Lanjutan: Rumus Jumlah dan Selisih (Kelas XI Peminatan Mat & IA K13 Revisi)

B. Rumus Trigonometri Jumlah dan Selisih Dua Sudut

Arsip lama

  • di sini 1  (dari contoh pengembangan soal materi KTSP kelas x smt 1)
  • di sini 2 (dari materi KTSP kelas XI smt 1)

Sebagai pengingat saja

\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}.

C. 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}.

Sedikit gambaran untuk rumus no. 3, yaitu :

\begin{array}{|cll|c|}\hline &\begin{aligned}\cos \left ( \alpha +\gamma \right )&=\cos \alpha \cos \beta -\sin \alpha \sin \gamma \\ \cos \left ( \alpha -\gamma \right )&=\cos \alpha \cos \beta +\sin \alpha \sin \gamma \\ \end{aligned}&&\\\cline{2-2} &&+&\textrm{Misalkan}\\ &\begin{aligned}\cos \left ( \alpha +\gamma \right )+\cos \left ( \alpha -\gamma \right )&=2\cos \alpha \cos \gamma \end{aligned}&&X=\alpha +\gamma \\ &&&Y=\alpha -\gamma\\ &\textbf{Proses perubahannya adalah sebagai berikut}&&\\\cline{1-3} &\underset{\textrm{perhatikan}}{\underbrace{\begin{matrix} \begin{array}{ccc}\\ &X=\alpha +\gamma &\\ &Y=\alpha -\gamma &+\\\cline{2-2}\\ &X+Y=2\alpha &\\ &2\alpha =X+Y&\\ &\alpha =\displaystyle \frac{X+Y}{2}& \end{array} & \textrm{dan} & \begin{array}{ccc}\\ &X=\alpha +\gamma &\\ &Y=\alpha -\gamma &-\\\cline{2-2}\\ &X-Y=2\gamma &\\ &2\gamma =X-Y&\\ &\gamma =\displaystyle \frac{X-Y}{2}& \end{array} \end{matrix}}} &&\\\cline{1-3} &\textbf{Sehingga rumus akan menjadi}&&\\ &\cos X+\cos Y =2\cos \displaystyle \frac{\left ( X+Y \right )}{2}\cos \frac{\left ( X- Y\right )}{2}&&\\\hline \end{array}.

Untuk rumus yang lain diperoleh dengan cara yang semisal ilustrasi di atas tentunya dengan penyesuaian.

D. Rumus Perkalian Trigonometri

\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}.

E. Persamaan Trigonometri Bentuk  a\sin x+b\cos x=k\cos \left ( x-\alpha \right ).

Perhatikanlah bentuk berikut

\begin{aligned}a\sin x+b\cos x&=k\cos \left ( x-\alpha \right )\\ a\sin x+b\cos x&=k\left ( \cos x\cos \alpha +\sin x \sin \alpha \right )\\ a\sin x+b\cos x&=k\cos x\cos \alpha +k\sin x \sin \alpha\\ \textrm{sehingga didap}&\textrm{at kesamaan yang menghasilkan}\\ \bullet \quad a&=k\cos \alpha \\ \bullet \quad b&=k\sin \alpha \\ \textrm{Selanjutnya jik}&\textrm{a dikuadratkan masing-masing lalu dijumlahkan}\\ a^{2}+b^{2}&=k^{2}\cos ^{2}\alpha +k^{2}\sin ^{2}\alpha \\ a^{2}+b^{2}&=k^{2}\left ( \sin ^{2}\alpha +\cos ^{2}\alpha \right )\\ a^{2}+b^{2}&=k^{2} \\ k^{2}&=a^{2}+b^{2}\\ k&=\sqrt{a^{2}+b^{2}} , \textrm{ambil}\: \: k\: \: \textrm{positif saja}\\ \textrm{Selanjutnya jik}&\textrm{a}\\ \displaystyle \frac{k\sin \alpha }{k\cos \alpha }&=\displaystyle \frac{b}{a}\\ \tan \alpha &=\displaystyle \frac{b}{a}\\ &\begin{array}{|c|c|c|c|}\hline \textrm{Nilai}\: \: a\: \: \textrm{dan}\: \: b&\displaystyle \frac{b}{a}&\tan \alpha &\textrm{Kuadran sudut}\: \: \alpha \\\hline a> 0\: \: \textrm{dan}\: \: b> 0&> 0&> 0&\textrm{I}\\ a< 0\: \: \textrm{dan}\: \: b> 0& < 0&< 0&\textrm{II}\\ a< 0\: \: \textrm{dan}\: \: b< 0&> 0&> 0&\textrm{III}\\ a> 0\: \: \textrm{dan}\: \: b< 0&< 0&< 0&\textrm{IV}\\\hline \end{array} \end{aligned}.

\begin{array}{|c|}\hline \begin{aligned}&\\ \textbf{CATATAN}:&\\ \textrm{Bentuk}\: &\\ &a\cos x^{\circ}+b\sin x^{\circ}\\ \textrm{dapat d}&\textrm{inyatakan dalam tipe}\\ &\begin{cases} k \cos \left ( x+\alpha \right )^{\circ} \\ k \cos \left ( x-\alpha \right )^{\circ} \quad\Rightarrow \qquad \textrm{yang sedang kita bahas di atas}\\ k \sin \left ( x+\alpha \right )^{\circ} \\ k \sin \left ( x-\alpha \right )^{\circ} \end{cases}\\ & \end{aligned}\\\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:

Sebagai pengingat

\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}.

\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}.

Jawab:

\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}.

\begin{array}{ll}\\ 4.&\textrm{Carilah nilai}\: \: k\: \: \textrm{dan}\: \: \: \alpha \: \: \left ( k>0,\: \: 0^{\circ}\leq \alpha \leq 360^{\circ} \right )\: \: \textrm{untuk tiap pasangan persaamaan berikut ini}!\\ &\begin{array}{llll}\\ \textrm{a}.&k\cos \alpha =2\: ;\: k\sin \alpha =2&\textrm{f}.&k\cos \alpha =-\sqrt{3}\: ;\: k\sin \alpha =1\\ \textrm{b}.&k\cos \alpha =2\: ;\: k\sin \alpha =-2&\textrm{g}.&k\cos \alpha =-\sqrt{3}\: ;\: k\sin \alpha =-1\\ \textrm{c}.&k\cos \alpha =-2\: ;\: k\sin \alpha =2&\textrm{h}.&k\cos \alpha =1\: ;\: k\sin \alpha =\sqrt{3}\\ \textrm{d}.&k\cos \alpha =-2\: ;\: k\sin \alpha =-2&\textrm{h}.&k\cos \alpha =1\: ;\: k\sin \alpha =-\sqrt{3}\\ \textrm{e}.&k\cos \alpha =3\: ;\: k\sin \alpha =\sqrt{3}&\textrm{h}.&k\cos \alpha =3\: ;\: k\sin \alpha =-\sqrt{3} \end{array} \end{array}.

Jawab:

\begin{array}{|c|c|c|c|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\begin{cases} b & =y=k\sin \alpha \\ a & =x=k\cos \alpha \end{cases}\\ & \end{aligned}}\\\hline \textrm{No}&\multicolumn{2}{|c|}{\tan \alpha =\displaystyle \frac{b}{a}}&k=\sqrt{b^{2}+a^{2}}\\\hline 4.\textrm{a}&\begin{aligned}\tan \alpha &=\displaystyle \frac{2}{2}=1\\ &(\textrm{kuadran 1}) \end{aligned}&\begin{aligned}\tan \alpha &=1=\tan 45^{\circ}\\ \alpha &=45^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{2^{2}+2^{2}}=\sqrt{2.2^{2}}=2\sqrt{2}\end{aligned}\\\hline 4.\textrm{b}&\begin{aligned}\tan \alpha &=\displaystyle \frac{-2}{2}=-1\\ &(\textrm{kuadran 4}) \end{aligned}&\begin{aligned}\tan \alpha &=-1=\tan \left ( -45^{\circ} \right )\\ &=\tan \left ( 360^{\circ}-45^{\circ} \right )=\tan 315^{\circ}\\ \alpha &=315^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(-2)^{2}+2^{2}}=\sqrt{2.2^{2}}=2\sqrt{2}\end{aligned}\\\hline 4.\textrm{c}&\begin{aligned}\tan \alpha &=\displaystyle \frac{2}{-2}=-1\\ &(\textrm{kuadran 2}) \end{aligned}&\begin{aligned}\tan \alpha &=-1=\tan \left ( -45^{\circ} \right )\\ &=\tan \left ( 180^{\circ}-45^{\circ} \right )=\tan 135^{\circ}\\ \alpha &=135^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{2^{2}+(-2)^{2}}=\sqrt{2.2^{2}}=2\sqrt{2}\end{aligned}\\\hline 4.\textrm{d}&\begin{aligned}\tan \alpha &=\displaystyle \frac{-2}{-2}=1\\ &(\textrm{kuadran 3}) \end{aligned}&\begin{aligned}\tan \alpha &=1=\tan \left ( 45^{\circ} \right )\\ &=\tan \left ( 180^{\circ}+45^{\circ} \right )=\tan 225^{\circ}\\ \alpha &=225^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{2.2^{2}}=2\sqrt{2}\end{aligned}\\\hline  \end{array}.

\begin{array}{|c|c|c|c|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\begin{cases} b & =y=k\sin \alpha \\ a & =x=k\cos \alpha \end{cases}\\ & \end{aligned}}\\\hline \textrm{No}&\multicolumn{2}{|c|}{\tan \alpha =\displaystyle \frac{b}{a}}&k=\sqrt{b^{2}+a^{2}}\\\hline 4.\textrm{e}&\begin{aligned}\tan \alpha &=\displaystyle \frac{\sqrt{3}}{3}\\ &(\textrm{kuadran 1}) \end{aligned}&\begin{aligned}\tan \alpha &=\displaystyle \frac{\sqrt{3}}{3}=\tan \left ( 30^{\circ} \right )\\ \alpha &=30^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(\sqrt{3})^{2}+(3)^{2}}=\sqrt{3+9}\\ &=\sqrt{12}=\sqrt{4.3}=\sqrt{2^{2}.3}=2\sqrt{3}\end{aligned}\\\hline 4.\textrm{f}&\begin{aligned}\tan \alpha &=\displaystyle \frac{1}{-\sqrt{3}}\\ &(\textrm{kuadran 2}) \end{aligned}&\begin{aligned}\tan \alpha &=-\displaystyle \frac{\sqrt{3}}{3}=\tan \left ( -30^{\circ} \right )\\ \tan \alpha &=\tan \left ( 180^{\circ}-30^{\circ} \right )\\ &=\tan 150^{\circ}\\ \alpha &=150^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(1)^{2}+\left ( -\sqrt{3} \right )^{2}}=\sqrt{1+3}\\ &=\sqrt{4}=\sqrt{2^{2}}=2\end{aligned}\\\hline 4.\textrm{g}&\begin{aligned}\tan \alpha &=\displaystyle \frac{-1}{-\sqrt{3}}\\ &(\textrm{kuadran 3}) \end{aligned}&\begin{aligned}\tan \alpha &=\displaystyle \frac{\sqrt{3}}{3}=\tan \left ( 30^{\circ} \right )\\ \tan \alpha &=\tan \left ( 180^{\circ}+30^{\circ} \right )\\ &=\tan 210^{\circ}\\ \alpha &=210^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(-1)^{2}+\left ( -\sqrt{3} \right )^{2}}=\sqrt{1+3}\\ &=\sqrt{4}=\sqrt{2^{2}}=2\end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ 5.&\textrm{Tunjukkan bahwa}:\\ &\begin{array}{llll}\\ \textrm{a}.&3\cos x^{\circ} -3\sin x^{\circ} =3\sqrt{2}\cos \left ( x^{\circ}-\displaystyle \frac{7}{4}\pi \right )\\ \textrm{b}.&3\cos x^{\circ} -3\sin x^{\circ} =3\sqrt{2}\cos \left ( x^{\circ}+\displaystyle \frac{1}{4}\pi \right )\\ \textrm{c}.&3\cos x^{\circ} -3\sin ^{\circ} =3\sqrt{2}\sin \left ( x^{\circ}-\displaystyle \frac{5}{4}\pi \right )\\ \textrm{d}.&3\cos x^{\circ} -3\sin x^{\circ} =3\sqrt{2}\sin \left ( x^{\circ}+\displaystyle \frac{3}{4}\pi \right )\\ \end{array} \end{array}.

Bukti:

\begin{aligned}\textrm{a}.\quad\textrm{Diketahui bahwa}:&\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos \left ( x^{\circ}-\alpha \right )^{\circ}\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos x^{\circ}\cos \alpha +k\sin x^{\circ}\sin \alpha \\ \textrm{diperoleh persam}&\textrm{aan}\begin{cases} b=y=k\sin \alpha & =-3\\ a=x=k\cos \alpha & =3 \end{cases}\\ k&=\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+(-3)^{2}}=\sqrt{2.3^{2}}=3\sqrt{2}\\ \tan \alpha &=\displaystyle \frac{y}{x}=\displaystyle \frac{-3}{3}=-1\quad (\textbf{kuadran 4})\\ &=\tan \left ( -45^{\circ} \right )=\tan \left ( -\displaystyle \frac{1}{4}\pi \right )=\tan \left (2 \pi -\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{7}{4}\pi \\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos \left ( x-\alpha \right )^{\circ}=3\sqrt{2}\cos \left ( x^{\circ}-\displaystyle \frac{7}{4}\pi \right )\qquad \blacksquare \end{aligned}.

\begin{aligned}\textrm{b}.\quad\textrm{Diketahui bahwa}:&\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos \left ( x^{\circ}+\alpha \right )^{\circ}\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos x^{\circ}\cos \alpha -k\sin x^{\circ}\sin \alpha \\ \textrm{diperoleh persam}&\textrm{aan}\begin{cases} b=y=k\sin \alpha & =3\\ a=x=k\cos \alpha & =3 \end{cases}\\ k&=\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+3^{2}}=\sqrt{2.3^{2}}=3\sqrt{2}\\ \tan \alpha &=\displaystyle \frac{y}{x}=\displaystyle \frac{3}{3}=1\quad (\textbf{kuadran 1})\\ &=\tan \left ( 45^{\circ} \right )=\tan \left ( \displaystyle \frac{1}{4}\pi \right )\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos \left ( x+\alpha \right )^{\circ}=3\sqrt{2}\cos \left ( x^{\circ}+\displaystyle \frac{1}{4}\pi \right )\qquad \blacksquare \end{aligned}.

\LARGE{\fbox{\LARGE{\fbox{LATIHAN SOAL}}}}.

Yang belum dibahas dan atau belum ditunjukkan buktinya, silahkan gunakan sebagai latihan mandiri

Sumber Referensi

  1. Kurnia, N.,  dkk. 2017. Jelajah Matematika 2 SMA Kelas XI Peminatan MIPA(Edisi Revisi 2016). Jakarta: Yudistira.
  2. 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.
  3. Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 5 Kelas 3. Jakarta: Erlangga.

 

Tentang ahmadthohir1089

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