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

$\begin{array}{l}\\ 6.& \text{Nilai dari}{\displaystyle \frac{\sin {49}^{\circ }}{\cos {41}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin {73}^{\circ }}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1& & & \text{d}.& 0,143\\ \text{b}.& {\displaystyle -0,321}& \text{c}.& 0& \text{e}.& {\displaystyle 0,321}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {49}^{\circ }}{\cos {41}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin {73}^{\circ }}}}\\ & ={\displaystyle \frac{\sin {49}^{\circ }}{\cos ({90}^{\circ }-{49}^{\circ })}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin ({90}^{\circ }-{17}^{\circ })}}}\\ & ={\displaystyle \frac{\sin {49}^{\circ }}{\sin {49}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\cos {17}^{\circ }}}}\\ & =1-1\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Nilai dari}\\ & p=r\sin \alpha \cos \beta \\ & q=r\sin \alpha \sin \beta \\ & s=r\cos \alpha \\ & \text{maka pernyataan berikut yang}\\ & \text{tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& p^2+t^2+s^2=r^2& & & \\ \text{b}.& p^2-t^2+s^2=r^2\\ \text{c}.& p^2+t^2-s^2=r^2& \\ \text{d}.& -p^2+t^2+s^2=r^2& \\ \text{e}.& -p^2-t^2+s^2=r^2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Saat}\\ & p^2+q^2\text{maka hasilnya adalah}\\ & \begin{array}{l}\\ p^2& =r^2{\sin }^2\alpha {\cos }^2\beta & \\ q^2& =r^2{\sin }^2\alpha {\sin }^2\beta & +\\ & =r^2{\sin }^2\alpha ({\cos }^2\beta +{\sin }^2\beta )\\ & =r^2{\sin }^2\alpha (1)\\ & =r^2{\sin }^2\alpha \end{array}\\ & \text{Dan saat}\\ & p^2+q^2+s^2\text{akan diperoleh hasil}\\ & \begin{array}{l}\\ p^2+q^2& =r^2{\sin }^2\alpha & \\ s^2& =r^2{\cos }^2\alpha & +\\ & =r^2{\sin }^2\alpha +r^2{\cos }^2\alpha \\ & =r^2({\sin }^2\alpha +{\cos }^2\alpha )\\ & =r^2(1)\\ & =r^2\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Nilai dari}\\ & {\displaystyle \frac{\cos ({90}^{\circ }+\theta )\sec (2\pi -\theta )\tan (\pi -\theta )}{\sec (\theta -2\pi )\sin ({540}^{\circ }+\theta )\cot (\theta -{90}^{\circ })}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1& & & \text{d}.& -\tan \theta \\ \text{b}.& {\displaystyle 0}& \text{c}.& 1& \text{e}.& {\displaystyle \tan \theta }\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Ingat kembali sudut-sudut}\\ & \text{yang berelasi dari kudran selain I}\\ & \text{ke kuadran I beserta tandanya}\\ & {\displaystyle \frac{\cos ({90}^{\circ }+\theta )\sec (2\pi -\theta )\tan (\pi -\theta )}{\sec (\theta -2\pi )\sin ({540}^{\circ }+\theta )\cot (\theta -{90}^{\circ })}}\\ & ={\displaystyle \frac{(-\sin \theta ).\sec \theta .(-\tan \theta )}{\sec \theta .(-\sin \theta ).(-\tan \theta )}}\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Diketahui bahwa}\\ & \sin \theta +\cos \theta ={\displaystyle \frac{1}{2}}\\ & \text{maka nilai dari}\\ & {\sin }^3\theta +{\cos }^3\theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}& & & \text{d}.& {\displaystyle \frac{5}{8}}\\ \\ \text{b}.& {\displaystyle \frac{3}{4}}& \text{c}.& {\displaystyle \frac{9}{15}}& \text{e}.& {\displaystyle \frac{11}{16}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \sin \theta +\cos \theta ={\displaystyle \frac{1}{2}}\\ & \iff {(\sin \theta +\cos \theta )}^2={\displaystyle \frac{1}{4}}\\ & \iff {\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta ={\displaystyle \frac{1}{4}}\\ & \iff 1+2\sin \theta \cos \theta ={\displaystyle \frac{1}{4}}\\ & \iff 2\sin \theta \cos \theta =-{\displaystyle \frac{3}{4}}\\ & \iff \sin \theta \cos \theta =-{\displaystyle \frac{3}{8}}\\ & \mathbf{\text{Selanjutnya}}\\ & {\sin }^3\theta +{\cos }^3\theta \\ & =(\sin \theta +\cos \theta )({\sin }^2\theta -\sin \theta \cos \theta +{\cos }^2\theta )\\ & =\left({\displaystyle \frac{1}{2}}\right)(1-(-{\displaystyle \frac{3}{8}}))\\ & ={\displaystyle \frac{1}{2}\times {\displaystyle \frac{11}{8}}}\\ & ={\displaystyle \frac{11}{16}}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jika diketahui}{\displaystyle \frac{3}{2}\pi <x<2\pi }\\ & \text{dan}\tan x=m,\\ & \text{maka nilai dari}\sin x\cos x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\frac{1}{m^2+1}}& & & \text{d}.& {\displaystyle -\frac{m}{m^2-1}}\\ \\ \text{b}.& {\displaystyle -\frac{m}{m^2+1}}& \text{c}.& {\displaystyle \frac{m}{m^2+1}}& \text{e}.& {\displaystyle \frac{m}{m^2-1}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}{\displaystyle \frac{3}{2}\pi <x<2\pi }\\ & \text{ini daerah Kwadran IV, akibatnya adalah nilai}\\ & \{\begin{array}{ll}\sin x& =-\\ \cos x& =+\\ \tan x& =-\end{array}\\ & \mathbf{\text{Selanjutnya ada pernyataan}}\tan x=m\\ & \text{ini artinya}\tan x={\displaystyle \frac{m}{1}}\\ & \mathbf{\text{Perhatikanlah ilustrasi gambar berikut}}\end{array}\end{array}$.

$.\begin{array}{rl}& \text{maka nilai dari}\\ & \sin x\cos x\left(\text{ingat yang diminta di Kwadran IV}\right)\\ & =(-{\displaystyle \frac{m}{\sqrt{m^2+1}}})\times (+{\displaystyle \frac{1}{\sqrt{m^2+1}}})\\ & =-{\displaystyle \frac{m}{m^2+1}}\end{array}$