PAS MATEMATIKA BLOG

Turunan (Kelas XII MIPA)

A. Pendahuluan

Mengenal laju perubahan untuk nilai suatu fungsi

$\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Laju Perubahan}}\\\hline \textrm{Laju erubahan rata-rata}&\textrm{Laju perubahan sesaat}\\\hline \begin{aligned}&\\ &\displaystyle \frac{\Delta y}{\Delta x}=\displaystyle \frac{f\left ( x_{2} \right )-f\left ( x_{1} \right )}{x_{2}-x_{1}}\\ & \end{aligned}&\begin{aligned}&\\ &\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(a+h)-f(a)}{h}\\ & \end{aligned}\\\hline \end{array}$

Perhatikanlah ilustrasi gambar berikut

Misalkan diketahui fungsi $y=f(x)$ terdefinisi untuk semua harga x di sekitar $x=k$ . Jika $\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$ ada, maka bentuk $\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$ disebut turunan dari fungsi $f(x)$ saat $x=k$ .

B. Notasi

Notasi turunan fungsi dilambangkan dengan ${f}'(k)$ dengan ${f}'(k)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$ .
Lambang ${f}'(k)$ dibaca f aksen k desebut turunan atau derivatif untuk fungsi $f(x)$ terhadap $x$ saat $x=k$ .
Jika limitnya ada, dapat dikatakan fungsi $f(x)$ diferensiabel(dapat didiferebsialkan) saat $x=k$ dan bentuk limitnya selanjutna dilambangkan dengan ${f}'(k)$ .
Misalkan fungsi $f(x)$ memiliki turunan ${f}'(x)$ . Jika ${f}'(k)$ tidak terdefinisi maka $f(x)$ tidak diferensiabel di $x=k$ .

Bentuk umum turunan fungsi ang selanjutnya disebut juga turunan pertama fungsi $y$ terhadap $x$ dapat dinotasikan dengan berbagai bentuk berikut yaitu:

$\color{blue}{y}\: '=\displaystyle \frac{dy}{dx}=\displaystyle \frac{d\left ( f(x) \right )}{dx}={f}'(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}$

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

$\begin{array}{ll}\\ 1.&\textrm{Jika} \: \: g(x)=3x-5, \textrm{hitunglah laju perubahan fungsi}\: \: g\: \: \textrm{di}\: \: x=2\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{g(x)=3x-5}\\\hline \textrm{Cara Pertama}&\textrm{Cara Kedua}\\\hline \begin{aligned} g(x)&=3x-5\\ g(2)&=3.2-5=1\\ {g}'(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{g(x)-g(2)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{(3x-5)-(1)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{3x-6}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle 3\\ &=3 \end{aligned}&\begin{aligned}g(2)\: \,&=1\\ g(2+&h)=3(2+h)-5=3h+1\\ g(2+&h)-g(2)=3h\\ {g}'(2)\: &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{g(2+h)-g(2)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle\frac{3h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 3\\ &=3\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{Jadi, laju perubahan fungsi}\: \: g\: \: \textrm{di}\: \: x=2\: \: \textrm{adalah 3}}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: f(x)=2017x^{2},\: \: \textrm{tentukanlah}\: \: {f}'(x)\: \: \textrm{dan}\: \: {f}'(1)\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline {f}'(x)&{f}'(1)\\\hline \begin{aligned}f(x)&=2017x^{2}\\ f(x+h)&=2017(x+h)^{2}\\ &=2017\left ( x^{2}+2xh+h^{2} \right )\\ &=2017x^{2}+4034xh+2017h^{2}\\ {f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (2017x^{2}+4034xh+2017h^{2} \right )-\left (2017x^{2} \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4034xh+2017h^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 4034x+2017h\\ &=4034x \end{aligned}&\begin{aligned}{f}'(x)&=4034x\\ \textrm{maka},&\\ {f}'(1)&=4034.1\\ &=4034\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui bahwa fungsi}\: \: f(x)=\displaystyle \frac{1}{x^{2}}\: \: \textrm{dengan daerah asal}\: \: \textrm{D}_{f}=\left \{ x\: |\: x\in \mathbb{R},\: \: \textrm{dan}\: \: x\neq 0 \right \}.\\ &\textrm{a}.\quad \textrm{Tunjukkan bahwa}\: \: {f}'(a)=\displaystyle -\frac{2}{a^{3}}\\ &\textrm{b}.\quad \textrm{jelaskanlah mengapa}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|l|l|}\hline \begin{aligned}{f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{(x+h)^{2}}-\displaystyle \frac{1}{x^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}-(x+h)^{2}}{\left (x(x+h) \right )^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-\left ( x^{2}+2xh+h^{2} \right )}{h\left ( x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2xh-h^{2}}{h\left (x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: -\displaystyle \frac{2x+h}{\left (x(x+h) \right )^{2}}\\ &=-\displaystyle \frac{2x}{x^{4}}\\ &=-\displaystyle \frac{2}{x^{3}} \end{aligned}&\begin{aligned}\textrm{Untuk}&\: \textrm{jawaban poin a dan b adalah sebagai berikut}\\ {f}'(x)&=-\displaystyle \frac{2}{x^{3}}\\ {f}'(a)&=-\displaystyle \frac{2}{a^{3}}\\ \textrm{maka},&\\ {f}'(0)&=-\displaystyle \frac{2}{0^{3}}\\ &=-\displaystyle \frac{2}{0}\\ &\\ &\textrm{karena penyebut berupa bilangan}\: \: 0\\ &\textrm{maka}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}$

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:

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