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.

Bukti:

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

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

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

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

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

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