Rumus Penjumlahan Pengurangan dan Perkalian Trigonometri

         Materi Rumus Perkalian, Penjumlahan, dan Pengurangan Trigonometri merupakan kelanjutan dari materi "Rumus Trigonometri untuk Jumlah dan Selisih Dua Sudut". Silahkan juga baca materi "Perbandingan Trigonometri Sudut-sudut Berelasi". Rumus Perkalian, Penjumlahan, dan Pengurangan Trigonometri ini biasanya akan banyak kita gunakan pada materi integral dan limit. Jadi, harus kita ingat rumus-rumus ini karena akan sangat berguna untuk materi lainnya dalam matematika.

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 :
$\begin{align} \sin A \cos B & = \frac{1}{2}[ \sin (A+B) + \sin (A- B) ] \\ \cos A \sin B & = \frac{1}{2}[ \sin (A+B) - \sin (A- B) ] \\ \cos A \cos B & = \frac{1}{2}[ \cos (A+B) + \cos (A- B) ] \\ \sin A \sin B & = - \frac{1}{2}[ \cos (A+B) - \cos (A- B) ] \end{align}$

Pembuktian Rumus Perkalian trigonometri untuk sinus dan cosinus :
*). 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 :
$\begin{align} \sin P + \sin Q & = 2 \sin \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) \\ \sin P - \sin Q & = 2 \cos \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) \\ \cos P + \cos Q & = 2 \cos \frac{1}{2}(P+Q) \cos \frac{1}{2}(P-Q) \\ \cos P - \cos Q & = -2 \sin \frac{1}{2}(P+Q) \sin \frac{1}{2}(P-Q) \\ \tan P + \tan Q & = \frac{2\sin(P+Q)}{\cos (P+Q) + \cos (P-Q) } \\ \tan P - \tan Q & = \frac{2\sin(P-Q)}{\cos (P+Q) + \cos (P-Q) } \end{align}$

Pembuktian rumus penjumlahan dan pengurangan trigonometri :
*). 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 : $\begin{align} s _ n = \frac{ \sin \left( a + \frac{n-1}{2} b \right) \sin \left( \frac{n}{2} b \right) }{\sin \frac{b}{2}} \end{align}$

Tags : trigonometri

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