PAS MATEMATIKA BLOG

Identitas Trigonometri

Untuk identitas trigonometri dapat disederhanakan sebagaimana terdapat dalam tabel berikut ini:

$\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Pythagoras}\: \: \begin{cases} \sin ^{2}\alpha +\cos ^{2}\alpha =1 \\ \sec ^{2}\alpha =\tan ^{2}\alpha +1 \\ \csc ^{2}\alpha =\cot ^{2}\alpha +1 \end{cases}}\\\hline \textrm{Setengah}&\textrm{Satu}\\\hline \begin{cases} \sin \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{2}} \\ \cos \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1+\cos \alpha }{2}} \\ \tan \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }} \end{cases}&\begin{cases} \sin \alpha =\displaystyle \frac{1}{\csc \alpha } \\ \cos \alpha =\displaystyle \frac{1}{\sec \alpha } \\ \tan \alpha =\displaystyle \frac{1}{\cot \alpha } \\ \tan \alpha =\displaystyle \frac{\sin \alpha }{\cos \alpha } \\ \cot \alpha =\displaystyle \frac{\cos \alpha }{\sin \alpha } \end{cases}\\\hline \textrm{Dua}&\textrm{Tiga}\\\hline \begin{cases} \sin 2\alpha =2\sin \alpha \cos \alpha \\ \cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha \\ \tan 2\alpha =\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \end{cases}&\begin{cases} \sin 3\alpha =3\sin \alpha -4\sin ^{3}\alpha \\ \cos 3\alpha =4\cos ^{3}\alpha -3\cos \alpha \\ \tan 3\alpha =\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \end{cases}\\\hline \end{array}$

Bahasan yang sebelumnya

$\begin{cases} \sin \left ( \alpha +\gamma \right ) & =\sin \alpha \cos \gamma +\cos \alpha \sin \gamma \\ \sin \left ( \alpha -\gamma \right ) & =\sin \alpha \cos \gamma -\cos \alpha \sin \gamma \\ \cos \left ( \alpha +\gamma \right ) & =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \\ \cos \left ( \alpha +\gamma \right ) & =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \\ \tan \left ( \alpha +\gamma \right ) & =\displaystyle \frac{\tan \alpha +\tan \gamma }{1-\tan \alpha \tan \gamma } \\ \tan \left ( \alpha -\gamma \right ) & =\displaystyle \frac{\tan \alpha -\tan \gamma }{1+\tan \alpha \tan \gamma } \end{cases}$ .

Yang akhirnya dapat ditunjukkan rumus jumlah dan selisih untuk sinus dan cosinus

$\begin{cases} \sin X+\sin Y & =2\sin \frac{1}{2}\left (X+Y \right )\cos \frac{1}{2}\left ( X-Y \right ) \\ \sin X-\sin Y & =2\cos \frac{1}{2}\left (X+Y \right )\sin \frac{1}{2}\left ( X-Y \right )\\ \cos X+\cos Y & =2\cos \frac{1}{2}\left (X+Y \right )\cos \frac{1}{2}\left ( X-Y \right ) \\ \cos X-\cos Y & =-2\sin \frac{1}{2}\left (X+Y \right )\sin \frac{1}{2}\left ( X-Y \right ) \end{cases}$

Selanjutnya untuk rumus perkalian sinus dan cosinus

$\begin{array}{|c|c|}\hline \textrm{Sejenis}&\textrm{Tidak Sejenis}\\\hline \begin{cases} 2\cos \alpha \cos \gamma &=\cos \left ( \alpha +\gamma \right )+\cos \left ( \alpha -\gamma \right )\\ - 2\sin \alpha \sin \gamma &=\cos \left ( \alpha +\gamma \right )-\cos \left ( \alpha -\gamma \right ) \end{cases}&\begin{cases} 2\sin \alpha \cos \gamma &=\sin \left ( \alpha +\gamma \right )+\sin \left ( \alpha -\gamma \right )\\ 2\cos \alpha \sin \gamma &=\sin \left ( \alpha +\gamma \right )-\sin \left ( \alpha -\gamma \right ) \end{cases}\\\hline \end{array}$

$\LARGE{\fbox{\LARGE{\fbox{CONTOH SOAL}}}}$

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkanlah}\\ &\begin{array}{ll}\\ \textrm{a}.&\sin 2\alpha =2\sin \alpha \cos \alpha \\ \textrm{b}.&\cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha =2\cos ^{2}\alpha -1=1-2\sin ^{2}\alpha \\ \textrm{c}.&\tan 2\alpha =\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\\ \textrm{d}.&\sin 3\alpha=3\sin \alpha -4\sin ^{3}\alpha \\ \textrm{e}.&\cos 3\alpha =4\cos ^{3}\alpha -3\cos \alpha \\ \textrm{f}.&\tan 3\alpha =\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \end{array} \end{array}$

Bukti

$\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \sin \left ( \alpha +\gamma \right )&=\sin \alpha \cos \gamma +\cos \alpha \sin \gamma \\ \textrm{dengan}\: &\textrm{mengubah}\: \: \gamma =\alpha\\ \sin \left ( \alpha +\alpha \right )&=\sin \alpha \cos \alpha +\cos \alpha \sin \alpha \\ \sin 2\alpha &=2\sin \alpha \cos \alpha \qquad \blacksquare \end{aligned}&\begin{aligned}\textrm{b}.\quad \cos \left ( \alpha +\gamma \right )&=\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \\ \textrm{dengan}\: &\textrm{mengubah}\: \: \gamma =\alpha\\ \cos \left ( \alpha +\alpha \right )&=\cos \alpha \cos \alpha -\sin \alpha \sin \alpha \\ \cos 2\alpha &=\cos ^{2}\alpha -\sin ^{2}\alpha \qquad \blacksquare \end{aligned}\\\cline{1-1} \begin{aligned}\textrm{c}.\quad \tan \left ( \alpha +\gamma \right )&=\displaystyle \frac{\tan \alpha +\tan \gamma }{1-\tan \alpha \tan \gamma }\\ \textrm{dengan}\: &\textrm{mengganti}\: \: \alpha =\gamma \\ \tan \left ( \alpha +\alpha \right )&=\displaystyle \frac{\tan \alpha +\tan \alpha }{1-\tan \alpha \tan \alpha }\\ \tan 2\alpha &=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\qquad \blacksquare \\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{ingat}&\: \textrm{bahwa}\\ &\begin{cases} 1 &=\sin ^{2}\alpha +\cos ^{2}\alpha \\ \sin ^{2}\alpha & =1-\cos ^{2}\alpha \\ \cos ^{2}\alpha &= 1-\sin ^{2}\alpha \end{cases}\\ \textrm{sehin}&\textrm{gga persamaan}\\ \cos 2\alpha &=\cos ^{2}\alpha -\sin ^{2}\alpha\\ &=\cos ^{2}\alpha -\left ( 1-\cos ^{2}\alpha \right )\\ &=2\cos ^{2}\alpha -1\qquad \blacksquare \\ &=2\left ( 1-\sin ^{2}\alpha \right )-1\\ &=2-2\sin ^{2}\alpha -1\\ &=1-2\sin ^{2}\alpha \qquad \blacksquare \end{aligned}\\\hline \end{array}$

$\begin{array}{|l|l|}\hline \begin{aligned}\textrm{e}.\quad \cos 3\alpha &=\cos \left ( 2\alpha +\alpha \right )\\ &=\cos 2\alpha \cos \alpha -\sin 2\alpha \sin \alpha \\ &=\left ( 2\cos ^{2}\alpha -1 \right )\cos \alpha -\left ( 2\sin \alpha \cos \alpha \right )\sin \alpha \\ &=2\cos ^{3}\alpha -\cos \alpha -2\sin ^{2}\alpha \cos \alpha \\ &=2\cos ^{3}\alpha -\cos \alpha -2\left ( 1-\cos ^{2}\alpha \right ) \cos \alpha\\ &=2\cos ^{3}\alpha -\cos \alpha -2\cos \alpha +2\cos ^{2}\alpha\\ &=4\cos ^{3}\alpha -3\cos \alpha \qquad \blacksquare\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{f}.\quad \tan 3\alpha &=\tan \left ( 2\alpha +\alpha \right )\\ &=\displaystyle \frac{\tan 2\alpha +\tan \alpha }{1-\tan 2\alpha \tan \alpha }\\ &=\displaystyle \frac{\left ( \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \right )+\tan \alpha }{1-\left ( \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \right ).\tan \alpha }\\ &=\displaystyle \frac{\displaystyle \frac{2\tan \alpha +\tan \alpha -\tan ^{3}\alpha }{1-\tan ^{2}\alpha }}{\displaystyle \frac{\left ( 1-\tan ^{2}\alpha \right )-2\tan ^{2}\alpha }{1-\tan ^{2}\alpha }}\\ &=\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha }\qquad \blacksquare \end{aligned} \\\hline \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Buktikanlah bahwa}\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sin \left ( 90^{\circ}-\beta \right )=\cos \beta &\textrm{i}.&\sin \left ( 270^{\circ}-\beta \right )=-\cos \beta&\textrm{q}.&\sin 15^{\circ}=\displaystyle \frac{\sqrt{3}-1}{2\sqrt{2}}\\ \textrm{b}.&\cos \left ( 90^{\circ}-\beta \right )=\sin \beta &\textrm{j}.&\cot \left ( 270^{\circ}-\beta \right )=\tan \beta&\textrm{r}.&\cos 15^{\circ}=\displaystyle \frac{1}{4}\sqrt{2}\left ( \sqrt{3}+1 \right )\\ \textrm{c}.&\tan \left ( 90^{\circ}-\beta \right )=\cot \beta &\textrm{k}.&\sin \left ( 360^{\circ}-\beta \right )=-\sin \beta&\textrm{s}.&\tan 15^{\circ}=2-\sqrt{3} \\ \textrm{d}.&\sin \left ( 180^{\circ}-\beta \right )=\sin \beta &\textrm{l}.&\cos \left ( 360^{\circ}-\beta \right )=\cos \beta&\textrm{t}.&\displaystyle \frac{\sin \left ( \alpha +\beta \right )}{\cos \alpha \cos \beta }=\tan \alpha +\tan \beta \\ \textrm{e}.&\cos \left ( 180^{\circ}-\beta \right )=-\cos \beta &\textrm{m}.&\cot \left ( 360^{\circ}-\beta \right )=-\cot \beta&\textrm{u}.&\displaystyle \frac{\sin \left ( \alpha -\beta \right )}{\sin \left ( \alpha +\beta \right ) }=\frac{\tan \alpha -\tan \beta }{\tan \alpha +\tan \beta }\\ \textrm{f}.&\cot \left ( 180^{\circ}-\beta \right )=-\cot \beta &\textrm{n}.&\sin \left ( -\beta \right )=-\sin \beta &\textrm{v}.&\displaystyle \frac{\cos \left ( \alpha +\beta \right )}{\cos\alpha \cos \beta }=1-\tan \alpha \tan \beta \\ \textrm{g}.&\sin \left ( 180^{\circ}+\beta \right )=-\sin \beta &\textrm{o}.&\cos \left ( -\beta \right )=\cos \beta &\textrm{w}.&\displaystyle \frac{\cos \left ( \alpha +\beta \right )}{\cos \left ( \alpha -\beta \right ) }=\frac{1-\tan \alpha \tan \beta }{1+\tan \alpha \tan \beta } \\ \textrm{h}.&\csc \left ( 180^{\circ}+\beta \right )=-\csc \beta &\textrm{p}.&\tan \left ( -\beta \right )=-\tan \beta &\textrm{x}.&\displaystyle \frac{\sin 3\gamma +\sin \gamma }{\cos 3\gamma +\cos \gamma }=\tan 2\gamma\\ \end{array} \end{array}$

Bukti

untuk pengingat kita

$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \cdots \alpha &0^{0}&30^{0}&45^{0}&60^{0}&90^{0}&180^{0}&270^{0}&360^{0}\\\hline \sin \alpha &0&\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0&-1&0\\\hline \cos \alpha &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1&0&1\\\hline \tan \alpha &0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&TD&0&TD&0\\\hline \end{array}$

maka

$\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \sin \left ( 90^{\circ}-\beta \right )&=\sin 90^{\circ}\cos \beta -\cos 90^{\circ}\sin \beta \\ &=1.\cos \beta -0.\sin \beta \\ &=\cos \beta\qquad \blacksquare \\ & \end{aligned}&\begin{aligned}\textrm{n}.\quad \sin \left ( -\beta \right )&=\sin \left ( 0^{\circ}-\beta \right )\\ &=\sin 0^{\circ}.\cos \beta -\cos 0^{\circ}\sin \beta \\ &=0-1.\sin \beta \\ &=-\sin \beta\qquad \blacksquare \end{aligned}\\\hline \begin{aligned}\textrm{h}.\quad \csc \left ( 180^{\circ}+\beta \right )&=\displaystyle \frac{1}{\sin \left ( 180^{\circ}+\beta \right )}\\ &=\displaystyle \frac{1}{\sin 180^{\circ}\cos \beta +\cos 180^{\circ}\sin \beta }\\ &=\displaystyle \frac{1}{0.\cos \beta +(-1).\sin \beta }\\ &=-\displaystyle \frac{1}{\sin \beta }\\ &=-\csc \beta\qquad \blacksquare \\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{s}.\quad\tan 15^{\circ}&=\tan \left ( 45^{\circ}-30^{\circ} \right )\\ &=\displaystyle \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}}\\ &=\displaystyle \frac{1-\displaystyle \frac{1}{\sqrt{3}}}{1+\displaystyle \frac{1}{\sqrt{3}}}\\ &=\left ( \displaystyle \frac{1-\displaystyle \frac{1}{\sqrt{3}}}{1+\displaystyle \frac{1}{\sqrt{3}}} \right )\times \left ( \displaystyle \frac{\sqrt{3}}{\sqrt{3}} \right )\\ &=\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}\times \left (\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}-1} \right )\\ &=\displaystyle \frac{3-2\sqrt{3}+1}{3-1}\\ &=\displaystyle \frac{4-2\sqrt{3}}{2}\\ &=2-\sqrt{3}\qquad \blacksquare \end{aligned}\\\hline \end{array}$

$\begin{array}{|l|l|}\hline \begin{aligned}\textrm{t}.\quad \displaystyle \frac{\sin \left ( \alpha +\beta \right )}{\cos \alpha \cos \beta }&=\displaystyle \frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta }\\ &=\displaystyle \frac{\sin \alpha \cos \beta }{\cos \alpha \cos \beta }+\displaystyle \frac{\cos \alpha \sin \beta }{\cos \alpha \cos \beta }\\ &=\displaystyle \frac{\sin \alpha }{\cos \alpha}+\displaystyle \frac{\sin \beta }{\cos \beta }\\ &=\tan \alpha +\tan \beta\qquad \blacksquare \end{aligned}&\begin{aligned}\textrm{v}.\quad \displaystyle \frac{\cos \left ( \alpha +\beta \right )}{\cos \alpha \cos \beta }&=\displaystyle \frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta }\\ &=\displaystyle \frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }-\displaystyle \frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }\\ &=1-\displaystyle \frac{\sin \alpha }{\cos \alpha}\times \displaystyle \frac{\sin \beta }{\cos \beta }\\ &=1-\tan \alpha \tan \beta\qquad \blacksquare \end{aligned}\\\hline \end{array}$

$\begin{array}{|l|l|}\hline \begin{aligned}\textrm{x}.\quad \displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }&=\displaystyle \frac{2\sin \displaystyle \frac{\left (3\alpha +\alpha \right )}{2}\cos \displaystyle \frac{\left ( 3\alpha -\alpha \right )}{2}}{2\cos \displaystyle \frac{\left ( 3\alpha +\alpha \right )}{2}\cos \displaystyle \frac{\left ( 3\alpha -\alpha \right )}{2}}\\ &=\displaystyle \frac{\sin 2\alpha }{\cos 2\alpha }\\ &=\tan 2\alpha \qquad \blacksquare\\ &\\ &\\ &\quad\qquad\qquad\textbf{atau}\Rightarrow\qquad \\ &\textrm{mungkin lumayan rumit}\\ &\\ & \end{aligned}&\begin{aligned}\textrm{x}.\quad \displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }&=\displaystyle \frac{3\sin \alpha -4\sin ^{3}\alpha +\sin \alpha }{4\cos ^{3}\alpha -3\cos \alpha +\cos \alpha }\\ &=\displaystyle \frac{4\sin \alpha -4\sin ^{3}\alpha }{4\cos ^{3}\alpha -2\cos \alpha }\\ &=\displaystyle \frac{4\sin \alpha \left ( 1-\sin ^{2}\alpha \right )}{2\cos \alpha \left ( 2\cos ^{2}\alpha -1 \right )}=\displaystyle \frac{4\sin \alpha }{2\cos \alpha }\times \displaystyle \frac{\cos ^{2}\alpha }{\cos 2\alpha }\\ &=\displaystyle \frac{2\tan \alpha \cos ^{2}\alpha }{\cos 2\alpha }\\ &=\displaystyle \frac{2\tan \alpha \cos ^{2}\alpha }{\cos^{2}\alpha -\sin ^{2}\alpha }\\ &=\displaystyle \frac{2\tan \alpha }{\displaystyle \frac{\cos ^{2}\alpha }{\cos ^{2}\alpha }-\displaystyle \frac{\sin ^{2}\alpha }{\cos ^{2}\alpha }}=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\\ &=\tan 2\alpha \qquad \blacksquare \end{aligned} \\\hline \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Sederhanakanlah bentuk berikut ini}\\ &\begin{array}{ll}\\ \textrm{a}.&\sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )\\ \textrm{b}.&\sin \left ( x+\displaystyle \frac{1}{4}m \right )+\sin \left ( x-\displaystyle \frac{1}{4}m \right )\\ \textrm{c}.&\cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )\\ \textrm{d}.&\cos \left ( \displaystyle \frac{5}{4}m+3x \right )+\cos \left ( \displaystyle \frac{5}{4}m-3x \right ) \end{array} \end{array}$

Solusi

$\begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad \sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )&=2\cos \displaystyle \frac{\left (2x+60^{\circ} \right )+\left ( 2x-60^{\circ} \right )}{2}\sin \displaystyle \frac{\left (2x+60^{\circ} \right )-\left ( 2x-60^{\circ} \right )}{2}\\ &=2\cos \displaystyle \frac{4x}{2}\sin \displaystyle \frac{120^{\circ}}{2}\\ &=2\cos 2x\sin 60^{\circ}\\ &=2\cos 2x.\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\sqrt{3}\cos 2x \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad \cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )&=-2\sin \displaystyle \frac{\left (2x+4y \right )+\left ( 2x-4y \right )}{2}\sin \displaystyle \frac{\left (2x+4y \right )-\left ( 2x-4y \right )}{2}\\ &=-2\sin \displaystyle \frac{4x}{2}\sin \displaystyle \frac{8y}{2}\\ &=-2\sin 2x\sin 4y\\ \end{aligned} \\\hline \end{array}$

Sumber Referensi

Kurnia, N., dkk. 2017. Jelajah Matematika 2 SMA Kelas XI Peminatan MIPA(Edisi Revisi 2016). Jakarta: Yudistira.
Sembiring, S., Zulkifli, M., Marsito, dan Rusdi, I. 2017. Matematika untuk siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam(Edisi Revisi 2016). Bandung: SEWU.
Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 5 Kelas 3. Jakarta: Erlangga.

Lanjuta Rumus trigonometri

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

Bukti

Perhatikanlah ΔAA’C dan ΔAA’B

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

Sedangkan untuk bukti

$\cos \left ( \alpha -\beta \right )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$

Perhatikanlah ilustrasi berikut juga

Jika kita ubah menjadi

maka

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

Sehingga

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

Dan untuk formula tangenm misalkan kita diminta untuk menunjukkan bukti berikut

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

maka sebagai buktinya adalah:

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

Rumus Trigonometri Jumlah dan Selisih Dua Sudut

Perhatikanlah gambar berikut
Untuk mendapatkan rumus $\cos \left ( \alpha \pm \beta \right )$ , perhatikan ilustrasi berikut ini

$\begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\begin{aligned}\textrm{Diketahui} \: \: &\textrm{bahwa : Pada lingkaran di atas}\\ &\begin{cases} A & =(r,0) \\ B & =\left ( r\cos \alpha ,r\sin \alpha \right ) \\ C & =\left ( r\cos (\alpha +\beta ),r\sin (\alpha +\beta ) \right ) \\ D & =\left ( r\cos \beta ,-r\sin \beta \right ) \end{cases}\\ \end{aligned}}\\\hline \textrm{Jarak}&\multicolumn{2}{|c|}{OA=OB=OC=OD=r}\\\hline &(AC)^{2}&\begin{aligned}&(AC)^{2}\\ &=\left ( x_{C}-x_{A} \right )^{2}+\left ( y_{C}-y_{A} \right )^{2}\\ &=\left ( r\cos (\alpha +\beta )-r \right )^{2}+\left ( r\sin (\alpha +\beta )-0 \right )^{2}\\ &=\left ( r^{2}\cos ^{2}(\alpha +\beta )-2r^{2}\cos (\alpha +\beta )+r^{2} \right )+r^{2}\sin ^{2}(\alpha +\beta )\\ &=r^{2}\left ( \cos ^{2}(\alpha +\beta )+\sin ^{2}(\alpha +\beta ) \right )+r^{2}-2r^{2}\cos (\alpha +\beta )\\ &=r^{2}+r^{2}-2r^{2}\cos (\alpha +\beta )\\ &=2r^{2}-2r^{2}\cos (\alpha +\beta ) \end{aligned} \\\cline{2-3} (AC)^{2}=(BD)^{2}&(BD)^{2}&\begin{aligned}&(BD)^{2}\\ &=\left ( x_{D}-x_{B} \right )^{2}+\left ( y_{D}-y_{B} \right )^{2}\\ &=\left ( r\cos \alpha - r\cos \beta \right )^{2}+\left ( r\sin \alpha +r\sin \beta \right )^{2}\\ &=...\\ &=...\\ &=...\\ &=2r^{2}-2r^{2}\cos \alpha \cos \beta +2r^{2}\sin \alpha \sin \beta \end{aligned}\\\cline{2-3} &\multicolumn{2}{|c|}{\begin{aligned}(AC)^{2}&=(BD)^{2}\\ 2r^{2}-2r^{2}\cos (\alpha +\beta )&=2r^{2}-2r^{2}\cos \alpha \cos \beta +2r^{2}\sin \alpha \sin \beta\\ \cos (\alpha +\beta )&=\cos \alpha \cos \beta-\sin \alpha \sin \beta \end{aligned}}\\\hline \multicolumn{3}{|c|}{\begin{aligned}\textrm{Jadi},&\: \cos \left ( \alpha +\beta \right )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \end{aligned}}\\\hline \end{array}$

Sedangkan untuk rumus $\sin \left ( \alpha \pm \beta \right )$

kita uraikan dengan ilustrasi gambar berikut ini

Perhatikanlah untuk segi empat tali busurnya

$\begin{aligned}AC\times BD&=BC\times AD+AB\times DC\\ 1\times BD&=\sin x\times \cos y+\cos x\times \sin y,\qquad\textnormal{diameter=1 satuan=AC=BE}\\ BD&=\sin x\times \cos y+\cos x\times \sin y,\qquad \textrm{karena}\: \: \angle BAD=\angle BED,\: \: \textrm{maka}\\ \sin (x+y)&=\sin x\times \cos y+\cos x\times \sin y ,\qquad \angle ABC=\angle ADC=\angle BDE=90^{0}\end{aligned}$

Selanjutnya

$\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{untuk}\: \: \alpha \: \textrm{dan}\: \beta }\\\hline \alpha \neq \beta &\alpha = \beta \\\hline \begin{cases} \textrm{sinus} & (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ \textrm{sinus} & (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ \textrm{cosinus} & (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ \textrm{cosinus} & (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \textrm{tangen} & (\alpha +\beta )=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ \textrm{tangen} & (\alpha -\beta )=\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta } \end{cases}&\begin{aligned}\sin 2\alpha &=2\sin \alpha \cos \alpha \\ \cos 2\alpha &=\cos ^{2}\alpha -\sin ^{2}\alpha \\ &=\begin{cases} \cos 2\alpha &=2\cos ^{2}\alpha -1 \\ \cos 2\alpha &=1-2\sin ^{2}\alpha \end{cases}\\ \tan 2\alpha &=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \end{aligned}\\\hline \end{array}$

Selanjutnya saat $\alpha =\beta$ disebut sudut ganda

Ukuran Penyebaran Data

Perhatikanlah Tabel berikut

$\begin{array}{|c|l|l|}\hline \textrm{No}&\qquad\qquad\quad\textrm{Istilah}&\qquad\qquad\textrm{Formula}\\\hline 1&\textrm{Jangkauan/\textit{range} (J)}&x_{maksimum}-x_{minimum}\\\hline 2&\textrm{Jangkauan kuartil/hamparan(H)}&Q_{3}-Q_{1}\\\hline 3.&\textrm{Simpangan kuartil}\: \left ( \textrm{Q}_{d} \right )\: \: atau&\\ &\textrm{Jangkauan semi antarkuartil}&\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\\hline 4&\textrm{Simpangan rata-rata (SR)}&\begin{cases} SR&=\displaystyle \frac{\sum_{i=1}^{n}\left | x_{i}-\bar{x} \right |}{n} \\ &atau \\ SR&=\displaystyle \frac{\sum_{i=1}^{n}f_{i}\left | x_{i}-\bar{x} \right |}{\sum_{i=1}^{n}f_{i}} \end{cases}\\\hline 5&\textrm{Ragam/Varians}\: \left ( \textrm{S}^{2} \right )&\begin{cases} S^{2}& =\displaystyle \frac{\sum_{i=1}^{n}\left ( x_{i}-\bar{x} \right )^{2}}{n} \\ & atau \\ S^{2}& =\displaystyle \frac{\sum_{i=1}^{n}f_{i}\left ( x_{i}-\bar{x} \right )^{2}}{\sum_{i=1}^{n}f_{i}} \end{cases}\\\hline 6&\textrm{Simpangan baku atau}&\begin{matrix} S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{\sum_{i=1}^{n}\left ( x_{i}-\bar{x} \right )^{2}}{n}}\\ atau \end{matrix} \\ &\textrm{Standar deviasi}&S=S^{2}=\sqrt{\displaystyle \frac{\sum_{i=1}^{n}f_{i}\left ( x_{i}-\bar{x} \right )^{2}}{\sum_{i=1}^{n}f_{i}}}\\\hline \end{array}$

Ukuran Letak Data

Ukuran Data Memusat

A. Data Tunggal

$\begin{array}{|ll|ll|}\hline \multicolumn{2}{|l|}{\textrm{Ukuran Pemusatan(Tendensi Sentral)}}&\multicolumn{2}{|l|}{\textrm{Ukuran Penyebaran(Dispersi)}}\\\hline \multicolumn{2}{|l|}{\textrm{Mean}\left ( \bar{x} \right )/\textrm{Rataan hitung}}&\multicolumn{2}{|l|}{\textrm{Jangkauan}\left ( J \right )/\textrm{Rentang}\left ( R \right )}\\ &\bar{x}=\displaystyle \frac{x_{1}+x_{2}+x_{3}+...+x_{n}}{n}&&J=R=x_{maks}-x_{min}\\\hline \multicolumn{2}{|l|}{\textrm{Median}\left ( M_{e} \right )/\textrm{Nilai datum tengah}}&\multicolumn{2}{|l|}{\textrm{Hamparan/Jangkauan antarkuartil}}\\ &\textrm{Data Ganjil}:\quad M_{e}=x_{\frac{n+1}{2}}&&H=Q_{3}-Q_{1}\\ &\textrm{Data Genap}:\quad M_{e}=\displaystyle \frac{1}{2}\left ( x_{\frac{n}{2}}+x_{\frac{n}{2}+1} \right )&&\\\hline \multicolumn{2}{|l|}{\textrm{Modus}\left (M_{o} \right )}&\multicolumn{2}{|l|}{\textrm{Simpangan kuartil}\left ( Q_{d} \right )}\\ &M_{o}:\: \textrm{Nilai datum dengan frekuensi terbesar} &&Q_{q}=\displaystyle \frac{1}{2}H\\\hline \multicolumn{2}{|l|}{\textrm{Kuartil}\left ( Q\right )}&\multicolumn{2}{|l|}{\textrm{Langkah}\left ( L \right )}\\ &\textrm{Data Ganjil}:\quad \begin{cases} Q_{1}= & x_{\frac{1}{4}(n+1)}\\ Q_{2}= & x_{\frac{2}{4}(n+1)}\\ Q_{3}= & x_{\frac{3}{4}(n+1)} \end{cases}&&L=\displaystyle \frac{3}{2}H\\ &\textrm{Data Genap}:\quad \begin{cases} Q_{1}= & x_{\frac{1}{4}n+\frac{1}{2}}\\ Q_{2}= & x_{\frac{2}{4}n+\frac{1}{2}}\\ Q_{3}= & x_{\frac{3}{4}n+\frac{1}{2}} \end{cases}&&\\\hline &&\multicolumn{2}{|l|}{\textrm{Pagar dalam dan Pagar luar}}\\\hline \end{array}$

$\begin{array}{|ll|ll|}\hline \multicolumn{2}{|l|}{\textrm{Ukuran Pemusatan(Tendensi Sentral)}}&\multicolumn{2}{|l|}{\textrm{Ukuran Penyebaran(Dispersi)}}\\\hline &&\multicolumn{2}{|l|}{\textrm{Pagar dalam dan Pagar luar}}\\ &&&\bullet \: \textrm{Pagar dalam}:Q_{1}-L\\ &&&\bullet \: \textrm{Pagar luar}:Q_{3}+L\\ &&&\begin{cases} \textrm{Data} & \textrm{normal } \\ &:Q_{1}-L\leq x_{i}\leq Q_{3}+L\\ \textrm{Data} & \textrm{tak normal(pencilan)}\\ &:\begin{cases} x_{i}<Q_{1}-L \\ x_{i}>Q_{3}+L \end{cases} \end{cases}\\\cline{3-4} &&\multicolumn{2}{|l|}{\textrm{Simpangan rata}\left ( SR \right )}\\ &&&SR=\displaystyle \frac{\sum \left | x_{i}-\bar{x} \right |}{n}\\\cline{3-4} &&\multicolumn{2}{|l|}{\textrm{Ragam/varians}\left ( s^{2} \right )}\\ &&&s^{2}=\displaystyle \frac{\sum \left ( x_{i}-\bar{x} \right )^{2}}{n}\\\cline{3-4} &&\multicolumn{2}{|l|}{\textrm{Simpangan baku}\left ( s=\sqrt{s^{2}} \right )}\\\hline \end{array}$ .

B. Data Berkelompok

$\begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\textbf{Data Deskriptif Berkelompok}}\\\hline \textrm{Rataan Hitung}\left ( \bar{\textrm{x}} \right )&\textrm{Modus}\left ( \textrm{M}_{o} \right )&\textrm{Kuartil}\left ( \textrm{Q} \right )\\\hline \bar{\textrm{x}}=\bar{x}_{s}+\displaystyle \frac{\sum f_{i}.d_{i}}{\sum f_{i}}&\textrm{M}_{o}=t_{p}+\displaystyle p\left ( \frac{d_{1}}{d_{1}+d_{2}} \right )&\textrm{Q}_{i}=t_{p}+\displaystyle p\left ( \frac{\displaystyle \frac{i.n}{4}-\sum f_{i}}{f_{b}} \right )\\\hline \multicolumn{3}{|c|}{\textbf{Keterangan}}\\\hline \begin{aligned}\bar{x}_{s}=&\textrm{rataan sementara}\\ x_{i}=&\textrm{titik tengahinterval}\\ &\textrm{kelas ke}-i\\ d_{i}=&x_{i}-\bar{x}_{s}\\ f_{i}=&\textrm{frekuensi kelas ke}-i\\ &\\ &\\ &\\ &\\ &\end{aligned}&\begin{aligned}t_{p}=&\textrm{tepi bawah kelas modus}\\ p=&\textrm{panjang interval kelas}\\ d_{1}=&f_{0}-f_{-1}\\ d_{2}=&f_{0}-f_{+1}\\ f_{0}=&\textrm{frekuensi kelas modus}\\ f_{-1}=&\textrm{frekuensi sebelum kelas modus}\\ f_{+1}=&\textrm{frekuensi setelah kelas modus}\\ &\\ &\\ & \end{aligned}&\begin{aligned}tp=&\textrm{tepi bawah kuartil ke}-i\\ p=&\textrm{panjang interval kelas}\\ n=&\textrm{banyaknya data}\\ \sum f_{i}=&\textrm{jumlah semua frekuensi}\\ &\textrm{sebelum kelas kurtil ke}-i\\ f_{q}=&\textrm{frekuensi kelas kuartil ke}-i\\ Q_{i}=&\textrm{kuartil ke}-i\\ Q_{1}=&\textrm{kuartil bawah}\\ Q_{2}=&\textrm{kuatil tengah/median}\\ Q_{3}=&\textrm{kuatil atas} \end{aligned}\\\hline \end{array}$

$\LARGE\fbox{\fbox{{CONTOH SOAL}}}$

$\begin{array}{ll}\\ \fbox{1}.&\textrm{Berikut yang}\: \: bukan\: \: \textrm{merupakan ukuran penyebaran data adalah} ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \textrm{rentang}&&\textrm{d}.\quad\textrm{kuartil}\\ \textrm{b}.\quad\textrm{varians}&\textrm{c}.\quad\textrm{jarak antarkuartil}&\textrm{e}.\quad\textrm{simpangan baku}\\ &&\\ &&(\textbf{SMBB TELKOM 2006})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textrm{d}.\quad \textrm{karena kuartil merupakan ukuran pemusatan data} \end{array}$

$\begin{array}{|c|c|}\hline \textrm{Daftar distribusi data tunggal}&\textrm{Daftar distribusi data berkelompok}\\\hline n\geq 30&n\geq 30\\\hline \begin{aligned}2.\: \: \textrm{Sebagai}&\: \textrm{misal data jumlah anak}\\ \textrm{dari}\: &\: \textrm{30 karyawan sebuah}\\ \textrm{peru}&\textrm{sahaan}\\ 3&\, 2\, 0\, 1\, 4\, 2\, 2\, 2\, 1\, 2\\ 3&\, 0\, 3\, 2\, 1\, 1\, 2\, 1\, 2\, 2\\ 2&\, 1\, 2\, 2\, 0\, 3\, 1\, 1\, \, 2\, 3 \end{aligned}&\begin{aligned}\textrm{Langkah}&-\textrm{langkah membuat daftar}\\ \textrm{dengan}\: &\textrm{mentukan hal-hal berikut}\\ 1.\: &Jangkauan(J)=x_{maks}-x_{min}\\ 2.\: &\textit{Banyak kelas}(k)=1+3,3\times \log n\\ 3.\: &\textit{Panjang kelas}(c)=\displaystyle \frac{J}{k}\\ 4.\: &\textit{Batas kelas pertama}=\textit{datum terkecil} \end{aligned}\\\hline \begin{array}{|c|c|}\hline \textrm{Juml anak}&\textrm{Frekuensi}\\\hline 0&3\\ 1&8\\ 2&13\\ 3&5\\ 4&1\\\hline \textrm{Jumlah}&30\\\hline \end{array}&\begin{aligned}3.\: \: \textrm{Buat}&\textrm{lah Daftar distribusi frekuensi}\\ &\textrm{untuk data berikut}\\ &79\, 68\, 60\, 73\, 62\, 78\, 56\, 64\, 68\, 53\, 72\, 67\, 64\, 62\\ &58\, 53\, 52\, 72\, 59\, 74\, 52\, 71\, 62\, 51\, 70\, 60\, 72\, 68\\ &74\, 67\, 70\, 70\, 57\, 55\, 62\, 52\, 61\, 77\, 63\, 63\\ & \end{aligned}\\\hline \end{array}$

$\color{blue}\textbf{Pembahasan untuk no.3}\\ \begin{aligned}\textrm{Dari}&\: \textrm{data di atas kita mendapatkan}\\ 1.\quad&J=x_{maks}-x_{min}=79-51=28\\ 2.\quad&k=1+3,3\times \log 40=1+3,3\times (1,602)=1+(5,2866)=6,2866\\ &\: \: \,\approx 6\qquad (\textrm{ada 6 kelas dari hasil pembulatan})\\ 3.\quad&c=\displaystyle \frac{J}{k}=\frac{28}{6}=4,\bar{66}\approx 5\quad (\textrm{pembulatan})\\ 4.\quad&x_{min}=51,\: \textrm{maka kelas interval pertama adalah}=51-55\\ 5.\quad&\textrm{Hasil yang diperoleh}\\ &\begin{array}{|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}\\\hline 51-55&7\\ 56-60&6\\ 61-65&9\\ 66-70&8\\ 71-75&7\\ 76-80&3\\\hline \textrm{Jumlah}&40\\\hline \end{array} \end{aligned}$

$\begin{array}{ll}\\ \fbox{4}.&\textrm{Dari no.3 di atas, buatkanlah data dalam tabel}\\ &\textrm{daftar distribusi frekuensi relatif}\\\\ &\textrm{Jawab}\\ &\begin{matrix} \begin{array}{|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}\\\hline 51-55&7\\ 56-60&6\\ 61-65&9\\ 66-70&8\\ 71-75&7\\ 76-80&3\\\hline \textrm{Jumlah}&40\\\hline \end{array} &\textrm{menjadi} &\begin{array}{|c|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}&\textrm{Frekuensi Relatif}\\\hline 51-55&7&0,175=17,5\%\\ 56-60&6&0,15=15\%\\ 61-65&9&0,225=22,5\%\\ 66-70&8&0,2=20\%\\ 71-75&7&0,175=17,5\%\\ 76-80&3&0,075=7,5\%\\\hline \textrm{Jumlah}&40&100\%\\\hline \end{array} \end{matrix} \end{array}$

$\begin{array}{ll}\\ \fbox{5}&\textrm{Pada tabel di atas buatkanlah ke daftar distribusi frekuensi kumulatif}\\ &\textrm{kurang dari}\\ \\ &\textrm{Pembahasan}:\\ &\begin{array}{|c|c|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}&\textrm{Nilai}&f_{k}\leq \\\hline 51-55&7&\leq 55,5&7\\ 56-60&6&\leq 60,5&13\\ 61-65&9&\leq 65,5&22\\ 66-70&8&\leq 70,5&30\\ 71-75&7&\leq 75,5&37\\ 76-80&3&\leq 80,5&40\\\hline \textrm{Jumlah}&40&&\\\hline \end{array}\Rightarrow \begin{array}{l}\\ \textbf{sebagai penjelasan}\\ \\ 13=7+6\\ 22=7+6+9\\ 30=7+6+9+8\\ 37=7+6+9+8+7\\ 40=7+6+9+8+7+3\\ \\ \\ \end{array} \end{array}$

$\begin{array}{ll}\\ \fbox{6}&\textrm{Pada tabel di atas buatkanlah ke daftar distribusi frekuensi kumulatif}\\ &\textrm{lebih dari}\\ \\ &\textrm{Pembahasan}:\\ &\begin{array}{|c|c|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}&\textrm{Nilai}&f_{k}\leq \\\hline 51-55&7&\geq 50,5&40\\ 56-60&6&\geq 55,5&33\\ 61-65&9&\geq 60,5&27\\ 66-70&8&\geq 65,5&18\\ 71-75&7&\geq 70,5&10\\ 76-80&3&\geq 75,5&3\\\hline \textrm{Jumlah}&40&&\\\hline \end{array}\Rightarrow \begin{array}{l}\\ \textbf{sebagai penjelasan}\\ 40\\ 33=40-7\\ 27=40-7-6\\ 18=40-7-6-9\\ 10=40-7-6-9-8\\ 3=40-7-6-9-8-7\\ \\ \\ \end{array} \end{array}$

Sumber Referensi

Johanes, Kastolan, dan Sulasim. 2004. Kompetensi Matematika SMA Kelas 2 Semester 1 Program Ilmu Sosial Kurikulum Berbasis Kompetensi 2004. Jakarta: Yudistira.
Kanginan, Marthen, Yuza Terzalgi. 2014. Matematikauntuk SMA-SMK/SMK Kelas XI. Bandung: SEWU.
Sobirin. 2006. Kompas Matematika: Strategi Praktis Menguasai Tes Matematika SMA Kelas 2 IPA. Jakarta: Kawan Pustaka.
Tampomas, Husein. 1999. Seribu Pena Matematika SMU Jilid 2 kelas 2. Jakarta: Erlangga.
Wirodikromo, Sartono. 2007. Matematika untuk SMA Kelas XI. Jakarta: Erlangga.

STATISTIKA (KELAS XII WAJIB K13)

A. Pendahuluan Statistika

$\color{blue}\begin{array}{|c|l|l|}\hline \textrm{No}&\quad\textrm{Istilah}&\qquad\qquad\qquad\qquad\qquad \textrm{Penjelasan}\\\hline 1.&\textrm{Statistika}&\textrm{Cabang ilmu tentang cara mengumpulkan, menyusun}\\ &&\textrm{penyajian, dan penganalisaan dari suatu data}\\\hline 2.&\textrm{Statistik}&\textrm{Data yang telah tersusun ke dalam daftar atau diagram}\\\hline 3.&\textrm{Populasi}&\textrm{Keseluruhan objek dari hasil penelitian yang memenuhi}\\ &&\textrm{syarat tertentu}\\\hline 4.&\textrm{Sampel}&\textrm{Bagian dari populasi yang dapat mewakili seluruh populasi}\\\hline \end{array}$

Sebagai tambahan penjelasan

$\color{blue}\begin{tabular}{|l|p{7.5cm}|}\hline Istilah&Pengertian dan atau Penjelasan\\\hline Statistika&Lihat pengertian di atas\\ Statistik&Hasil pengolahan data.\\ Statistika deskriptif&Statistika baik yang berkenaan dengan kegiatan pengumpulan, penyajian, penyederhanaan atau penganalisaan, serta penentuan khusus dari suatu data tanpa penarikan suatu kesimpulan.\\ populasi&Keseluruhan objek yang akan diteliti.\\ Sampel (Contoh)&Bagian dari populasi yang diamati.\\ Data&Kumpulan dari datum.\\ Datum&Informasi atau catatan keterangan dari penelitian.\\ Data kualitatif&Data yang menunjukkan sifat atau kondisi objek.\\ Data kuantitatif&Data yang menunjukkan jumlah objek.\\ Data ukuran (Data kontinu)&Data yang diperoleh dengan cara mengukur besaran objek.\\ Data cacahan (Data diskrit)&Data yang diperoleh dengan cara mencacah, membilang atau menghitung banyak objek.\\\hline \end{tabular}$

B. Pengumpulan Data

Pengumpulan data dapat dilakukan dengan metode wawancara, pengamatan langsung(observasi) dan bisa juga dengan menggunakan angket.
Setiap keterangan yang diperoleh dalam pengamatan dinamakan datum, dan dari sekumpulan datum inilah nantinya yang disebut data.

C. Penyajian Data

Dalam statika data statistik dapat disajikan dalam berbagai bentuk, mmenyesuaikan jenisnya data. Data statistik dapat berupa daftar bilangan yang memiliki kondisi tertentu sebagai misal berupa data tunggal. Selain data statistik dapat dinyatakan dalam daftar bilangan, data juga bisa dinyatakan dalam bentuk tabel/daftar distribusi frekuensi, atupun diagram.

$\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Penyajian Data}}\\\hline \textrm{Bentuk Diagram}&\textrm{BentukDaftar Distribusi Frekuensi}\\\hline \begin{aligned}\blacklozenge &\: \textrm{Diagram garis}\\ \blacklozenge &\: \textrm{Diagram batang daun}\\ \blacklozenge &\: \textrm{Diagram kotak garis}\\ &\\ & \end{aligned}&\begin{aligned}\blacklozenge &\: \textrm{Daftar distribusi data tunggal}\\ \blacklozenge &\: \textrm{Daftar distribusi frekuensi data berkelompok}\\ \blacklozenge &\: \textrm{Daftar distribusi frekuensi relatif}\\ \blacklozenge &\: \textrm{Daftar distribusi kumulatif}\\ \blacklozenge &\: \textrm{Histogram, poligon frekuensi, dan ogif }\end{aligned}\\\hline \end{array}$ .

Selanjutnya untuk lebih lengkapnya silahkan baca di sini