Lanjutan Contoh Soal 6

\fbox{51}. Perhatikanlah gambar di bawah ini. Tunjukkan bahwa  \sin \left ( \alpha +\beta \right )=\sin \alpha \cos \beta +\cos \alpha \sin \beta dengan cara menyatakan luas segitiga sebagai jumlah dari dua segitiga lainnya.

248

Bukti:

249

Perhatikanlah ΔAA’C dan ΔAA’B

\begin{aligned}\displaystyle \frac{AC}{\sin 90^{0}}&=\displaystyle \frac{CA'}{\sin \alpha }=\displaystyle \frac{AA'}{\sin \angle C}\\ AA'&=AC.\sin \angle C\\ &=AC.\sin \left ( 90^{0}-\alpha \right )\\ &=AC.\cos \alpha\\ &\textnormal{dengan cara yang kurang lebih sama akan diperoleh juga}\\ AA'&=AB.\cos \beta\\ &\textnormal{selanjutnya kita tentukan luas seperti perinyah soal}\\ \left [ ABC \right ]&=\left [ AA'C \right ]+\left [ AA'B \right ]\\ \displaystyle \frac{1}{2}.AB.AC.\sin \left ( \alpha +\beta \right )&=\displaystyle \frac{1}{2}.AC.AA'.\sin \alpha +\displaystyle \frac{1}{2}.AB.AA'.\sin \beta \\ \sin \left ( \alpha +\beta \right )&=\displaystyle \frac{AC.AA'.\sin \alpha }{AB.AC}+\displaystyle \frac{AB.AA'.\sin \beta }{AB.AC}\\ &=\displaystyle \frac{AA'}{AB}.\sin \alpha +\displaystyle \frac{AA'}{AC}.\sin \beta \\ &=\displaystyle \frac{\left ( AB.\cos \beta \right )}{AB}.\sin \alpha +\displaystyle \frac{\left ( AC.\cos \alpha \right )}{AC}.\sin \beta \\ \sin \left ( \alpha +\beta \right )&=\sin \alpha .\cos \beta +\cos \alpha .\sin \beta \quad \blacksquare \end{aligned}   .

\fbox{52}. Tunjukkan bahwa

\begin{array}{lll}\\ &a.&\sin \left ( 2\alpha \right )=2\sin \alpha \cos \alpha\\ &b.&\sin \alpha =2\sin \left ( \displaystyle \frac{1}{2}\alpha \right ).\cos \left ( \displaystyle \frac{1}{2}\alpha \right ) \end{array}.

Bukti:

\begin{aligned}\textnormal{a.}\quad \sin \left ( \alpha +\beta \right )&=\sin \alpha \cos \beta + \cos \alpha \sin \beta ,&\textnormal{dan jika}\quad \beta =\alpha ,&&\textnormal{maka}\\ \sin \left ( \alpha +\alpha \right )&=\sin \alpha \cos \alpha +\cos \alpha \sin \alpha \\ \sin 2\alpha &=2\sin \alpha \cos \alpha\qquad \blacksquare \\ \textnormal{b.}\qquad\quad \sin 2\alpha &=2\sin \alpha \cos \alpha ,&\textnormal{dan jika}\quad \alpha =\displaystyle \frac{1}{2}\alpha ,&&\textnormal{maka}\\ \sin 2\left ( \displaystyle \frac{1}{2}\alpha \right )&=2\sin \left ( \displaystyle \frac{1}{2}\alpha \right )\cos \left ( \displaystyle \frac{1}{2}\alpha \right )\\ \sin \alpha &=2\sin \left ( \displaystyle \frac{1}{2}\alpha \right )\cos \left ( \displaystyle \frac{1}{2}\alpha \right )\qquad \blacksquare \end{aligned}\\     .

\fbox{53}. Tunjukkan bahwa Pada ΔABC dengan AD sebagai garis bagi dalam  \angle BAC serta titik D pada BC , maka akan berlaku

AD=\left (\displaystyle \frac{2bc}{b+c} \right )\cos \displaystyle \frac{1}{2}\angle A

Bukti:

Perhatikanlah ilustrasi berikut

250

AD adalah garis bagi yaitu sebuah garis lurus yang di tarik dari suatu titik sudut (A) ke sisi di hadapannya (BC) dan membagi sudut tersebut (A) menjadi dua dan sama besarnya.

\begin{aligned}\left [ ABC \right ]&=\left [ ACD \right ]+\left [ ABD \right ]\\ \displaystyle \frac{1}{2}.b.c.\sin \angle A&=\displaystyle \frac{1}{2}.AD.b.\sin \displaystyle \frac{1}{2}\angle A+\displaystyle \frac{1}{2}.AD.c.\sin \displaystyle \frac{1}{2}\angle A\\ b.c.\left ( 2.\sin \displaystyle \frac{1}{2}\angle A.\cos \displaystyle \frac{1}{2}\angle A \right )&=\left ( b+c \right ).AD.\sin \displaystyle \frac{1}{2}\angle A\\ 2bc.\cos \displaystyle \frac{1}{2}\angle A&=\left ( b+c \right ).AD\\ AD&=\left ( \displaystyle \frac{2bc}{b+c} \right ).\cos \displaystyle \frac{1}{2}\angle A\qquad \blacksquare \end{aligned}.

\fbox{54}. Perhatikanlah gambar berikut

252

Tunjukkan bahwa  \cos \left ( \alpha -\beta \right )=\cos \alpha \cos \beta +\sin \alpha \sin \beta  dengan cara menyatakan luas segitiga sebagai luas dari segitiga lainnya.

Bukti:

Perhatikan ilustrasi berikut

253.4

\begin{array}{|c|c|c|c|c|c|}\hline \textrm{No}&\textrm{Segitiga}&\textrm{Aturan}\: \textrm{Sinus}&\textrm{No}&\textrm{Segitiga}&\textrm{Aturan}\: \textrm{Sinus}\\\hline 1.&\triangle ACD&\displaystyle \frac{CD}{\sin \angle A }=\displaystyle \frac{AC}{\sin \angle D}=\displaystyle \frac{AD}{\sin \angle C}&2.&\triangle BCD&\displaystyle \frac{CD}{\sin \angle B}=\displaystyle \frac{BC}{\sin \angle D}=\displaystyle \frac{BD}{\sin \angle C}\\ \cline{3-3}\cline{6-6} &&\begin{aligned}CD&=\displaystyle \frac{AC}{\sin 90^{0}}\times \sin \angle A\\ &=\displaystyle \frac{b}{1}\times \sin \alpha \\ &=b\: \sin \alpha\\ &\textnormal{gunakan dalil Pythagoras untuk mencari AD,}\\ AD&=b\: \cos \alpha \end{aligned}&&&\begin{aligned}BD&=\displaystyle \frac{BC}{\sin 90^{0}}\times \sin \angle C\\ &=\displaystyle \frac{a}{1}\times \sin \beta \\ &=a\: \sin \beta \\ &\textnormal{gunakan juga dalil Pythagoras, maka}\\ CD&=a\: \cos \beta \end{aligned}\\\hline \end{array}.

Sehingga

\begin{aligned}\left [ ABC \right ]&=\left [ ACD \right ]+\left [ BCD \right ]\\ \displaystyle \frac{1}{2}ab\sin \left ( 90^{0}-\alpha +\beta \right )&=\displaystyle \frac{1}{2}ab\cos \alpha \cos \beta +\displaystyle \frac{1}{2}ab\sin \alpha \sin \beta \\ \sin \left ( 90^{0}-\left ( \alpha -\beta \right ) \right )&=\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \cos \left ( \alpha -\beta \right )&=\cos \alpha \cos \beta +\sin \alpha \sin \beta.\qquad \blacksquare \end{aligned}.

\fbox{55}. Tunjukkan bahwa

\begin{array}{lll}\\ &a.&\cos \left ( \alpha +\beta \right )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &b.&\cos \left ( 2\alpha \right )=2\cos ^{2}\alpha -1=1-2\sin ^{2}\alpha \\ &c.&\cos \alpha =2\cos ^{2}\displaystyle \frac{1}{2}\alpha -1=1-2\sin ^{2}\displaystyle \frac{1}{2}\alpha \end{array}.

Bukti:

\begin{aligned}\textnormal{a.}\quad \cos \left ( \alpha -\beta \right )&=\cos \alpha \cos \beta +\sin \alpha \sin \beta, &\textnormal{ganti}\: \beta\: \: \textrm{dengan}-\beta \\ \cos \left ( \alpha -\left ( -\beta \right ) \right )&=\cos \alpha \cos \left ( -\beta \right )+\sin \alpha \sin \left ( -\beta \right )\\ \cos \left ( \alpha +\beta \right )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta.\qquad \blacksquare \\ \textnormal{b.}\quad \cos \left ( \alpha +\beta \right )&=\cos \alpha \cos \beta -\sin \alpha \sin \beta &\textnormal{jika}\: \beta =\alpha \\ \cos \left ( \alpha +\alpha \right )&=\cos \alpha \cos \alpha -\sin \alpha \sin \alpha \\ \cos \left ( 2\alpha \right )&=\cos ^{2}\alpha -\sin ^{2}\alpha \\ &=\cos ^{2}\alpha -\left ( 1-\cos ^{2}\alpha \right )\\ &=2\cos ^{2}\alpha -1.\qquad \blacksquare &\textnormal{atau}\\ &=\cos ^{2}\alpha -\sin ^{2}\alpha \\ &=\left ( 1-\sin ^{2}\alpha \right )-\sin ^{2}\alpha \\ &=1-2\sin ^{2}\alpha .\qquad \blacksquare\\ \textnormal{c.}\qquad\quad \cos 2\alpha&=2\cos ^{2}\alpha -1,&\textnormal{dengan mengganti}\: \alpha =\displaystyle \frac{1}{2}\alpha \\ \cos 2\left ( \displaystyle \frac{1}{2}\alpha \right )&=2\cos ^{2}\displaystyle \frac{1}{2}\alpha -1\\ \cos \alpha &=2\cos ^{2}\displaystyle \frac{1}{2}\alpha -1.\qquad \blacksquare &\textnormal{atau}\\ \cos 2\alpha &=1-2\sin ^{2}\alpha, &\textnormal{dengan cara yang sama seperti di atas}\\ \cos \alpha &=1-2\sin ^{2}\displaystyle \frac{1}{2}\alpha.\qquad \blacksquare \end{aligned}    .

\fbox{56}. Tunjukkan bahwa

\begin{array}{lll}\\ &a.&\tan \left ( \alpha +\beta \right )=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ &b.&\tan \left ( \alpha -\beta \right )=\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta } \end{array} .

Bukti:

\begin{aligned}\textnormal{a.}\quad \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\sin \left ( \alpha +\beta \right )}{\cos \left ( \alpha +\beta \right )}\\ &=\displaystyle \frac{\left (\sin \alpha \cos \beta +\cos \alpha \sin \beta \right )\times \left ( \displaystyle \frac{1}{\cos \alpha \cos \beta } \right ) }{\left (\cos \alpha \cos \beta -\sin \alpha \sin \beta \right ) \times \left ( \displaystyle \frac{1}{\cos \alpha \cos \beta } \right )}\\ &=\displaystyle \frac{\displaystyle \frac{\sin \alpha }{\cos \alpha }+\displaystyle \frac{\sin \beta }{\cos \beta }}{1-\displaystyle \frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}\\ &=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }.\qquad \blacksquare \\ \textnormal{b.}\quad \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }&\textnormal{dengan mengganti}\: \beta =-\beta \: \: \textrm{maka,}\\ &=\displaystyle \frac{\tan \alpha +\tan \left ( -\beta \right )}{1-\tan \alpha \tan \left ( -\beta \right )}\\ \tan \left ( \alpha -\beta \right )&=\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }.\qquad \blacksquare \end{aligned}       .

\fbox{57}. Tentukanlah nilai

\begin{array}{ll}\\ a.&\sin 15^{0}\\ b.&\cos 15^{0}\\ c.&\tan 15^{0} \end{array}        .

Jawab:

Perhatikanlah bukti soal no. 55, sehingga kita dapat menyelesaikannya sebagai berikut:

\begin{aligned}\textnormal{a.}\quad\quad \cos \left (2\alpha \right )&=1-2\sin ^{2}\alpha, &\textnormal{pilih untuk}\quad \alpha =15^{0} \\ \cos \left (2\times 15^{0} \right )&=1-2\sin^{2} 15^{0}\\ \cos 30^{0}&=1-2\sin ^{2}15^{0}\\ \displaystyle \frac{1}{2}\sqrt{3}&=1-2\sin ^{2}15^{0}\\ \sqrt{3}&=2-4\sin ^{2}15^{0}\\ \sin ^{2}15^{0}&=\displaystyle \frac{2-\sqrt{3}}{4}\\ \sin 15^{0}&=\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}\\ \textnormal{b.}\quad\quad \sin ^{2}\alpha +&\cos ^{2}\alpha =1,&\textnormal{rumus identitas}\\ \sin^{2}15^{0}+&\cos ^{2}15^{0}=1\\ \cos ^{2}15^{0}&=1-\sin ^{2}15^{0}\\ &=\displaystyle 1-\left ( \frac{2-\sqrt{3}}{4} \right ),&\textnormal{lihat jawaban poin a)}\\ &=\displaystyle \frac{2+\sqrt{3}}{4}\\ \cos 15^{0}&=\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}\\ \textnormal{c.}\qquad\quad \tan \alpha &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ \tan 15^{0}&=\displaystyle \frac{\sin 15^{0}}{\cos 15^{0}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}}{\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}}\\ &=\sqrt{\displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}}\times \sqrt{\frac{2-\sqrt{3}}{2-\sqrt{3}}},&\textnormal{sekawan dari penyebut}\\ &=\sqrt{\displaystyle \frac{4-2.2\sqrt{3}+3}{4-3}}\\ &=\sqrt{7-4\sqrt{3}} \end{aligned}    .

Sumber Referensi

  1. Kanginan, Marthen. 2007. Matematika untuk Kelas X Semester 2 Sekolah Menengah Atas. Bandung: GRAFINDO MEDIA PRATAMA.
  2. Noormandiri. 2004. Matematika SMA untuk Kelas XI Program Ilmu Alam  2^A . Jakarta: Erlangga

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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