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

$\begin{array}{l}\\ 26.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\sin (x-y)}& & & \text{d}.& \cos (x-y)\\ \text{b}.& {\displaystyle -\tan (x-y)}& & & \text{e}.& {\displaystyle \tan (x-y)}\\ \text{c}.& {\displaystyle \sin (x+y)}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}}\\ & ={\displaystyle \frac{-2\sin (x+y)\sin (x-y)}{2\sin (x+y)\cos (x-y)}}\\ & =-\tan (x-y)\end{array}\end{array}$.

$\begin{array}{l}\\ 27.& \text{Nilai dari}\\ & 8\cos 82,5^{\circ }\sin 37,5^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4(\sqrt{3}+\sqrt{2})}& & & \text{d}.& 2(\sqrt{3}-\sqrt{2})\\ \text{b}.& {\displaystyle 4(\sqrt{3}-\sqrt{2})}& & & \text{e}.& {\displaystyle \sqrt{3}-\sqrt{2}}\\ \text{c}.& {\displaystyle 2(\sqrt{3}+\sqrt{2})}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& 8\cos 82,5^{\circ }\sin 37,5^{\circ }\\ & =4\times 2\cos 82,5^{\circ }\sin 37,5^{\circ }\\ & =4\times (\sin (82,5^{\circ }+37,5^{\circ })-\sin (82,5^{\circ }-37,5^{\circ }))\\ & =4\times (\sin {120}^{\circ }-\sin {45}^{\circ })\\ & =4\times (\sin ({180}^{\circ }-{60}^{\circ })-\sin {45}^{\circ })\\ & =4\times (\sin {60}^{\circ }-\sin {45}^{\circ })\\ & =4\times \left({\displaystyle \frac{1}{2}\sqrt{3}-{\displaystyle \frac{1}{2}\sqrt{2}}}\right)\\ & =2(\sqrt{3}-\sqrt{2})\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& \text{Bentuk lain dari}\\ & -2\cos 5A.\cos 7A\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\cos 6A-\cos A}& & & \\ \text{b}.& {\displaystyle -\cos 6A+\cos A}& & & \\ \text{c}.& {\displaystyle \cos 12A-\cos 2A}\\ \text{d}.& -\cos 12A+\cos 2A\\ \text{e}.& {\displaystyle -\cos 12A-\cos 2A}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& -2\cos 5A.\cos 7A\\ & =-(2\cos 5A.\cos 7A)\\ & =-(\cos 12A+\cos (-2A))\\ & =-(\cos 12A+\cos 2A)\\ & =-\cos 12A-\cos 2A\end{array}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Bentuk sederhana dari}\\ & 4\sin \left({\displaystyle \frac{1}{4}\pi +x}\right)\cos \left({\displaystyle \frac{1}{4}\pi -x}\right)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 2+2\sin 2x}& & & \text{d}.& 2+2\sin x\\ \text{b}.& {\displaystyle 2+\sin 2x}& & & \text{e}.& 2+\sin x\\ \text{c}.& {\displaystyle 2\sin 2x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& 4\sin \left({\displaystyle \frac{1}{4}\pi +x}\right)\cos \left({\displaystyle \frac{1}{4}\pi -x}\right)\\ & =2(\sin \left({\displaystyle \frac{1}{2}\pi }\right)+\sin (2x))\\ & =2(1+\sin 2x)\\ & =2+2\sin 2x\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \text{Nilai dari}\\ & \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{3}{8}}}& & & \text{d}.& {\displaystyle \frac{3}{8}}\\ \\ \text{b}.& {\displaystyle -\frac{1}{8}}& & & \text{e}.& {\displaystyle \frac{5}{8}}\\ \text{c}.& {\displaystyle \frac{1}{8}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & =\sqrt{3}\sin {80}^{\circ }\sin {20}^{\circ }(-\sin {40}^{\circ })\\ & =-\sqrt{3}\sin {80}^{\circ }\sin {40}^{\circ }\sin {20}^{\circ }\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{2}(\cos {60}^{\circ }-\cos {20}^{\circ })})\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{4}+{\displaystyle \frac{\cos {20}^{\circ }}{2}}})\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{2}\sqrt{3}\sin {80}^{\circ }\cos {20}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {100}^{\circ }+\sin {60}^{\circ })}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {80}^{\circ }+{\displaystyle \frac{1}{2}\sqrt{3}})}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }+{\displaystyle \frac{1}{8}\sqrt{9}}}}\\ & ={\displaystyle \frac{3}{8}}\end{array}\end{array}$.