Rumus Trigonometri cosinus untuk jumlah dan selisih dua sudut
Rumus Jumlah dan selisih dua sudut untuk cosinus adalah :
$ \begin{align} \cos ( \alpha + \beta ) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \cos ( \alpha - \beta ) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{align} $
$ \begin{align} \cos ( \alpha + \beta ) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \cos ( \alpha - \beta ) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{align} $
Untuk membuktikan rumus cosinus, ada dua cara yaitu :
Cara I : Menggunakan konsep jarak dua titik
Perhatikan gambar berikut,
$\clubsuit $ Gambar di atas adalah lingkaran yang berpusat di O dan berjari-jari $ r $. Dari gambar tersebut, diperoleh $ OC=OB=OD=OA = r \, $ dan koordinat titik-titik kutubnya yaitu titik A, titik B, titik C, dan titik D, adalah $ A(r, 0), B(r \cos \alpha, r \sin \alpha ), C(r \cos(\alpha + \beta ), r \sin(\alpha + \beta )) $ , dan $ D(r \cos \beta , -r \sin \beta ) $.
$\clubsuit $ Konsep jarak (AB) dua titik A($x_1,y_1$) dan B($x_2,y_2$) :
$ \begin{align} AB & = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 } \\ AB^2 & = (x_2-x_1)^2 + (y_2-y_1)^2 \end{align} $
$ \clubsuit $ Identitas trigonometri : $ \sin ^2 A + \cos ^2 A = 1 $
$ \clubsuit $ Jarak AC : $ A(r, 0) \, $ dan $ C (r \cos(\alpha + \beta ) , r \sin(\alpha + \beta )) $
$ \begin{align} AC^2 & = [r \cos(\alpha + \beta ) - r]^2 + [r \sin(\alpha + \beta ) - 0 ]^2 \\ & = [r \cos(\alpha + \beta ) - r]^2 + [r \sin(\alpha + \beta ) - 0 ]^2 \\ & = r^2 \cos ^2 (\alpha + \beta ) - 2r^2 \cos(\alpha + \beta ) + r^2 + r^2 \sin ^2 (\alpha + \beta ) \\ & = r^2 [\cos ^2 (\alpha + \beta ) + \sin ^2 (\alpha + \beta ) ]- 2r^2 \cos(\alpha + \beta ) + r^2 \\ & = r^2 [1 ]- 2r^2 \cos(\alpha + \beta ) + r^2 \\ AC^2 & = 2r^2 - 2r^2 \cos(\alpha + \beta ) \end{align} $
$ \clubsuit $ Jarak DB : $ D(r \cos \beta , -r \sin \beta ) \, $ dan $ B(r \cos \alpha, r \sin \alpha ) $
$ \begin{align} DB^2 & = [r \cos \alpha - r \cos \beta]^2 + [r \sin \alpha - ( -r \sin \beta ) ]^2 \\ & = [r \cos \alpha - r \cos \beta]^2 + [r \sin \alpha + r \sin \beta ]^2 \\ & = (r^2 \cos ^2 \alpha - 2r^2 \cos \alpha \cos \beta + r^2 \cos ^2 \beta ) + ( r^2 \sin ^2 \alpha + 2r^2 \sin \alpha \sin \beta + r^2 \sin ^2 \beta ) \\ & = r^2 (\cos ^2 \alpha + \sin ^2 \alpha ) + r^2 ( \cos ^2 \beta + \sin ^2 \beta) -2r^2 ( \cos \alpha \cos \beta - \sin \alpha \sin \beta ) \\ & = r^2 (1 ) + r^2 ( 1 ) -2r^2 ( \cos \alpha \cos \beta - \sin \alpha \sin \beta ) \\ DB^2 & = 2r^2 -2r^2 ( \cos \alpha \cos \beta - \sin \alpha \sin \beta ) \end{align} $
$ \clubsuit $ Panjang AC sama dengan panjang DB
$ \begin{align} AC & = DB \\ AC^2 & = DB^2 \\ 2r^2 - 2r^2 \cos(\alpha + \beta ) & = 2r^2 -2r^2 ( \cos \alpha \cos \beta - \sin \alpha \sin \beta ) \\ \cos(\alpha + \beta ) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \end{align} $
Sehingga Terbukti : $ \cos(\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $
$ \clubsuit $ Membuktikan rumus $ \cos(\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $
Konsep sudut negatif : $ \sin (-A) = - \sin A \, $ dan $ \cos ( -A) = \cos A $
Menggunakan rumus jumlah dua sudut : $ \begin{align} \cos(\alpha + \beta ) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \end{align} $
$ \begin{align} \cos(\alpha - \beta ) & = \cos(\alpha + (- \beta) ) \\ & = \cos \alpha \cos (-\beta) - \sin \alpha \sin (- \beta ) \\ & = \cos \alpha \cos \beta - \sin \alpha . (- \sin \beta) \\ & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{align} $
Sehingga terbukti : $ \cos(\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $
Cara II : Menggunakan aturan cosinus pada segitiga :
Perhatikan gambar berikut,
$ \spadesuit $ Identitas trigonometri : $ \sin ^2 A + \cos ^2 A = 1 $
$ \spadesuit $ Jarak titik P dan Q :
$ \begin{align} PQ^2 & = (x_2-x_1)^2 + (y_2-y_1)^2 \\ PQ^2 & = (\cos a - \cos b)^2 + (\sin a - \sin b)^2 \\ & = (\cos ^2 a - 2\cos a \cos b + \cos ^2 a) + (\sin ^2 a - 2\sin a \sin b + \sin ^2 a) \\ & = ( \sin ^2 a + \cos ^2 a ) + (\sin ^2 b + \cos ^2 b ) - 2(\cos a \cos b + \sin a \sin b) \\ & = (1) + (1 ) - 2(\cos a \cos b + \sin a \sin b) \\ PQ^2 & = 2 - 2(\cos a \cos b + \sin a \sin b) \end{align} $
$ \spadesuit $ Aturan cosinus pada segitiga POQ
substitusi juga $ PQ^2 = 2 - 2(\cos a \cos b + \sin a \sin b) $
$ \begin{align} PQ^2 & = PO^2 + QO^2 - 2.PO.QO .\cos (a-b) \\ PQ^2 & = 1^2 + 1^2 - 2.1.1 . \cos (a-b) \\ 2 - 2 \cos (a-b) & = PQ^2 \, \, \, \, \, \text{(substitusi } PQ^2 ) \\ 2 - 2 \cos (a-b) & = 2 - 2(\cos a \cos b + \sin a \sin b) \\ \cos (a-b) & = \cos a \cos b + \sin a \sin b \end{align} $
sehingga terbukti : $ \cos (a-b) = \cos a \cos b + \sin a \sin b $
Contoh :
1). Tentukan nilai trigonometri berikut :
a). $ \cos 75^\circ $
b). $ \cos 15^\circ $
c). $ \cos 105^\circ $
Penyelesaian :
a). Nilai $ \cos 75^\circ $
Gunakan rumus : $ \cos (a+b) = \cos a \cos b - \sin a \sin b $
$ \begin{align} \cos 75^\circ & = \cos (45^\circ + 30^\circ) \\ & = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ \\ & = \frac{1}{2}\sqrt{2} . \frac{1}{2}\sqrt{3} - \frac{1}{2}\sqrt{2} . \frac{1}{2} \\ & = \frac{1}{4}\sqrt{2} (\frac{1}{2}\sqrt{3} - 1) \end{align} $
Sehingga nilai $ \cos 75^\circ = \frac{1}{4}\sqrt{2} (\sqrt{3} - 1) $
b). Nilai $ \cos 15^\circ $
Gunakan rumus : $ \cos (a-b) = \cos a \cos b + \sin a \sin b $
$ \begin{align} \cos 15^\circ & = \cos (45^\circ - 30^\circ) \\ & = \cos 45^\circ \cos 30^\circ -+ \sin 45^\circ \sin 30^\circ \\ & = \frac{1}{2}\sqrt{2} . \frac{1}{2}\sqrt{3} + \frac{1}{2}\sqrt{2} . \frac{1}{2} \\ & = \frac{1}{4}\sqrt{2} (\frac{1}{2}\sqrt{3} + 1) \end{align} $
Sehingga nilai $ \cos 15^\circ = \frac{1}{4}\sqrt{2} (\sqrt{3} + 1) $
c). Nilai $ \cos 105^\circ $
Gunakan rumus : $ \cos (a+b) = \cos a \cos b - \sin a \sin b $
$ \begin{align} \cos 105^\circ & = \cos (60^\circ + 45^\circ) \\ & = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ \\ & = \frac{1}{2} . \frac{1}{2}\sqrt{2} - \frac{1}{2}\sqrt{3} . \frac{1}{2} \sqrt{2} \\ & = \frac{1}{4}\sqrt{2} (1- \sqrt{3}) \end{align} $
Sehingga nilai $ \cos 105^\circ = \frac{1}{4}\sqrt{2} (1- \sqrt{3}) $
Rumus Trigonometri sinus untuk jumlah dan selisih dua sudut
Rumus Jumlah dan selisih dua sudut untuk sinus adalah :
$ \begin{align} \sin ( \alpha + \beta ) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \sin ( \alpha - \beta ) & = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \end{align} $
$ \begin{align} \sin ( \alpha + \beta ) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \sin ( \alpha - \beta ) & = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \end{align} $
*). Sebelumnya telah dipelajari sudut komplemen pada materi "Perbandingan Trigonometri Sudut-sudut Berelasi" .
Sudut komplemen : $ \sin A = \cos (90^\circ - A ) \, $ dan $ \cos A = \sin (90^\circ - A) $
*). Menenrapkan sudut komplemen dan rumus cosinus untuk jumlah dan selisih dua sudut :
$ \begin{align} \sin ( \alpha + \beta ) & = \cos [90^\circ - ( \alpha + \beta )] \\ & = \cos [90^\circ - \alpha - \beta ] \\ & = \cos [(90^\circ - \alpha) - \beta ] \\ & = \cos (90^\circ - \alpha) \cos \beta + \sin (90^\circ - \alpha) \sin \beta \\ \sin ( \alpha + \beta ) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \end{align} $
Jadi, terbukti : $ \sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $
*). Pembuktian rumus sinus : $ \sin ( \alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta $
$ \begin{align} \sin ( \alpha - \beta ) & = \sin ( \alpha +(-\beta)) \\ & = \sin \alpha \cos (-\beta ) + \cos \alpha \sin ( - \beta ) \\ & = \sin \alpha \cos \beta + \cos \alpha . (- \sin \beta ) \\ \sin ( \alpha - \beta ) & = \sin \alpha \cos \beta - \cos \alpha \sin \beta \end{align} $
Jadi, terbukti : $ \sin ( \alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta $
Contoh :
2). Tentukan nilai trigonometri beriktu :
a). $ \sin 75^\circ $
b). $ \sin 15^\circ $
penyelesaian :
a). Nilai $ \sin 75^\circ $
gunakan rumus : $ \sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $
$ \begin{align} \sin 75^\circ & = \sin ( 45^\circ + 30^\circ ) \\ & = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \\ & = \frac{1}{2}\sqrt{2}. \frac{1}{2}\sqrt{3} + \frac{1}{2}\sqrt{2} . \frac{1}{2} \\ & = \frac{1}{4}\sqrt{2}(\sqrt{3} + 1) \end{align} $
Sehingga nilai $ \sin 75^\circ = \frac{1}{4}\sqrt{2}(\sqrt{3} + 1) $
b). Nilai $ \sin 15^\circ $
gunakan rumus : $ \sin ( \alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta $
$ \begin{align} \sin 15^\circ & = \sin ( 45^\circ - 30^\circ ) \\ & = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \\ & = \frac{1}{2}\sqrt{2}. \frac{1}{2}\sqrt{3} - \frac{1}{2}\sqrt{2} . \frac{1}{2} \\ & = \frac{1}{4}\sqrt{2}(\sqrt{3} - 1) \end{align} $
Sehingga nilai $ \sin 15^\circ = \frac{1}{4}\sqrt{2}(\sqrt{3} - 1) $
Rumus Trigonometri Tan untuk jumlah dan selisih dua sudut
Rumus Jumlah dan selisih dua sudut untuk Tan adalah :
$ \begin{align} \tan ( \alpha + \beta ) & = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta } \\ \tan ( \alpha - \beta ) & = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta } \\ \end{align} $
$ \begin{align} \tan ( \alpha + \beta ) & = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta } \\ \tan ( \alpha - \beta ) & = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta } \\ \end{align} $
*) gunakan rumus sin dan cos jumlah serta selisih sudut, dan $ \tan A = \frac{\sin A}{\cos A} $
$ \begin{align} \tan ( \alpha + \beta ) & = \frac{ \sin ( \alpha + \beta ) }{\cos ( \alpha + \beta )} \\ & = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} \\ & = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} . \frac{\frac{1}{\cos \alpha \cos \beta}}{\frac{1}{\cos \alpha \cos \beta}} \\ & = \frac{\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} \\ & = \frac{\frac{\sin \alpha \cos \beta }{\cos \alpha \cos \beta} + \frac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{ \cos \alpha \cos \beta}{\cos \alpha \cos \beta} - \frac{ \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} \\ & = \frac{\frac{\sin \alpha }{\cos \alpha } + \frac{ \sin \beta}{ \cos \beta}}{1 - \frac{ \sin \alpha }{\cos \alpha }\frac{ \sin \beta}{ \cos \beta}} \\ \tan ( \alpha + \beta ) & = \frac{ \tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta } \end{align} $
Sehingga terbukti : $ \tan ( \alpha + \beta ) = \frac{ \tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta } $
*). Pembuktian rumus : $ \tan ( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta } $
*). sudut negatif : $ \tan (-A) = - \tan A $
$ \begin{align} \tan ( \alpha - \beta ) & = \tan ( \alpha + (- \beta )) \\ & = \frac{ \tan \alpha + \tan (-\beta )}{1 - \tan \alpha \tan (-\beta ) } \\ & = \frac{ \tan \alpha - \tan \beta }{1 - \tan \alpha . (- \tan \beta ) } \\ & = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha \tan \beta } \end{align} $
Sehingga terbukti : $ \tan ( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta } $
Contoh :
3). Tentukan nilai trigonometri dari :
a). $ \tan 75^\circ $
b). $ \tan 15^\circ $
Penyelesaian :
a). $ \tan 75^\circ $
gunakan : $ \tan ( \alpha + \beta ) = \frac{ \tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta } $
$ \begin{align} \tan 75^\circ & = \tan ( 45^\circ + 30^\circ ) \\ & = \frac{ \tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ } \\ & = \frac{ 1 + \frac{1}{3} \sqrt{3} }{1 - 1.\frac{1}{3} \sqrt{3} } \\ & = \frac{ 1 + \frac{1}{3} \sqrt{3} }{1 - \frac{1}{3} \sqrt{3} } . \frac{3}{3} \\ & = \frac{ 3 + \sqrt{3} }{3 - \sqrt{3} } \\ & = \frac{ 3 + \sqrt{3} }{3 - \sqrt{3} } . \frac{ 3 + \sqrt{3} }{3 + \sqrt{3} } \\ & = \frac{ 9 + 6\sqrt{3} + 3 }{9 - 3 } \\ & = \frac{ 12 + 6\sqrt{3} }{6 } \\ \tan 75^\circ & = 2 + \sqrt{3} \end{align} $
Jadi, nilai $ \tan 75^\circ = 2 + \sqrt{3} $
b). $ \tan 15^\circ $
gunakan : $ \tan ( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta } $
$ \begin{align} \tan 15^\circ & = \tan ( 45^\circ - 30^\circ ) \\ & = \frac{ \tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ } \\ & = \frac{ 1 - \frac{1}{3} \sqrt{3} }{1 + 1.\frac{1}{3} \sqrt{3} } \\ & = \frac{ 1 - \frac{1}{3} \sqrt{3} }{1 + \frac{1}{3} \sqrt{3} } . \frac{3}{3} \\ & = \frac{ 3 - \sqrt{3} }{3 + \sqrt{3} } \\ & = \frac{ 3 - \sqrt{3} }{3 + \sqrt{3} } . \frac{ 3 - \sqrt{3} }{3 - \sqrt{3} } \\ & = \frac{ 9 - 6\sqrt{3} + 3 }{9 - 3 } \\ & = \frac{ 12 - 6\sqrt{3} }{6 } \\ \tan 15^\circ & = 2 - \sqrt{3} \end{align} $
Jadi, nilai $ \tan 15^\circ = 2 - \sqrt{3} $
4). Jika diketahui $ \sin 5^\circ = x \, $ , tentukan nilai dari :
a). $ \sin 50^\circ $
b). $ \cos 65^\circ $
c). $ \tan 25^\circ $
Penyelesaian :
*). Menentukan nilai $ \cos 5^\circ \, $ dan $ \tan 5^\circ $
Diketahui $ \sin 5^\circ = x \rightarrow \sin 5^\circ = \frac{de}{mi} = \frac{x}{1} $
artinya sisi depan adalah $ x \, $ dan sisi miring adalah 1, dengan pythagoras diperoleh sisi samping adalah $ \sqrt{1-x^2} $ .
sehingga nilai : $ \cos 5^\circ = \frac{sa}{mi} = \frac{\sqrt{1-x^2}}{1} = \sqrt{1-x^2} \, $ dan $ \tan 5^\circ = \frac{de}{sa} = \frac{x}{\sqrt{1-x^2}} $
a). Nilai $ \sin 50^\circ $
$ \begin{align} \sin 50^\circ & = \sin (45^\circ + 5^\circ) \\ & = \sin 45^\circ \cos 5^\circ + \cos 45^\circ \sin 5^\circ \\ & = \frac{1}{2} \sqrt{2}. \sqrt{1-x^2} + \frac{1}{2} \sqrt{2}. x \\ & = \frac{1}{2} \sqrt{2}( \sqrt{1-x^2} + x ) \end{align} $
jadi, nilai $ \sin 50^\circ = \frac{1}{2} \sqrt{2}( \sqrt{1-x^2} + x ) $
b). Nilai $ \cos 65^\circ $
$ \begin{align} \cos 65^\circ & = \cos (60^\circ + 5^\circ) \\ & = \cos 60^\circ \cos 5^\circ - \sin 60^\circ \sin 5^\circ \\ & = \frac{1}{2}. \sqrt{1-x^2} - \frac{1}{2} \sqrt{3} . x \\ & = \frac{1}{2} ( \sqrt{1-x^2} - \sqrt{3} x ) \end{align} $
jadi, nilai $ \cos 65^\circ = \frac{1}{2} ( \sqrt{1-x^2} - \sqrt{3} x ) $
c). Nilai $ \tan 25^\circ $
$ \begin{align} \tan 25^\circ & = \tan (30^\circ - 5^\circ ) \\ & = \frac{\tan 30^\circ - \tan 5^\circ }{1 + \tan 30^\circ \tan 5^\circ } \\ & = \frac{\frac{1}{3} \sqrt{3} - \frac{x}{\sqrt{1-x^2}} }{1 + \frac{1}{3} \sqrt{3} . \frac{x}{\sqrt{1-x^2}} } \\ & = \frac{\frac{1}{3} \sqrt{3} - \frac{x}{\sqrt{1-x^2}} }{1 + \frac{1}{3} \sqrt{3} \frac{x}{\sqrt{1-x^2}} } . \frac{3}{3} \\ & = \frac{ \sqrt{3} - \frac{3x}{\sqrt{1-x^2}} }{3 + \frac{\sqrt{3} x}{\sqrt{1-x^2}} } \end{align} $
jadi, nilai $ \tan 25^\circ = \frac{ \sqrt{3} - \frac{3x}{\sqrt{1-x^2}} }{3 + \frac{\sqrt{3} x}{\sqrt{1-x^2}} } $
