Contoh Soal Trigonometri (XI MIPA K13 Revisi)

\color{blue}\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}.

\color{magenta}\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 \color{black}\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 \color{black}\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 kita kita tampilkan tabel sudut ostomewa berikut

\color{red}\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}.

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