Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Pertama

$\begin{array}{l}\\ 1.& \text{Nilai}{105}^{\circ }\text{jika dinyatakan ke radian}\\ & \text{adalah}....\text{radian}\\ \\ & \text{a}.{\displaystyle \frac{1}{3}\pi }\\ \\ & \text{b}.{\displaystyle \frac{5}{6}\pi }\\ \\ & \text{c}.{\displaystyle \frac{5}{12}\pi }\\ \\ & \text{d}.{\displaystyle \frac{7}{12}\pi }\\ \\ & \text{e}.{\displaystyle \frac{9}{12}\pi }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ {180}^{\circ }& =\pi radian\\ 1^{\circ }& ={\displaystyle \frac{\pi }{180}radian}\\ 105\times 1^{\circ }& =105\times {\displaystyle \frac{\pi }{180}radian}\\ {105}^{\circ }& ={\displaystyle \frac{7}{12}\pi radian}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Nilai}\tan {240}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \sqrt{3}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ \\ & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle -\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {240}^{\circ }& =\tan ({180}^{\circ }+{60}^{\circ })\\ & =\tan {60}^{\circ }\\ & =\sqrt{3}\\ \mathbf{\text{catatan}}& :\text{ingat sudut berelasi}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Pada gambar di atas perbandingan}\\ & \sin \theta \text{adalah}....\\ & \text{a}.\sqrt{{\displaystyle \frac{a^2-d^2}{f^2+g^2}}}\\ & \text{b}.\sqrt{{\displaystyle \frac{a^2+b^2}{f^2+g^2}}}\\ & \text{c}.\sqrt{{\displaystyle \frac{a^2-b^2}{f^2-g^2}}}\\ & \text{d}.\sqrt{{\displaystyle \frac{a^2+b^2}{f^2-g^2}}}\\ & \text{e}.\sqrt{{\displaystyle \frac{a^2-b^2}{f^2+g^2}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Dari so}& \text{al diketahui bahwa}\\ \sin \theta & ={\displaystyle \frac{c}{e}={\displaystyle \frac{\sqrt{a^2-b^2}}{\sqrt{f^2+g^2}}}}\\ & =\sqrt{{\displaystyle \frac{a^2-b^2}{f^2+g^2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai dari}({\cos }^2{17}^{\circ }-{\sin }^2{73}^{\circ })\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0& & & \text{d}.& 1\\ \text{b}.& {\displaystyle \frac{1}{3}}& \text{c}.& {\displaystyle \frac{2}{\sqrt{3}}}& \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& ({\cos }^2{17}^{\circ }-{\sin }^2{73}^{\circ })\\ & =({\cos }^2{17}^{\circ }-{(\sin {73}^{\circ })}^2)\\ & =({\cos }^2{17}^{\circ }-{(\sin ({90}^{\circ }-{17}^{\circ }))}^2)\\ & =({\cos }^2{17}^{\circ }-{\cos }^2{17}^{\circ })\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Jika diketahui}\\ & {\displaystyle \frac{x{\csc }^2{30}^{\circ }{\sec }^2{45}^{\circ }}{8{\cos }^2{45}^{\circ }{\sin }^2{60}^{\circ }}={\tan }^2{60}^{\circ }-{\tan }^2{30}^{\circ },}\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2& & & \text{d}.& 1\\ \text{b}.& {\displaystyle -1}& \text{c}.& 0& \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{x{\csc }^2{30}^{\circ }{\sec }^2{45}^{\circ }}{8{\cos }^2{45}^{\circ }{\sin }^2{60}^{\circ }}={\tan }^2{60}^{\circ }-{\tan }^2{30}^{\circ }}\\ & {\displaystyle \frac{x\left(4\right)\left({\displaystyle \frac{4}{2}}\right)}{8\left({\displaystyle \frac{2}{4}}\right)\left({\displaystyle \frac{3}{4}}\right)}=3-\left({\displaystyle \frac{1}{3}}\right)}\\ & {\displaystyle \frac{8x}{3}={\displaystyle \frac{8}{3}}}\\ & x=1\end{array}\end{array}$