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.