Lanjutan 4 Persamaan Trigonometri

B. 2 Persamaan Trigonometri Bentuk Kuadrat

Persamaan trigonometri terkadang juga terdapat dalam bentuk kuadrat, sehingga penyelesaiannya menyesuaikan dengan persamaan kuadrat tersebut yaitu proses faktorisasi, atau melengkapkan kudrat sempurna,dan atau dengan rumus ABC.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \sin x-2{\sin }^{2}x=0\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & \sin x-2{\sin }^{2}x=0(\mathbf{\text{lalu difaktorkan}})\\ & \sin x(1-2\sin x)=0\\ & \sin x=0\text{atau}1-2\sin x=0\\ & \text{selanjutnya}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\sin x& =0\\ \sin x& =\sin 0^{\circ }\\ x& =0^{\circ }+k{.360}^{\circ }\\ & \text{atau}\\ x& ={180}^{\circ }+k{.360}^{\circ }\\ \text{saat}& k=0\\ x& =0^{\circ }\text{dan}{180}^{\circ }\\ \text{saat}& k=1\\ x& ={360}^{\circ }\text{dan}{540}^{\circ }\end{array}& \begin{array}{rl}\sin x& ={\displaystyle \frac{1}{2}}\\ \sin x& =\sin {30}^{\circ }\\ x& ={30}^{\circ }+k{.360}^{\circ }\\ & \text{atau}\\ x& =({180}^{\circ }-{30}^{\circ })+k{.360}^{\circ }\\ & ={150}^{\circ }+k{.360}^{\circ }\\ \text{saat}& k=0\\ x& ={30}^{\circ }\text{dan}{150}^{\circ }\\ \text{saat}& k=1\\ x& ={390}^{\circ }\text{dan}{510}^{\circ }\end{array}\\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{0^{\circ },{30}^{\circ },{150}^{\circ },{180}^{\circ },{360}^{\circ }\}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah himpunan penyelesaian dari}\\ & 2{\tan }^2\theta -\sec \theta +1=0\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & {\tan }^2\theta -\sec \theta +1=0\\ & 2({\sec }^2\theta -1)-\sec \theta +1=0\\ & 2{\sec }^2\theta -\sec \theta -1=0(\mathbf{\text{lalu difaktorkan}})\\ & (2\sec \theta +1)(\sec \theta -1)=0\\ & (2\sec \theta +1)=0\text{atau}(\sec \theta -1)=0\\ & \sec \theta =-{\displaystyle \frac{1}{2}\text{atau}\sec \theta =1}\\ & {\displaystyle \frac{1}{\cos \theta }=-\frac{1}{2}\text{atau}{\displaystyle \frac{1}{\cos \theta }=1}}\\ & \cos \theta =-2(\mathbf{\text{tidak mungkin}})\text{atau}\cos \theta =1\\ & \text{selanjutnya}\\ & \cos \theta =1\iff \cos \theta =\cos 0^{\circ }\\ & \iff \theta =\pm 0^{\circ }+k{.360}^{\circ }\iff \theta =k{.360}^{\circ }\\ & k=0\Rightarrow x=0^{\circ }\\ & k=1\Rightarrow x={360}^{\circ }\\ & k=2\Rightarrow x={720}^{\circ }\text{tidak memenuhi}\\ & \mathbf{\text{HP}}=\{0^{\circ },{360}^{\circ }\}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah himpunan penyelesaian dari}\\ & 5{\cos }^2\beta +3\cos \beta =2\text{untuk}{0}^{\circ }\le \beta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & 5{\cos }^2\beta +3\cos \beta =2\\ & 5{\cos }^2\beta +3\cos \beta -2=0(\mathbf{\text{lalu difaktorkan}})\\ & (5\cos \beta -2)(\cos \beta +1)=0\\ & (5\cos \beta -2)=0\text{atau}(\cos \beta +1)=0\\ & 5\cos \beta -2=0\text{atau}\cos \beta +1=0\\ & \cos \beta ={\displaystyle \frac{2}{5}\text{atau}\cos \beta =-1}\\ & \cos \beta =\cos 66,4^{\circ }\text{atau}\cos \beta ={180}^{\circ }\\ & \text{selanjutnya}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\beta & =\pm 66,4^{\circ }+k{.360}^{\circ }\\ k=0& \Rightarrow \beta =66,4^{\circ }\text{atau}\\ & \beta =-66,4^{\circ }(\text{tm})\\ k=1& \Rightarrow \beta =426,4^{\circ }(\text{tm})\\ & \text{atau}\beta =293,6^{\circ }\\ k=2& \Rightarrow \beta \text{tidak ada }\\ & \text{yang memenuhi}\end{array}& \begin{array}{rl}\beta & =\pm {180}^{\circ }+k{.360}^{\circ }\\ k=0& \Rightarrow \beta ={180}^{\circ }\text{atau}\\ & \beta =-{180}^{\circ }(\text{tm})\\ k=1& \Rightarrow \beta ={540}^{\circ }(\text{tm})\\ & \text{atau}\beta ={180}^{\circ }\\ k=2& \Rightarrow \beta \text{tidak ada }\\ & \text{yang memenuhi}\end{array}\\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{66,4^{\circ },{180}^{\circ },293,6^{\circ }\}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah himpunan penyelesaian dari}\\ & 2{\sin }^2\gamma +3\cos \gamma =3\text{untuk}{0}^{\circ }\le \gamma \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & 2{\sin }^2\gamma +3\cos \gamma =3\\ & 2(1-{\cos }^2\gamma )+3\cos \gamma -3=0\\ & 2-2{\cos }^2\gamma +3\cos \gamma -3=0\\ & -2{\cos }^2\gamma +3\cos \gamma -1=0\\ & 2{\cos }^2\gamma -3\cos \gamma +1=0(\mathbf{\text{lalu difaktorkan}})\\ & (2\cos \gamma -1)(\cos \gamma -1)=0\\ & (2\cos \gamma -1)=0\text{ atau}(\cos \gamma -1)=0\\ & \cos \gamma ={\displaystyle \frac{1}{2}\text{atau}\cos \gamma =1}\\ & \cos \gamma =\cos {60}^{\circ }\text{atau}\cos \gamma =0^{\circ }\\ & \text{selanjutnya}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\gamma & =\pm {60}^{\circ }+k{.360}^{\circ }\\ k=0& \Rightarrow \gamma ={60}^{\circ }\text{atau}\\ & \gamma =-{60}^{\circ }(\text{tm})\\ k=1& \Rightarrow \gamma ={420}^{\circ }(\text{tm})\\ & \text{atau}\gamma ={300}^{\circ }\\ k=2& \Rightarrow \gamma \text{tidak ada }\\ & \text{yang memenuhi}\end{array}& \begin{array}{rl}\gamma & =\pm 0^{\circ }+k{.360}^{\circ }\\ \gamma & =0^{\circ }+k{.360}^{\circ }\\ k=0& \Rightarrow \gamma =0^{\circ }\\ k=1& \Rightarrow \gamma ={360}^{\circ }\\ k=2& \Rightarrow \gamma \text{tidak ada}\\ & \text{yang memenuhi}\\ & ⋮\end{array}\\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{0^{\circ },{60}^{\circ },{300}^{\circ },{360}^{\circ }\}\end{array}$

B. 3 Persamaan Trigonometri Bentuk a sin x + b cos x 

Selain bentuk sederhana seperti yang telah diuraikan pada materi sebelumnya (lihat di sini), terdapat persamaan trigonometri bentuk  $a\sin x+b\cos x$. Bentuk $a\sin x+b\cos x$ ini dalam penyelesaiannya diubah ke dalam bentuk  $k\cos (x-\alpha )$. Adapun untuk menemukan pembuktian dari kesamaan rumus ini, Anda harus mempelajari materi rumus trigonometri jumlah dan selisih dua sudut.

$\begin{array}{rl}a\sin x& +b\cos x=k\cos (x-\theta )\\ \text{denga}& \text{n}:\\ & k=\sqrt{a^2+b^2}\\ & \tan \theta ={\displaystyle \frac{a}{b}}\\ & (a>0\text{dan}b>0,\text{maka}\theta \text{di kuadran I})\\ & (a>0\text{dan}b<0,\text{maka}\theta \text{di kuadran II})\\ & (a<0\text{dan}b<0,\text{maka}\theta \text{di kuadran III})\\ & (a<0\text{dan}b>0,\text{maka}\theta \text{di kuadran IV})\\ \\ & \text{dengan}a\text{pada sumbu Y dan}\\ & b\text{pada sumbu X}\end{array}$

$\begin{array}{rl}& \mathbf{\text{Dan ingat juga tabel nilai tangen}}\\ & \mathbf{\text{berikut}}\\ & \begin{array}{|cccccc|}\hline \theta & 0^{\circ }& {30}^{\circ }& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }\\ & & & & & \\ \tan \theta & 0& {\displaystyle \frac{1}{3}\sqrt{3}}& 1& \sqrt{3}& \mathbf{\text{TD}}\\ & & & & & \\ \theta & {120}^{\circ }& {135}^{\circ }& {150}^{\circ }& {180}^{\circ }& \\ & & & & & \\ \tan \theta & -\sqrt{3}& -1& -{\displaystyle \frac{1}{3}\sqrt{3}}& 0& \\ & & & & & \\ \hline\end{array}\end{array}$.

$\begin{array}{rl}& \begin{array}{|cccccc|}\hline \theta & {180}^{\circ }& {210}^{\circ }& {225}^{\circ }& {240}^{\circ }& {270}^{\circ }\\ & & & & & \\ \tan \theta & 0& {\displaystyle \frac{1}{3}\sqrt{3}}& 1& \sqrt{3}& \mathbf{\text{TD}}\\ & & & & & \\ \theta & {300}^{\circ }& {315}^{\circ }& {345}^{\circ }& {360}^{\circ }& \\ & & & & & \\ \tan \theta & -\sqrt{3}& -1& -{\displaystyle \frac{1}{3}\sqrt{3}}& 0& \\ & & & & & \\ \hline\end{array}\end{array}$.

Untuk lebih lanjut tentang bukti dan lain sebagainya akan dipelajari di subbab berikutnya setelah materi persamaan trigonometri ini.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \sin x+\sqrt{3}\cos x=2\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & \sin x+\sqrt{3}\cos x=2(\mathbf{\text{ingat}}:a=1,b=\sqrt{3})\\ & \sin x+\sqrt{3}\cos x=k\cos (x-\theta )=2\\ & \{\begin{array}{ll}k& =\sqrt{1^2+{\left(\sqrt{3}\right)}^2}=\sqrt{4}=2\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{1}{\sqrt{3}}={\displaystyle \frac{1}{3}\sqrt{3}\Rightarrow \theta ={30}^{\circ }}}}\end{array}\\ & \text{sudut}\theta \text{di kuadran I, karena}a,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& \sin x+\sqrt{3}\cos x=k\cos (x-\theta )=2\\ & \iff 2\cos (x-{30}^{\circ })=2\\ & \iff \cos (x-{30}^{\circ })=1\\ & \iff \cos (x-{30}^{\circ })=\cos 0^{\circ }\\ & \iff x-{30}^{\circ }=\pm 0^{\circ }+k{.360}^{\circ }\\ & \iff x={30}^{\circ }\pm 0^{\circ }+k{.360}^{\circ }\\ & \iff x={30}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x={30}^{\circ }(\text{memenuhi})\\ & k=1\Rightarrow x={390}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\left\{{30}^{\circ }\right\}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \sin x-\sqrt{3}\cos x=\sqrt{2}\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & \sin x-\sqrt{3}\cos x=\sqrt{2}(\mathbf{\text{ingat}}:a=1,b=-\sqrt{3})\\ & \sin x-\sqrt{3}\cos x=k\cos (x-\theta )=\sqrt{2}\\ & \{\begin{array}{ll}k& =\sqrt{1^2+{(-\sqrt{3})}^2}=\sqrt{4}=2\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{1}{-\sqrt{3}}=-{\displaystyle \frac{1}{3}\sqrt{3}\Rightarrow \theta ={150}^{\circ }}}}\end{array}\\ & \text{sudut}\theta \text{di kuadran II, karena}a>0,b<0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& \sin x-\sqrt{3}\cos x=k\cos (x-\theta )=\sqrt{2}\\ & \iff 2\cos (x-{150}^{\circ })=\sqrt{2}\\ & \iff \cos (x-{150}^{\circ })={\displaystyle \frac{\sqrt{2}}{2}={\displaystyle \frac{1}{2}\sqrt{2}}}\\ & \iff \cos (x-{150}^{\circ })=\cos {45}^{\circ }\\ & \iff x-{150}^{\circ }=\pm {45}^{\circ }+k{.360}^{\circ }\\ & \iff x={150}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x={150}^{\circ }+{45}^{\circ }={195}^{\circ }(\text{mm})\text{atau}\\ & x={150}^{\circ }-{45}^{\circ }={105}^{\circ }(\text{mm})\\ & k=1\Rightarrow x={150}^{\circ }\pm {45}^{\circ }+{360}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\{{105}^{\circ },{195}^{\circ }\}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \sqrt{6}\sin x+\sqrt{2}\cos x=2\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & \sqrt{6}\sin x+\sqrt{2}\cos x=2(\mathbf{\text{ingat}}:a=\sqrt{6},b=\sqrt{2})\\ & \sqrt{6}\sin x+\sqrt{2}\cos x=k\cos (x-\theta )=2\\ & \{\begin{array}{ll}k& =\sqrt{{\left(\sqrt{6}\right)}^2+{\left(\sqrt{2}\right)}^2}=\sqrt{8}=2\sqrt{2}\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{\sqrt{6}}{\sqrt{2}}=\sqrt{3}\Rightarrow \theta ={60}^{\circ }}}\end{array}\\ & \text{sudut}\theta \text{di kuadran I, karena}a>0,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& \sqrt{6}\sin x+\sqrt{2}\cos x=k\cos (x-\theta )=2\\ & \iff 2\sqrt{2}\cos (x-{60}^{\circ })=2\\ & \iff \cos (x-{60}^{\circ })={\displaystyle \frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}={\displaystyle \frac{1}{2}\sqrt{2}}}\\ & \iff \cos (x-{60}^{\circ })=\cos {45}^{\circ }\\ & \iff x-{60}^{\circ }=\pm {45}^{\circ }+k{.360}^{\circ }\\ & \iff x={60}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x={60}^{\circ }+{45}^{\circ }={105}^{\circ }(\text{mm})\text{atau}\\ & x={60}^{\circ }-{45}^{\circ }={15}^{\circ }(\text{mm})\\ & k=1\Rightarrow x={60}^{\circ }\pm {45}^{\circ }+{360}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\{{15}^{\circ },{105}^{\circ }\}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \cos x-\sqrt{3}\sin x=1\text{untuk}{0}^{\circ }\le \theta \le {360}^{\circ }\\ \\ & \text{Jawab}:\\ & -\sqrt{3}\sin x+\cos x=1(\mathbf{\text{ingat}}:a=-\sqrt{3},b=1)\\ & -\sqrt{3}\sin x+\cos x=k\cos (x-\theta )=1\\ & \{\begin{array}{ll}k& =\sqrt{{(-\sqrt{3})}^2+{\left(1\right)}^2}=\sqrt{4}=2\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{-\sqrt{3}}{1}=-\sqrt{3}\Rightarrow \theta ={300}^{\circ }}}\end{array}\\ & \text{sudut}\theta \text{di kuadran IV, karena}a<0,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& -\sqrt{3}\sin x+\cos x=k\cos (x-\theta )=1\\ & \iff 2\cos (x-{300}^{\circ })=1\\ & \iff \cos (x-{300}^{\circ })={\displaystyle \frac{1}{2}}\\ & \iff \cos (x-{300}^{\circ })=\cos {60}^{\circ }\\ & \iff x-{300}^{\circ }=\pm {60}^{\circ }+k{.360}^{\circ }\\ & \iff x={300}^{\circ }\pm {60}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x={300}^{\circ }+{60}^{\circ }={360}^{\circ }=0^{\circ }(\text{mm})\text{atau}\\ & x={300}^{\circ }-{60}^{\circ }={240}^{\circ }(\text{mm})\\ & k=1\Rightarrow x={300}^{\circ }\pm {60}^{\circ }+{360}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\{0^{\circ },{240}^{\circ },{360}^{\circ }\}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiasyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Noormandiri, B. K. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
3. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
4. Sukino. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Lanjutan 3 Persamaan Trigonometri

 B. 1 Persamaan Trigonometri Sederhana

Dalam penyelesaian persamaan trigonometri sederhana dapat digunakan salah satu rumus berikut, yaitu:

$\begin{array}{rl}(1).\sin x& =\sin {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k{.360}^{\circ }\\ \text{atau}\\ x=({180}^{\circ }-{\alpha }^{\circ })+k{.360}^{\circ }\end{array}\\ (2).\cos x& =\cos {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k{.360}^{\circ }\\ \text{atau}\\ x=-{\alpha }^{\circ }+k{.360}^{\circ }\end{array}\\ (3).\tan x& =\tan {\alpha }^{\circ }\\ & x={\alpha }^{\circ }+k{.180}^{\circ }\end{array}$.

Jika sudutnya dinyatakan dalam phi radian $(\pi \text{dibaca}:phi)$, maka persamaan trigonometri sederhananya adalah:

$\begin{array}{rl}(1).\sin x& =\sin {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k.2\pi \\ \text{atau}\\ x=(\pi -{\alpha }^{\circ })+k.2\pi \end{array}\\ (2).\cos x& =\cos {\alpha }^{\circ }\\ & \{\begin{array}{c}x={\alpha }^{\circ }+k.2\pi \\ \text{atau}\\ x=-{\alpha }^{\circ }+k.2\pi \end{array}\\ (3).\tan x& =\tan {\alpha }^{\circ }\\ & x={\alpha }^{\circ }+k.\pi \end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah akar-akar persamaan trigonometri}\\ & \text{berikut dan tentukan pula himpunan}\\ & \text{penyelesaiannya untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \text{a}.\sin x=\sin {50}^{\circ }\\ & \text{b}.\cos x=\cos {50}^{\circ }\\ & \text{c}.\tan x=\tan {50}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}.\text{a}.\sin x& =\sin {50}^{\circ }\\ x& =\{\begin{array}{ll}{50}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{50}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{50}^{\circ }(\text{memenuhi})& \\ {130}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{50}^{\circ },{130}^{\circ }\}\end{array}\\ & \begin{array}{rl}.\text{b}.\cos x& =\cos {50}^{\circ }\\ x& =\{\begin{array}{ll}{50}^{\circ }& +k{.360}^{\circ }\\ -{50}^{\circ }& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{50}^{\circ }(\text{memenuhi})& \\ -{50}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{50}^{\circ }+{360}^{\circ }={410}^{\circ }(\text{tidak memenuhi})& \\ -{50}^{\circ }+{360}^{\circ }={310}^{\circ }(\text{memenuhi})& \end{array}\\ \text{HP}& =\{{50}^{\circ },{310}^{\circ }\}\end{array}\\ & \begin{array}{rl}.\text{c}.\tan x& =\tan {50}^{\circ }\\ x& ={50}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={50}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={50}^{\circ }+{180}^{\circ }={230}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={50}^{\circ }+{360}^{\circ }={410}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{50}^{\circ },{230}^{\circ }\}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah himpunan penyelesaian dari }\\ & \text{persamaan-persamaan trigonometri berikut}\\ & \text{ini untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \begin{array}{l}\\ \text{a}.& \sin x={\displaystyle \frac{1}{2}}& \text{f}.& \tan x=-{\displaystyle \frac{1}{3}\sqrt{3}}& \text{k}.& \sin 2x={\displaystyle \frac{1}{2}}\\ \text{b}.& \cos x={\displaystyle \frac{1}{2}\sqrt{3}}& \text{g}& 2\cos x=-\sqrt{3}& \text{l}.& \cos 2x=-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{c}.& \tan x=\sqrt{3}& \text{h}& 3\tan x=\sqrt{3}& \text{m}.& \tan 2x=\sqrt{3}\\ \text{d}.& \sin x=-1& \text{i}.& \sin x=\sin {46}^{\circ }& \text{n}.& \sin (2x-{30}^{\circ })=\sin {45}^{\circ }\\ \text{e}.& \cos x=-{\displaystyle \frac{1}{2}\sqrt{2}}& \text{j}.& \cos x=\cos {93}^{\circ }& \text{o}.& \sin (2x+{60}^{\circ })=\sin {90}^{\circ }\end{array}\end{array}$

$.\text{Jawab}:$

$\begin{array}{rl}.\text{a}.\sin x& ={\displaystyle \frac{1}{2}}\\ \sin x& =\sin {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{30}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ {150}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{30}^{\circ },{150}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{b}.\cos x& ={\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos x& =\cos {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ -{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ -{30}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tidak memenuhi})& \\ -{30}^{\circ }+{360}^{\circ }={330}^{\circ }(\text{memenuhi})& \end{array}\\ \text{HP}& =\{{30}^{\circ },{330}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{c}.\tan x& =\sqrt{3}\\ \tan x& =\tan {60}^{\circ }\\ x& ={60}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={60}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={60}^{\circ }+{180}^{\circ }={240}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={60}^{\circ }+{360}^{\circ }={420}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{60}^{\circ },{240}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{d}.\sin x& =-1\\ \sin x& =\sin {270}^{\circ }\\ x& =\{\begin{array}{ll}{270}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{270}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}{270}^{\circ }& \text{memenuhi}\\ -{90}^{\circ }& \text{tidak memenuhi}\end{array}\\ k=1& \text{tidak memenuhi semuanya}\\ \text{HP}& =\left\{{270}^{\circ }\right\}\end{array}$.

$\begin{array}{rl}.\text{k}.\sin 2x& ={\displaystyle \frac{1}{2}}\\ \sin 2x& =\sin {30}^{\circ }\\ 2x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{30}^{\circ })& +k{.360}^{\circ }\end{array}\\ \text{sehin}& \text{gga}\\ x& =\{\begin{array}{ll}{15}^{\circ }& +k{.180}^{\circ }\\ ({90}^{\circ }-{15}^{\circ })& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{15}^{\circ }(\text{memenuhi})& \\ {75}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{15}^{\circ }+{180}^{\circ }={195}^{\circ }(\text{memenuhi})& \\ {75}^{\circ }+{180}^{\circ }={255}^{\circ }(\text{memenuhi})& \end{array}\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{15}^{\circ },{75}^{\circ },{195}^{\circ },{255}^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{l}.\cos 2x& =-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos 2x& =-\cos {30}^{\circ }=\cos ({180}^{\circ }-{30}^{\circ })=\cos {150}^{\circ }\\ 2x& =\{\begin{array}{ll}{150}^{\circ }& +k{.360}^{\circ }\\ -{150}^{\circ }& +k{.360}^{\circ }\end{array}\\ \text{sehin}& \text{gga}\\ x& =\{\begin{array}{ll}{75}^{\circ }& +k{.180}^{\circ }\\ -{75}^{\circ }& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{75}^{\circ }(\text{memenuhi})& \\ -{75}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{75}^{\circ }+{180}^{\circ }={255}^{\circ }(\text{memenuhi})& \\ -{75}^{\circ }+{180}^{\circ }={105}^{\circ }(\text{memenuhi})& \end{array}\\ k=2& \\ x& =\{\begin{array}{ll}{75}^{\circ }+{360}^{\circ }={435}^{\circ }(\text{tidak memenuhi})& \\ -{75}^{\circ }+{360}^{\circ }={285}^{\circ }(\text{memenuhi})& \end{array}\\ k=3& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{75}^{\circ },{105}^{\circ },{255}^{\circ },{285}^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{m}.\tan 2x& =\sqrt{3}\\ \tan 2x& =\tan {60}^{\circ }\\ 2x& ={60}^{\circ }+k{.180}^{\circ }\\ \text{sehin}& \text{gga}\\ x& ={30}^{\circ }+k{.90}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={30}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={30}^{\circ }+{90}^{\circ }={120}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={30}^{\circ }+{180}^{\circ }={210}^{\circ }\text{memenuhi}\\ k=3& \\ x& ={30}^{\circ }+{270}^{\circ }={300}^{\circ }\text{memenuhi}\\ k=4& \\ x& ={30}^{\circ }+{360}^{\circ }={390}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{30}^{\circ },{120}^{\circ },{210}^{\circ },{300}^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{n}.\sin (2x-{30}^{\circ })& =\sin {45}^{\circ }\\ (2x-{30}^{\circ })& =\{\begin{array}{ll}{45}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{45}^{\circ })& +k{.360}^{\circ }\end{array}\\ 2x& =\{\begin{array}{ll}{45}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\\ {135}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ x& =\{\begin{array}{ll}37,5^{\circ }& +k{.180}^{\circ }\\ 82,5^{\circ }& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }& \\ 82,5^{\circ }& \end{array}\\ k=1& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }+{180}^{\circ }& =217,5^{\circ }\\ 82,5^{\circ }+{180}^{\circ }& =262,5^{\circ }\end{array}\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{37,5^{\circ },82,5^{\circ },217,5^{\circ },262,5^{\circ }\}\end{array}$.

$\begin{array}{rl}.\text{o}.\sin (2x+{60}^{\circ })& =\sin {90}^{\circ }\\ (2x+{60}^{\circ })& =\{\begin{array}{ll}{90}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{90}^{\circ })& +k{.360}^{\circ }\end{array}\\ 2x& =\{\begin{array}{ll}{90}^{\circ }-{60}^{\circ }& +k{.360}^{\circ }\\ {90}^{\circ }-{60}^{\circ }& +k{.360}^{\circ }\end{array}\\ x& ={15}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}\\ x& ={15}^{\circ }\\ k=1& \text{diperoleh}\\ x& ={15}^{\circ }+{180}^{\circ }={195}^{\circ }\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{15}^{\circ },{195}^{\circ }\}\end{array}$

Lanjutan 2 Persamaan Trigonometri

 A. 2  Relasi Sudut

Mengingatkan kembali materi tentang nilai sudut diberbagai kuadran yang selanjutnya berkaitan erat dengan relasi sudutnya dari kuadran selain satu diubah ke kuadran satu supaya mudah menentukan nilai trigonometri.

Untuk tanda perbandingan trigonometrinya berkaitan dengan relasi sudutnya adalah disajikan sebagaimana dalam bagan berikut

$\begin{array}{cccccc}\text{Nilai yang positif}& & & & & \\ \text{hanya}\mathbf{\text{sinus}}& & & & \text{Semua nilai trigon}& \mathbf{\text{positif}}\\ & & & & & \\ \text{Nilai yang positif}& & & & \text{Nilai yang positif}& \\ \text{hanya}\mathbf{\text{tangen}}& & & & \text{hanya}\mathbf{\text{cosinus}}& \end{array}$.

atau

$\begin{array}{cccccc}\begin{array}{l}\\ \{\begin{array}{ll}\sin & =+\\ \cos & =-\\ \tan & =-\\ \csc & =+\\ \sec & =-\\ \cot & =-\end{array}& \end{array}& & & & & \begin{array}{l}\\ \{\begin{array}{ll}\sin & =+\\ \cos & =+\\ \tan & =+\\ \csc & =+\\ \sec & =+\\ \cot & =+\end{array}& \end{array}\\ & & & & \\ & & & & & \\ & & & & & \\ \begin{array}{l}\\ \{\begin{array}{ll}\sin & =-\\ \cos & =-\\ \tan & =+\\ \csc & =-\\ \sec & =-\\ \cot & =+\end{array}& \end{array}& & & & & \begin{array}{l}\\ \{\begin{array}{ll}\sin & =-\\ \cos & =+\\ \tan & =-\\ \csc & =-\\ \sec & =+\\ \cot & =-\end{array}& \end{array}\end{array}$.

Adapun penjabaran sudut-sudut yang berelasi sebagaimana ilustrasi bagan berikut, yaitu:

$\begin{array}{cccccc}\text{Kuadran II}& & & & \text{Kuadran I}& \\ ({180}^{\circ }-\alpha )& & & & \text{Semua nilai trigon}& \mathbf{\text{positif}}\\ & & & & & \\ \text{Kuadran III}& & & & \text{Kuadran IV}& \\ ({180}^{\circ }+\alpha )& & & & ({360}^{\circ }-\alpha )& \end{array}$

Ketentuan perubahan trigonometri berkaitan dengan sudut berelasi adalah sebagaimana tabel berikut:

KUADRAN PERTAMA

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran I}\\ & 0^{\circ }<\alpha <{90}^{\circ }\\ & =({90}^{\circ }-\alpha )\end{array}& \{\begin{array}{ll}\sin & =\cos \\ \cos & =\sin \\ \tan & =\cot \\ \csc & =\sec \\ \sec & =\csc \\ \cot & =\tan \end{array}& \begin{array}{rl}\sin ({90}^{\circ }-\alpha )& =\cos \alpha \\ \cos ({90}^{\circ }-\alpha )& =\sin \alpha \\ \tan ({90}^{\circ }-\alpha )& =\cot \alpha \\ \csc ({90}^{\circ }-\alpha )& =\sec \alpha \\ \sec ({90}^{\circ }-\alpha )& =\csc \alpha \\ \cot ({90}^{\circ }-\alpha )& =\tan \alpha \end{array}\\ \hline\end{array}$.

KUADRAN KEDUA

ada 2 pilihan yaitu:

pertama

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran II}\\ & {90}^{\circ }<\alpha <{180}^{\circ }\\ & =({90}^{\circ }+\alpha )\end{array}& \{\begin{array}{ll}\sin & =\cos \\ \cos & =\sin \\ \tan & =\cot \\ \csc & =\sec \\ \sec & =\csc \\ \cot & =\tan \end{array}& \begin{array}{rl}\sin ({90}^{\circ }+\alpha )& =\cos \alpha \\ \cos ({90}^{\circ }+\alpha )& =-\sin \alpha \\ \tan ({90}^{\circ }+\alpha )& =-\cot \alpha \\ \csc ({90}^{\circ }+\alpha )& =\sec \alpha \\ \sec ({90}^{\circ }+\alpha )& =-\csc \alpha \\ \cot ({90}^{\circ }+\alpha )& =-\tan \alpha \end{array}\\ \hline\end{array}$.

kedua

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Tidak Ada Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran II}\\ & {90}^{\circ }<\alpha <{180}^{\circ }\\ & =({180}^{\circ }-\alpha )\end{array}& \{\begin{array}{ll}\sin & =\sin \\ \cos & =\cos \\ \tan & =\tan \\ \csc & =\csc \\ \sec & =\sec \\ \cot & =\cot \end{array}& \begin{array}{rl}\sin ({180}^{\circ }-\alpha )& =\sin \alpha \\ \cos ({180}^{\circ }-\alpha )& =-\cos \alpha \\ \tan ({180}^{\circ }-\alpha )& =-\tan \alpha \\ \csc ({180}^{\circ }-\alpha )& =\csc \alpha \\ \sec ({180}^{\circ }-\alpha )& =-\sec \alpha \\ \cot ({180}^{\circ }-\alpha )& =-\cot \alpha \end{array}\\ \hline\end{array}$.

KUADRAN KETIGA

ada 2 pilihan juga yaitu:

pertama

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Tidak Ada Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran III}\\ & {180}^{\circ }<\alpha <{270}^{\circ }\\ & =({180}^{\circ }+\alpha )\end{array}& \{\begin{array}{ll}\sin & =\sin \\ \cos & =\cos \\ \tan & =\tan \\ \csc & =\csc \\ \sec & =\sec \\ \cot & =\cot \end{array}& \begin{array}{rl}\sin ({180}^{\circ }+\alpha )& =-\sin \alpha \\ \cos ({180}^{\circ }+\alpha )& =-\cos \alpha \\ \tan ({180}^{\circ }+\alpha )& =\tan \alpha \\ \csc ({180}^{\circ }+\alpha )& =-\csc \alpha \\ \sec ({180}^{\circ }+\alpha )& =-\sec \alpha \\ \cot ({180}^{\circ }+\alpha )& =\cot \alpha \end{array}\\ \hline\end{array}$.

kedua

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran III}\\ & {180}^{\circ }<\alpha <{270}^{\circ }\\ & =({270}^{\circ }-\alpha )\end{array}& \{\begin{array}{ll}\sin & =\cos \\ \cos & =\sin \\ \tan & =\cot \\ \csc & =\sec \\ \sec & =\csc \\ \cot & =\tan \end{array}& \begin{array}{rl}\sin ({270}^{\circ }-\alpha )& =-\cos \alpha \\ \cos ({270}^{\circ }-\alpha )& =-\sin \alpha \\ \tan ({270}^{\circ }-\alpha )& =\cot \alpha \\ \csc ({270}^{\circ }-\alpha )& =-\sec \alpha \\ \sec ({270}^{\circ }-\alpha )& =-\csc \alpha \\ \cot ({270}^{\circ }-\alpha )& =\tan \alpha \end{array}\\ \hline\end{array}$.

KUADRAN KEEMPAT

ada 2 pilihan juga yaitu:

pertama

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran IV}\\ & {270}^{\circ }<\alpha <{360}^{\circ }\\ & =({270}^{\circ }+\alpha )\end{array}& \{\begin{array}{ll}\sin & =\cos \\ \cos & =\sin \\ \tan & =\cot \\ \csc & =\sec \\ \sec & =\csc \\ \cot & =\tan \end{array}& \begin{array}{rl}\sin ({270}^{\circ }+\alpha )& =-\cos \alpha \\ \cos ({270}^{\circ }+\alpha )& =\sin \alpha \\ \tan ({270}^{\circ }+\alpha )& =-\cot \alpha \\ \csc ({270}^{\circ }+\alpha )& =-\sec \alpha \\ \sec ({270}^{\circ }+\alpha )& =\csc \alpha \\ \cot ({270}^{\circ }+\alpha )& =-\tan \alpha \end{array}\\ \hline\end{array}$.

kedua

$\begin{array}{|ccl|}\hline \text{Posisi}& \text{Tidak Ada Perubahan}& \text{Relasi Sudut}\\ \begin{array}{rl}& \text{Kuadran IV}\\ & {270}^{\circ }<\alpha <{360}^{\circ }\\ & =({360}^{\circ }-\alpha )\end{array}& \{\begin{array}{ll}\sin & =\sin \\ \cos & =\cos \\ \tan & =\tan \\ \csc & =\csc \\ \sec & =\sec \\ \cot & =\cot \end{array}& \begin{array}{rl}\sin ({360}^{\circ }-\alpha )& =-\sin \alpha \\ \cos ({360}^{\circ }-\alpha )& =\cos \alpha \\ \tan ({360}^{\circ }-\alpha )& =-\tan \alpha \\ \csc ({360}^{\circ }-\alpha )& =-\csc \alpha \\ \sec ({360}^{\circ }-\alpha )& =\sec \alpha \\ \cot ({360}^{\circ }-\alpha )& =-\cot \alpha \end{array}\\ \hline\end{array}$.

 A. 3  Sudut Negatif dan Sudut lebih Besar dari  ${360}^{\circ }$

$\begin{array}{rl}\text{a}.& \{\begin{array}{ll}\sin (-A)& =-\sin A\\ \cos (-A)& =\cos A\\ \tan (-A)& =-\tan A\end{array}\\ \text{b}.& \{\begin{array}{ll}\csc (-A)& =-\csc A\\ \sec (-A)& =\sec A\\ \cot (-A)& =-\cot A\end{array}\\ \text{c}.& \{\begin{array}{ll}\sin (n{.360}^{\circ }+A)& =\sin A\\ \cos (n{.360}^{\circ }+A)& =\cos A\\ \tan (n{.360}^{\circ }+A)& =\tan A\end{array},n\in \mathbb{N}\end{array}$.

Catatan : $0^{\circ }$=${360}^{\circ }$=${720}^{\circ }$=${1080}^{\circ }$=$n{.360}^{\circ }$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai}\\ & \text{a}.\sin {120}^{\circ }\\ & \text{b}.\cos {240}^{\circ }\\ & \text{c}.\tan {315}^{\circ }\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.\sin {120}^{\circ }& =\sin ({180}^{\circ }-{60}^{\circ })=\sin {60}^{\circ }\\ & ={\displaystyle \frac{1}{2}\sqrt{3},\text{atau}}\\ & =\sin ({90}^{\circ }+{30}^{\circ })=\cos {30}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{b}.\cos {240}^{\circ }& =\cos ({180}^{\circ }+{60}^{\circ })=-\cos {60}^{\circ }\\ & =-{\displaystyle \frac{1}{2},\text{atau}}\\ & =\cos ({270}^{\circ }-{30}^{\circ })=-\sin {30}^{\circ }=-\frac{1}{2}\\ \text{c}.\tan {315}^{\circ }& =\tan ({360}^{\circ }-{45}^{\circ })=-\tan {45}^{\circ }\\ & =-1,\text{atau}\\ & =\tan ({270}^{\circ }+{45}^{\circ })=-\cot {45}^{\circ }=-1\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Buktikan bahwa}\\ \\ & \text{a}.{\displaystyle \frac{\cos ({90}^{\circ }-B)}{\sec B}+\frac{\sin ({90}^{\circ }-B)}{\csc B}=2\sin B\cos B}\\ \\ & \text{b}.\tan C+\tan ({90}^{\circ }-C)=\sec C.\sec ({90}^{\circ }-C)\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{\cos ({90}^{\circ }-B)}{\sec B}+\frac{\sin ({90}^{\circ }-B)}{\csc B}}\\ & ={\displaystyle \frac{\sin B}{\sec B}+\frac{\cos B}{\csc B}}\\ & ={\displaystyle \frac{\sin B}{{\displaystyle \frac{1}{\cos B}}}+\frac{\cos B}{{\displaystyle \frac{1}{\sin B}}}}\\ & =\sin B\cos B+\sin B\cos B\\ & =2\sin B\cos B◼\end{array}\\ & \begin{array}{rl}\text{b}.& \tan C+\tan ({90}^{\circ }-C)\\ & =\tan C+\cot C\\ & ={\displaystyle \frac{\sin C}{\cos C}+\frac{\cos C}{\sin C}}\\ & ={\displaystyle \frac{{\sin }^{2}C+{\cos }^{2}C}{\sin C\cos C}={\displaystyle \frac{1}{\sin C\cos C}}}\\ & ={\displaystyle \frac{1}{\cos C}.\frac{1}{\sin C}}\\ & =\sec C.\csc C\\ & =\sec C.\sec ({90}^{\circ }-C)◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai}\\ & \text{a}.\tan (A-{90}^{\circ })\sin (-A)\\ & \text{b}.\cos {540}^{\circ }+\sin {690}^{\circ }\\ & \text{c}.\sin {2021}^{\circ }+\cos {2021}^{\circ }\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \tan (A-{90}^{\circ })\sin (-A)\\ & =\tan (-({90}^{\circ }-A))(-\sin A)\\ & =-\tan ({90}^{\circ }-A)(-\sin A)\\ & =\tan ({90}^{\circ }-A)(\sin A)\\ & =\cot A.\sin A\\ & ={\displaystyle \frac{\cos A}{\sin A}.\sin A}\\ & =\cos A\end{array}\\ & \begin{array}{rl}\text{b}.& \cos {540}^{\circ }+\sin {690}^{\circ }\\ & =\cos ({360}^{\circ }+{180}^{\circ })+\sin ({720}^{\circ }-{30}^{\circ })\\ & =\cos (0^{\circ }+{180}^{\circ })+\sin (0^{\circ }-{30}^{\circ })\\ & =\cos {180}^{\circ }+\sin (-{30}^{\circ })\\ & =\cos {180}^{\circ }-\sin {30}^{\circ }\\ & =-1-{\displaystyle \frac{1}{2}}\\ & =-{\displaystyle \frac{3}{2}}\end{array}\\ & \begin{array}{rl}\text{c}.& \sin {2021}^{\circ }+\cos {2021}^{\circ }\\ & =\sin ({5.360}^{\circ }+{221}^{\circ })+\cos ({5.360}^{\circ }+{221}^{\circ })\\ & =\sin (0^{\circ }+{221}^{\circ })+\cos (0^{\circ }+{221}^{\circ })\\ & =\sin {221}^{\circ }+\cos {221}^{\circ }\\ & =\sin ({180}^{\circ }+{41}^{\circ })+\cos ({180}^{\circ }+{41}^{\circ })\\ & =-\sin {41}^{\circ }-\cos {41}^{\circ }\end{array}\end{array}$

Lanjutan Persamaan Trigonometri

$\begin{array}{rl}& \text{f. Menentukan Nilai Perbandingan Trigonometri}\\ & \text{pada Segitiga Siku-Siku}\end{array}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui}\tan \theta ={\displaystyle \frac{a}{x}}\\ & \text{Tentukanlah nilai}{\displaystyle \frac{x}{\sqrt{a^2+x^2}}}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah gambar segitiga AOX berikut}\end{array}$

$.\begin{array}{rl}& \text{Dengan rumus Pythagoras dapatr ditentukan}\\ & \text{panjang ruas}\text{AX, yaitu}:\\ & A{O}^2+O{X}^2=A{X}^2\\ & \text{atau}\\ & A{X}^2=A{O}^2+O{X}^2\\ & AX=\sqrt{A{O}^2+O{X}^2}\\ & =\sqrt{x^2+a^2},\\ & \text{maka}\\ & \bullet \sin \theta ={\displaystyle \frac{a}{\sqrt{x^2+a^2}}}\\ & \bullet \cos \theta ={\displaystyle \frac{x}{\sqrt{x^2+a^2}}}\\ & \text{Jadi, nilai}{\displaystyle \frac{x}{\sqrt{x^2+a^2}}=\cos \theta }\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}\sin \beta +\cos \beta ={\displaystyle \frac{6}{5},\text{tentukanlah}}\\ & \text{a}.\sin \beta \cos \beta \\ & \text{b}.{\sin }^3\beta +{\cos }^3\beta \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \sin \beta +\cos \beta ={\displaystyle \frac{6}{5}}\\ & \text{saat masing-masing ruas dikuadratkan,}\\ & \text{maka}\\ & {(\sin \beta +\cos \beta )}^2={\left({\displaystyle \frac{6}{5}}\right)}^2\\ & {\sin }^2\beta +2\sin \beta \cos \beta +{\cos }^2\beta ={\displaystyle \frac{36}{25}}\\ & {\sin }^2\beta +{\cos }^2\beta +2\sin \beta \cos \beta ={\displaystyle \frac{36}{25}}\\ & 1+2\sin \beta \cos \beta ={\displaystyle \frac{36}{25}}\\ & 2\sin \beta \cos \beta ={\displaystyle \frac{36}{25}-1}\\ & 2\sin \beta \cos \beta ={\displaystyle \frac{36-25}{25}=\frac{11}{25}}\\ & \sin \beta \cos \beta ={\displaystyle \frac{11}{50}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\sin }^3\beta +{\cos }^3\beta \\ & =(\sin \beta +\cos \beta )({\sin }^2\beta +{\cos }^2\beta -\sin \beta \cos \beta )\\ & =(\sin \beta +\cos \beta )(1-\sin \beta \cos \beta )\\ & =\left({\displaystyle \frac{6}{5}}\right).(1-{\displaystyle \frac{11}{50}})\\ & =\left({\displaystyle \frac{6}{5}}\right).\left({\displaystyle \frac{50-11}{50}}\right)\\ & =\left({\displaystyle \frac{6}{5}}\right).\left({\displaystyle \frac{39}{50}}\right)\\ & ={\displaystyle \frac{3\times 39}{5\times 25}=\frac{117}{125}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}\tan \alpha ={\displaystyle \frac{1}{\sqrt{7}},\text{tentukanlah}}\\ & \left({\displaystyle \frac{{\csc }^2\alpha -{\sec }^2\alpha }{{\csc }^2\alpha +{\sec }^2\alpha }}\right)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}:\\ \tan \alpha & ={\displaystyle \frac{1}{\sqrt{7}},\text{dan ingat juga bahwa}}\\ {\sec }^2\alpha & ={\tan }^2\alpha +1={\left({\displaystyle \frac{1}{\sqrt{7}}}\right)}^2+1=\frac{1}{7}+1=\frac{8}{7}\\ \text{Demik}& \text{ian juga},\cot \alpha ={\displaystyle \frac{1}{\tan \alpha }={\displaystyle \frac{1}{\left(\frac{1}{\sqrt{7}}\right)}=\sqrt{7},}}\\ \text{maka},& {\csc }^2\alpha ={\cot }^2\alpha +1={\left(\sqrt{7}\right)}^2+1=7+1=8\\ \text{Selanj}& \text{utnya}\\ & \left({\displaystyle \frac{{\csc }^2\alpha -{\sec }^2\alpha }{{\csc }^2\alpha +{\sec }^2\alpha }}\right)=\left({\displaystyle \frac{8-{\displaystyle \frac{8}{7}}}{8+{\displaystyle \frac{8}{7}}}}\right)\\ & ={\displaystyle \frac{{\displaystyle \frac{56-8}{7}}}{{\displaystyle \frac{56+8}{7}}}}\\ & ={\displaystyle \frac{48}{64}}\\ & ={\displaystyle \frac{3}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Jika}\beta \text{sudut lancip dan}\cos \beta ={\displaystyle \frac{3}{5},}\\ & \text{tentukan nilai dari}{\displaystyle \frac{\sin \beta \tan \beta -1}{2{\tan }^2\beta }}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \\ \cos \beta & ={\displaystyle \frac{3}{5}\Rightarrow {\sin }^2\beta +{\cos }^2\beta =1}\\ {\sin }^2\beta & +{\cos }^2\beta =1\\ \sin \beta & =\sqrt{1-{\cos }^2\beta }=\sqrt{1-{\left({\displaystyle \frac{3}{5}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{9}{25}}}=\sqrt{{\displaystyle \frac{16}{25}}}={\displaystyle \frac{4}{5}}\\ \text{Sehingga}& \tan \beta ={\displaystyle \frac{\sin \beta }{\cos \beta }=\frac{{\displaystyle \frac{4}{5}}}{{\displaystyle \frac{3}{5}}}=\frac{4}{3}}\\ & {\displaystyle \frac{\sin \beta \tan \beta -1}{2{\tan }^2\beta }={\displaystyle \frac{{\displaystyle \frac{4}{5}\times \frac{4}{3}-1}}{2{\left({\displaystyle \frac{4}{3}}\right)}^2}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{16}{15}-1}}{{\displaystyle \frac{32}{9}}}={\displaystyle \frac{{\displaystyle \frac{1}{15}}}{{\displaystyle \frac{32}{9}}}={\displaystyle \frac{9}{32\times 15}}}}\\ & ={\displaystyle \frac{3}{32\times 5}}\\ & ={\displaystyle \frac{3}{160}}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

Persamaan Trigonometri

 $\text{A. 1 Identitas Trigonometri}$.

A. 1. 1  Nilai Trigonometri Sudut

$\text{a. Perbandingan Trigonometri dalam Segitiga Siku-Siku}$.

Perhatikanlah ilustrasi sebuah segitiga siku-siku sama kaki berikut

Diketahui pula bahwa :

$\begin{array}{c}\bullet \sin {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}}\\ \bullet \cos {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}}\\ \bullet \tan {45}^{\circ }=1\end{array}$.

$\begin{array}{c}\bullet \csc {45}^{\circ }={\displaystyle \sqrt{2}}\\ \bullet \sec {45}^{\circ }={\displaystyle \sqrt{2}}\\ \bullet \cot {45}^{\circ }=1\end{array}$.

Berikut ilustrasi segitiga dengan sudut istimewa yang lain yaitu ${30}^{\circ }$ dan  ${60}^{\circ }$.

$\begin{array}{|cc|}\hline \begin{array}{c}\bullet \sin {30}^{\circ }={\displaystyle \frac{1}{2}}\\ \bullet \cos {30}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \bullet \tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}}\\ \bullet \sin {60}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \bullet \cos {60}^{\circ }={\displaystyle \frac{1}{2}}\\ \bullet \tan {30}^{\circ }={\displaystyle \sqrt{3}}\end{array}& \begin{array}{c}\bullet \csc {30}^{\circ }={\displaystyle 2}\\ \bullet \sec {30}^{\circ }={\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}\\ \bullet \cot {30}^{\circ }={\displaystyle \sqrt{3}}\\ \bullet \csc {60}^{\circ }={\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}\\ \bullet \sec {60}^{\circ }={\displaystyle 2}\\ \bullet \cot {30}^{\circ }={\displaystyle \frac{1}{3}\sqrt{3}}\end{array}\\ \hline\end{array}$

Perhatikan segitiga ABC siku-siku di C berikut

Perhatikanlah segitiga OAB berikut

$\begin{array}{rl}\text{a}.& \sin \alpha ={\displaystyle \frac{y}{r}}\\ \text{b}.& \cos \alpha ={\displaystyle \frac{x}{r}}\\ \text{c}.& \tan \alpha ={\displaystyle \frac{y}{x}}\\ \text{d}.& \csc \alpha ={\displaystyle \frac{r}{y}}\\ \text{e}.& \sec \alpha ={\displaystyle \frac{r}{x}}\\ \text{f}.& \cot \alpha ={\displaystyle \frac{x}{y}}\end{array}$.

A. 1. 2  Identitas Trigonometri Dasar

$\text{a. Dalil Pythagoras Segitiga Siku-Siku}$.

$\begin{array}{c}\\ \begin{array}{rl}& \text{Dalil/rumus Pythagoras}\\ & a^2+b^2=c^2\\ & \text{atau}\\ & c=\sqrt{a^2+b^2}\end{array}& \begin{array}{rl}& \sin \mathrm{\angle }ACB={\displaystyle \frac{a}{c}}\\ & \cos \mathrm{\angle }ACB={\displaystyle \frac{b}{c}}\\ & \tan \mathrm{\angle }ACB={\displaystyle \frac{a}{b}={\displaystyle \frac{\sin \mathrm{\angle }ACB}{\cos \mathrm{\angle }ACB}}}\\ & \csc \mathrm{\angle }ACB={\displaystyle \frac{c}{a}}\\ & \sec \mathrm{\angle }ACB={\displaystyle \frac{c}{b}}\\ & \cot \mathrm{\angle }ACB={\displaystyle \frac{b}{a}={\displaystyle \frac{\cos \mathrm{\angle }ACB}{\sin \mathrm{\angle }ACB}}}\end{array}\end{array}$

$\text{b. Identitas trigonometri pada segitiga siku-siku}$.

$\begin{array}{rl}& \text{Dalil/rumus Pythagoras}\\ & a^2+b^2=c^2\\ & \text{Perhatikan lagi gambar di poin c di atas}\\ & \begin{array}{|cl|}\hline 1.& \text{Rumus saat dibagi dengan}{c}^2\\ & {\displaystyle \frac{a^2}{c^2}+{\displaystyle \frac{b^2}{c^2}={\displaystyle \frac{c^2}{c^2}\iff {\displaystyle \frac{a^2}{c^2}+{\displaystyle \frac{b^2}{c^2}=1}}}}}\\ & \\ & \text{menjadi}{\left({\displaystyle \frac{a}{c}}\right)}^2+{\left({\displaystyle \frac{b}{c}}\right)}^2=1\\ & {\sin }^2\mathrm{\angle }ACB+{\cos }^2\mathrm{\angle }ACB=1\\ 2& \text{Rumus saat dibagi dengan}{b}^2\\ & {\displaystyle \frac{a^2}{b^2}+{\displaystyle \frac{b^2}{b^2}={\displaystyle \frac{c^2}{b^2}\iff {\displaystyle \frac{a^2}{b^2}+1={\displaystyle \frac{c^2}{b^2}}}}}}\\ & \\ & \text{menjadi}{\left({\displaystyle \frac{a}{b}}\right)}^2+1={\left({\displaystyle \frac{c}{b}}\right)}^2\\ & {\tan }^2\mathrm{\angle }ACB+1={\sec }^2\mathrm{\angle }ACB\\ 3& \text{Rumus saat dibagi dengan}{a}^2\\ & {\displaystyle \frac{a^2}{a^2}+{\displaystyle \frac{b^2}{a^2}={\displaystyle \frac{c^2}{a^2}\iff 1+{\displaystyle \frac{b^2}{a^2}={\displaystyle \frac{c^2}{a^2}}}}}}\\ & \\ & \text{menjadi}1+{\left({\displaystyle \frac{b}{a}}\right)}^2={\left({\displaystyle \frac{c}{a}}\right)}^2\\ & 1+{\cot }^2\mathrm{\angle }ACB={\csc }^2\mathrm{\angle }ACB\\ \hline\end{array}\end{array}$

$\text{c. Tabel trigonometri nilai sudut istimewa}$.

$\begin{array}{|ccccccc|}\hline \alpha & 0^{\circ }& 30& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }& {180}^{\circ }\\ \sin \alpha & 0& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}\sqrt{3}}& 1& 0\\ \cos \alpha & 1& {\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}}& 0& -1\\ \tan \alpha & 0& {\displaystyle \frac{1}{3}\sqrt{3}}& 1& \sqrt{3}& \text{TD}& 0\\ \hline\end{array}$.

$\text{d. Aturan sinus pada segitiga sebarang}$.

${\displaystyle \frac{BC}{\sin \mathrm{\angle }A}={\displaystyle \frac{AC}{\sin \mathrm{\angle }B}={\displaystyle \frac{AB}{\sin \mathrm{\angle }C}}}}$

$\text{e. Aturan cosinus pada segitiga sebarang}$.

Perhatikanlah gmabar pada poin e di atas, aturan cosinusnya adalah:

$\begin{array}{rl}\bullet & \cos \mathrm{\angle }A={\displaystyle \frac{b^2+c^2-a^2}{2bc}}\\ \bullet & \cos \mathrm{\angle }B={\displaystyle \frac{a^2+c^2-a^2}{2ac}}\\ \bullet & \cos \mathrm{\angle }C={\displaystyle \frac{a^2+b^2-a^2}{2ab}}\end{array}$.

$\begin{array}{rl}& \text{Macam-Macam Identitas Trigonometri Dasar}\\ & 1.\csc \alpha ={\displaystyle \frac{1}{\sin \alpha }5.\tan \alpha ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}}\\ & 2.\sec \alpha ={\displaystyle \frac{1}{\cos \alpha }6.{\tan }^2\alpha +1={\sec }^2\alpha }\\ & 3.\cot \alpha ={\displaystyle \frac{1}{\tan \alpha }7.{\cot }^2\alpha +1={\csc }^2\alpha }\\ & 4.\cot \alpha ={\displaystyle \frac{\cos \alpha }{\sin \alpha }8.{\sin }^2\alpha +{\cos }^2=1}\end{array}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa}\\ & \tan \alpha ={\displaystyle \frac{\sin \alpha \cos \alpha }{1-{\sin }^2\alpha }}\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}\tan \alpha & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}\\ & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }\times \frac{\cos \alpha }{\cos \alpha }}\\ & ={\displaystyle \frac{\sin \alpha \cos \alpha }{{\cos }^2\alpha }}\\ & ={\displaystyle \frac{\sin \alpha \cos \alpha }{1-{\sin }^2\alpha }◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1}{\sqrt{{\tan }^2\beta }}\times \sin \beta =\cos \beta }\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{\sqrt{{\tan }^2\beta }}\times \sin \beta }& ={\displaystyle \frac{1}{\tan \beta }\times \sin \beta }\\ & ={\displaystyle \frac{\cos \beta }{\sin \beta }\times \sin \beta }\\ & =\cos \beta ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{{\cos }^2\gamma }{1-\sin \gamma }=1+\sin \gamma }\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{{\cos }^2\gamma }{1-\sin \gamma }}& ={\displaystyle \frac{1-{\sin }^2\gamma }{1-\sin \gamma }}\\ & ={\displaystyle \frac{(1-\sin \gamma )(1+\sin \gamma )}{1-\sin \gamma }}\\ & =1+\sin \gamma ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }={\cos }^2\theta -{\sin }^2\theta }\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }}& ={\displaystyle \frac{1-{\tan }^2\theta }{{\sec }^2\theta }={\displaystyle \frac{1-{\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }}}{{\displaystyle \frac{1}{{\cos }^2\theta }}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta }}}{{\displaystyle \frac{1}{{\cos }^2\theta }}}}\\ & ={\cos }^2\theta -{\sin }^2\theta ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tunjukkan bahwa}\\ & {\cos }^4\alpha -{\sin }^4\alpha =1-2{\sin }^2\alpha \\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\cos }^4\alpha -{\sin }^4\alpha & ={({\cos }^2\alpha )}^2-{({\sin }^2\alpha )}^2\\ & =({\cos }^2\alpha -{\sin }^2\alpha )({\cos }^2\alpha +{\sin }^2\alpha )\\ & =({\cos }^2\alpha -{\sin }^2\alpha )\times 1\\ & ={\cos }^2\alpha -{\sin }^2\alpha \\ & =(1-{\sin }^2\alpha )-{\sin }^2\alpha \\ & =1-2{\sin }^2\alpha ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{\sin \beta \sec \beta }{{\sin }^2\beta -{\tan }^2\beta }=-{\displaystyle \frac{\cos \beta }{{\sin }^3\beta }}}\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{\sin \beta \sec \beta }{{\sin }^2\beta -{\tan }^2\beta }}& ={\displaystyle \frac{\sin \beta \left({\displaystyle \frac{1}{\cos \beta }}\right)}{{\sin }^2\beta -{\displaystyle \frac{{\sin }^2\beta }{{\cos }^2\beta }}}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{\sin \beta }{\cos \beta }}\right)}{{\sin }^2\beta (1-{\displaystyle \frac{1}{{\cos }^2\beta }})}\times \frac{{\cos }^2\beta }{{\cos }^2\beta }}\\ & ={\displaystyle \frac{\sin \beta \cos \beta }{{\sin }^2\beta ({\cos }^2\beta -1)}}\\ & ={\displaystyle \frac{\cos \beta }{\sin \beta (-{\sin }^2\beta )}}\\ & =-{\displaystyle \frac{\cos \beta }{{\sin }^3\beta }◼}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Noormandiri, B. K. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
2. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA