Latihan Soal 7 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{l}\\ 61.& \text{Nilai}{105}^{\circ }\text{jika dinyatakan ke radian}\\ & \text{adalah}....\text{radian}\\ \\ & \text{a}.{\displaystyle \frac{1}{3}\pi }\\ \\ & \text{b}.{\displaystyle \frac{5}{6}\pi }\\ \\ & \text{c}.{\displaystyle \frac{5}{12}\pi }\\ \\ & \text{d}.{\displaystyle \frac{7}{12}\pi }\\ \\ & \text{e}.{\displaystyle \frac{9}{12}\pi }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ {180}^{\circ }& =\pi radian\\ 1^{\circ }& ={\displaystyle \frac{\pi }{180}radian}\\ 105\times 1^{\circ }& =105\times {\displaystyle \frac{\pi }{180}radian}\\ {105}^{\circ }& ={\displaystyle \frac{7}{12}\pi radian}\end{array}\end{array}$.

$\begin{array}{l}\\ 62.& \text{Nilai}\tan {240}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \sqrt{3}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ \\ & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle -\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {240}^{\circ }& =\tan ({180}^{\circ }+{60}^{\circ })\\ & =\tan {60}^{\circ }\\ & =\sqrt{3}\\ \mathbf{\text{catatan}}& :\text{ingat sudut berelasi}\end{array}\end{array}$.

$\begin{array}{l}\\ 63.& \text{Perhatikanlah gambar berikut}\end{array}$.

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

$\begin{array}{l}\\ 64.& \text{Nilai dari}({\cos }^2{17}^{\circ }-{\sin }^2{73}^{\circ })\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0& & & \text{d}.& 1\\ \text{b}.& {\displaystyle \frac{1}{3}}& \text{c}.& {\displaystyle \frac{2}{\sqrt{3}}}& \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& ({\cos }^2{17}^{\circ }-{\sin }^2{73}^{\circ })\\ & =({\cos }^2{17}^{\circ }-{(\sin {73}^{\circ })}^2)\\ & =({\cos }^2{17}^{\circ }-{(\sin ({90}^{\circ }-{17}^{\circ }))}^2)\\ & =({\cos }^2{17}^{\circ }-{\cos }^2{17}^{\circ })\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 65.& \text{Jika diketahui}\\ & {\displaystyle \frac{x{\csc }^2{30}^{\circ }{\sec }^2{45}^{\circ }}{8{\cos }^2{45}^{\circ }{\sin }^2{60}^{\circ }}={\tan }^2{60}^{\circ }-{\tan }^2{30}^{\circ },}\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2& & & \text{d}.& 1\\ \text{b}.& {\displaystyle -1}& \text{c}.& 0& \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{x{\csc }^2{30}^{\circ }{\sec }^2{45}^{\circ }}{8{\cos }^2{45}^{\circ }{\sin }^2{60}^{\circ }}={\tan }^2{60}^{\circ }-{\tan }^2{30}^{\circ }}\\ & {\displaystyle \frac{x\left(4\right)\left({\displaystyle \frac{4}{2}}\right)}{8\left({\displaystyle \frac{2}{4}}\right)\left({\displaystyle \frac{3}{4}}\right)}=3-\left({\displaystyle \frac{1}{3}}\right)}\\ & {\displaystyle \frac{8x}{3}={\displaystyle \frac{8}{3}}}\\ & x=1\end{array}\end{array}$.

$\begin{array}{l}\\ 66.& \text{Nilai dari}{\displaystyle \frac{\sin {49}^{\circ }}{\cos {41}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin {73}^{\circ }}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1& & & \text{d}.& 0,143\\ \text{b}.& {\displaystyle -0,321}& \text{c}.& 0& \text{e}.& {\displaystyle 0,321}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {49}^{\circ }}{\cos {41}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin {73}^{\circ }}}}\\ & ={\displaystyle \frac{\sin {49}^{\circ }}{\cos ({90}^{\circ }-{49}^{\circ })}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin ({90}^{\circ }-{17}^{\circ })}}}\\ & ={\displaystyle \frac{\sin {49}^{\circ }}{\sin {49}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\cos {17}^{\circ }}}}\\ & =1-1\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 67.& \text{Nilai dari}\\ & p=r\sin \alpha \cos \beta \\ & q=r\sin \alpha \sin \beta \\ & s=r\cos \alpha \\ & \text{maka pernyataan berikut yang}\\ & \text{tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& p^2+t^2+s^2=r^2& & & \\ \text{b}.& p^2-t^2+s^2=r^2\\ \text{c}.& p^2+t^2-s^2=r^2& \\ \text{d}.& -p^2+t^2+s^2=r^2& \\ \text{e}.& -p^2-t^2+s^2=r^2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Saat}\\ & p^2+q^2\text{maka hasilnya adalah}\\ & \begin{array}{l}\\ p^2& =r^2{\sin }^2\alpha {\cos }^2\beta & \\ q^2& =r^2{\sin }^2\alpha {\sin }^2\beta & +\\ & =r^2{\sin }^2\alpha ({\cos }^2\beta +{\sin }^2\beta )\\ & =r^2{\sin }^2\alpha (1)\\ & =r^2{\sin }^2\alpha \end{array}\\ & \text{Dan saat}\\ & p^2+q^2+s^2\text{akan diperoleh hasil}\\ & \begin{array}{l}\\ p^2+q^2& =r^2{\sin }^2\alpha & \\ s^2& =r^2{\cos }^2\alpha & +\\ & =r^2{\sin }^2\alpha +r^2{\cos }^2\alpha \\ & =r^2({\sin }^2\alpha +{\cos }^2\alpha )\\ & =r^2(1)\\ & =r^2\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 68.& \text{Nilai dari}\\ & {\displaystyle \frac{\cos ({90}^{\circ }+\theta )\sec (2\pi -\theta )\tan (\pi -\theta )}{\sec (\theta -2\pi )\sin ({540}^{\circ }+\theta )\cot (\theta -{90}^{\circ })}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1& & & \text{d}.& -\tan \theta \\ \text{b}.& {\displaystyle 0}& \text{c}.& 1& \text{e}.& {\displaystyle \tan \theta }\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Ingat kembali sudut-sudut}\\ & \text{yang berelasi dari kudran selain I}\\ & \text{ke kuadran I beserta tandanya}\\ & {\displaystyle \frac{\cos ({90}^{\circ }+\theta )\sec (2\pi -\theta )\tan (\pi -\theta )}{\sec (\theta -2\pi )\sin ({540}^{\circ }+\theta )\cot (\theta -{90}^{\circ })}}\\ & ={\displaystyle \frac{(-\sin \theta ).\sec \theta .(-\tan \theta )}{\sec \theta .(-\sin \theta ).(-\tan \theta )}}\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 69.& \text{Diketahui bahwa}\\ & \sin \theta +\cos \theta ={\displaystyle \frac{1}{2}}\\ & \text{maka nilai dari}\\ & {\sin }^3\theta +{\cos }^3\theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}& & & \text{d}.& {\displaystyle \frac{5}{8}}\\ \\ \text{b}.& {\displaystyle \frac{3}{4}}& \text{c}.& {\displaystyle \frac{9}{15}}& \text{e}.& {\displaystyle \frac{11}{16}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \sin \theta +\cos \theta ={\displaystyle \frac{1}{2}}\\ & \iff {(\sin \theta +\cos \theta )}^2={\displaystyle \frac{1}{4}}\\ & \iff {\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta ={\displaystyle \frac{1}{4}}\\ & \iff 1+2\sin \theta \cos \theta ={\displaystyle \frac{1}{4}}\\ & \iff 2\sin \theta \cos \theta =-{\displaystyle \frac{3}{4}}\\ & \iff \sin \theta \cos \theta =-{\displaystyle \frac{3}{8}}\\ & \mathbf{\text{Selanjutnya}}\\ & {\sin }^3\theta +{\cos }^3\theta \\ & =(\sin \theta +\cos \theta )({\sin }^2\theta -\sin \theta \cos \theta +{\cos }^2\theta )\\ & =\left({\displaystyle \frac{1}{2}}\right)(1-(-{\displaystyle \frac{3}{8}}))\\ & ={\displaystyle \frac{1}{2}\times {\displaystyle \frac{11}{8}}}\\ & ={\displaystyle \frac{11}{16}}\end{array}\end{array}$.

$\begin{array}{l}\\ 70.& \text{Jika diketahui}{\displaystyle \frac{3}{2}\pi <x<2\pi }\\ & \text{dan}\tan x=m,\\ & \text{maka nilai dari}\sin x\cos x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\frac{1}{m^2+1}}& & & \text{d}.& {\displaystyle -\frac{m}{m^2-1}}\\ \\ \text{b}.& {\displaystyle -\frac{m}{m^2+1}}& \text{c}.& {\displaystyle \frac{m}{m^2+1}}& \text{e}.& {\displaystyle \frac{m}{m^2-1}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}{\displaystyle \frac{3}{2}\pi <x<2\pi }\\ & \text{ini daerah Kwadran IV, akibatnya adalah nilai}\\ & \{\begin{array}{ll}\sin x& =-\\ \cos x& =+\\ \tan x& =-\end{array}\\ & \mathbf{\text{Selanjutnya ada pernyataan}}\tan x=m\\ & \text{ini artinya}\tan x={\displaystyle \frac{m}{1}}\\ & \mathbf{\text{Perhatikanlah ilustrasi gambar berikut}}\end{array}\end{array}$.

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