tan Q & = \frac{2\sin(P+Q)}{\cos (P+Q) + \cos (P-Q) } \\ \tan 105^\circ + \tan 15 ^\circ & = \frac{2\sin(105^\circ +15 ^\circ )}{\cos (105^\circ + 15 ^\circ ) + \cos (105^\circ - 15 ^\circ) } \\ & = \frac{2\sin(120^\circ )}{\cos (120 ^\circ ) + \cos (90 ^\circ) } \\ & = \frac{2\sin(180^\circ - 60^\circ )}{\cos (180^\circ - 60^\circ ) + \cos (90 ^\circ) } \\ & = \frac{2\sin( 60^\circ )}{ - \cos (60^\circ ) + \cos (90 ^\circ) } \\ & = \frac{2 . \frac{1}{2} \sqrt{3} }{ - \frac{1}{2} + 0 } \\ & = \frac{\sqrt{3} }{ - \frac{1}{2} } \\ & = -2\sqrt{3} \end{align} $
Jadi, nilai $ \tan 105^\circ + \tan 15 ^\circ = -2\sqrt{3} $
3). Tentukan nilai dari :
a). $ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ $
b). $ \sin 84^\circ \tan 42 ^\circ + \cos 84^\circ $
Penyelesaian :
a). Misalkan nilai $ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = x $
artinya kita mencari nilai $ x \, $ .
*). Gunakan sudut rangkap sinus : $ \sin 2A = 2\sin A \cos A $
Kedua ruas dikalikan $ 2\sin 20^\circ \, $ dan rumus $ 2\sin A \cos A = \sin 2A $
$ \begin{align} x & = \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ \\ 2\sin 20^\circ. x & = 2\sin 20^\circ . \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ \\ 2\sin 20^\circ. x & = (2\sin 20^\circ \cos 20^\circ ) \cos 40^\circ \cos 60^\circ \cos 80^\circ \\ 2\sin 20^\circ. x & = (\sin 2 \times 20^\circ ) \cos 40^\circ \cos 60^\circ \cos 80^\circ \\ 2\sin 20^\circ. x & = (\sin 40^\circ ) \cos 40^\circ \cos 60^\circ \cos 80^\circ \\ 2\sin 20^\circ. x & = \frac{1}{2}(2 \sin 40^\circ \cos 40^\circ ) \cos 60^\circ \cos 80^\circ \\ 2\sin 20^\circ. x & = \frac{1}{2}( \sin 2 \times 40^\circ ) \cos 60^\circ \cos 80^\circ \\ 2\sin 20^\circ. x & = \frac{1}{2}( \sin 80^\circ ) \cos 60^\circ \cos 80^\circ \\ 2\sin 20^\circ. x & = \frac{1}{2}. \frac{1}{2}( 2\sin 80^\circ \cos 80^\circ ) \cos 60^\circ \\ 2\sin 20^\circ. x & = \frac{1}{4}( \sin 2 \times 80^\circ ) \cos 60^\circ \\ 2\sin 20^\circ. x & = \frac{1}{4}( \sin 160^\circ ) \cos 60^\circ \\ 2\sin 20^\circ. x & = \frac{1}{4} \sin (180^\circ - 20^\circ ) \cos 60^\circ \\ 2\sin 20^\circ. x & = \frac{1}{4} \sin ( 20^\circ ) . \frac{1}{2} \\ 2\sin 20^\circ. x & = \frac{1}{8} \sin ( 20^\circ ) \\ x & = \frac{ \frac{1}{8} \sin ( 20^\circ ) }{ 2\sin 20^\circ} \\ x & = \frac{1}{16} \end{align} $
Jadi, nilai $ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16} $
b). Nilai $ \sin 84^\circ \tan 42 ^\circ + \cos 84^\circ $
*). Gunakan : $ \sin 2 A = 2\sin A \cos A \, $ dan $ \tan A = \frac{\sin A}{\cos A } $
serta $ \cos 2A = 1 - 2\sin ^2 A $
*). Menenylesaikan soal :
$ \begin{align} \sin 84^\circ \tan 42 ^\circ + \cos 84^\circ & = \sin 2 \times 42^\circ \tan 42 ^\circ + \cos 2 \times 42^\circ \\ & = 2\sin 42^\circ \cos 42^\circ . \frac{\sin 42 ^\circ}{\cos 42 ^\circ} + (1 - 2\sin ^2 42^\circ ) \\ & = 2\sin ^2 42^\circ + (1 - 2\sin ^2 42^\circ ) \\ & = 1 \end{align} $
Jadi, nilai $ \sin 84^\circ \tan 42 ^\circ + \cos 84^\circ = 1 $ .
4). Tentukan jumlah $ n \, $ suku pertama dari deret
$ \sin a + \sin (a + b) + \sin (a+2b) + \sin (a + 3b) + ... + \sin (a + (n-1)b) $
Pnyelesaian :
*). Soal ini adalah jumlah deret dengan suku-suku berbentuk trigonometri.
*). Jumlah $ n \, $ suku pertama ($ s_n$) maksudnya :
$ s_n = \sin a + \sin (a + b) + \sin (a+2b) + \sin (a + 3b) + ... + \sin (a + (n-1)b) $
*). Kita gunakan rumus :
$ \sin A \sin B = -\frac{\cos (A+B) - \cos (A - B)} \, $ atau $ 2\sin A \sin B = \cos (A- B) - \cos (A + B ) $
*). Semua suku kita kalilikan dengan $ 2 \sin \frac{b}{2} \, $ , kemudian dijumlahkan semua.
$ \begin{array}{cccccc} 2\sin a \sin \frac{b}{2} & = & \cos (a - \frac{b}{2} ) & - & \cos ( a + \frac{b}{2} ) & \\ 2\sin (a + b) \sin \frac{b}{2} & = & \cos (a + \frac{b}{2} ) & - & \cos ( a + \frac{3b}{2} ) & \\ 2\sin (a + 2b) \sin \frac{b}{2} & = & \cos (a + \frac{3b}{2} ) & - & \cos ( a + \frac{5b}{2} ) & \\ \vdots & & \vdots & & \vdots & \\ 2\sin (a + (n-1)b) \sin \frac{b}{2} & = & \cos (a + (n - \frac{3}{2})b ) & - & \cos ( a + (n - \frac{1}{2})b ) & + \\ \hline \\ 2 \sin \frac{b}{2} s_n & = & \cos (a - \frac{b}{2} ) & - & \cos ( a + (n - \frac{1}{2})b ) & \end{array} $
*). Gunakan rumus : $ \cos A - \cos B = -2 \sin \frac{1}{2}(A + B) \sin \frac{1}{2}(A-B) $
$ \begin{align} 2 \sin \frac{b}{2} s_n & = \cos (a - \frac{b}{2} ) - \cos ( a + (n - \frac{1}{2})b ) \\ & = -2 \sin \frac{1}{2} \left( (a - \frac{b}{2} ) + ( a + (n - \frac{1}{2})b ) \right) \sin \frac{1}{2} \left( (a - \frac{b}{2} ) - ( a + (n - \frac{1}{2})b ) \right) \\ 2 \sin \frac{b}{2} s_n & = 2 \sin \left( a + \frac{n-1}{2} b \right) \sin \left( \frac{n}{2} b \right) \\ \sin \frac{b}{2} s_n & = \sin \left( a + \frac{n-1}{2} b \right) \sin \left( \frac{n}{2} b \right) \\ s_n & = \frac{ \sin \left( a + \frac{n-1}{2} b \right) \sin \left( \frac{n}{2} b \right) }{\sin \frac{b}{2}} \end{align} $
Jadi, jumlah $ n \, $ suku pertamanya adalah : $ \begin{align} s _ n = \frac{ \sin \left( a + \frac{n-1}{2} b \right) \sin \left( \frac{n}{2} b \right) }{\sin \frac{b}{2}} \end{align} $
