Latihan Soal 6 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{l}\\ 51.& \text{Nilai dari}{\displaystyle \frac{1}{{\sec }^{2}A}+\frac{1}{{\csc }^{2}A}=....}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ e.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{{\sec }^{2}A}+\frac{1}{{\csc }^{2}A}}\\ & ={\cos }^{2}A+{\sin }^2=1\end{array}\end{array}$

$\begin{array}{l}\\ 52.& \text{Nilai dari}{\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}=....}\\ & \begin{array}{l}\\ \text{a}.& \cot B\times \cot C\\ \text{b}.& {\displaystyle \tan B\times \tan C}\\ \text{c}.& {\displaystyle \sec B\times \csc C}\\ \text{d}.& {\displaystyle \tan B\times \cot C}\\ e.& {\displaystyle \tan B\times \csc C}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}}\\ & ={\displaystyle \frac{\tan B+\tan C}{{\displaystyle \frac{1}{\tan B}+\frac{1}{\tan C}}}}\\ & ={\displaystyle \frac{\tan B+\tan C}{\left({\displaystyle \frac{\tan B+\tan C}{\tan B\times \tan C}}\right)}}\\ & =\tan B\times \tan C\end{array}\end{array}$

$\begin{array}{l}\\ 53.& \text{Nilai dari}\\ & {\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=....}\\ & \begin{array}{l}\\ \text{a}.& 2\tan A\\ \text{b}.& 2\cot A\\ \text{c}.& {\displaystyle 2\sec A}\\ \text{d}.& {\displaystyle 2\csc A}\\ e.& {\displaystyle 2\tan A.\sec A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}}\\ & =\tan A\left({\displaystyle \frac{1}{{\displaystyle \frac{1}{\cos A}-1}}+\frac{1}{{\displaystyle \frac{1}{\cos A}+1}}}\right)\\ & ={\displaystyle \frac{\sin A}{\cos A}\left({\displaystyle \frac{\cos A}{1-\cos A}+\frac{\cos A}{1+\cos A}}\right)}\\ & ={\displaystyle \frac{\sin A}{1-\cos A}+\frac{\sin A}{1+\cos A}}\\ & ={\displaystyle \frac{\sin A(1+\cos A)+\sin A(1-\cos A)}{(1-\cos A)(1+\cos A)}}\\ & ={\displaystyle \frac{2\sin A}{1-{\cos }^2}}\\ & ={\displaystyle \frac{2\sin A}{{\sin }^{2}A}}\\ & ={\displaystyle \frac{2}{\sin A}}\\ & =2\csc A\end{array}\\ \\ & \text{Sebagai catatanya}\\ & \text{Anda bisa gunakan cara yang lain}\end{array}$

$\begin{array}{l}\\ 54.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & \sin (2x-{20}^{\circ })=-\cos (3x+{50}^{\circ })\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{30}^{\circ }\\ \text{b}.& -{25}^{\circ }\\ \text{c}.& {20}^{\circ }\\ \text{d}.& {25}^{\circ }\\ e.& {30}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sin (2x-{20}^{\circ })& =-\cos (3x+{50}^{\circ })\\ \sin ({20}^{\circ }-2x)& =\cos (3x+{50}^{\circ })\\ \sin A& =\cos B,\text{artinya}\\ A+B& ={90}^{\circ },\text{maka}\\ ({20}^{\circ }-2x)+(3x+{50}^{\circ })& ={90}^{\circ }\\ x+{70}^{\circ }& ={90}^{\circ }\\ x& ={90}^{\circ }-{70}^{\circ }\\ & ={20}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 55.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & \tan (2x+{60}^{\circ })=\cot ({90}^{\circ }-3x)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {20}^{\circ }\\ \text{b}.& {30}^{\circ }\\ \text{c}.& {40}^{\circ }\\ \text{d}.& {50}^{\circ }\\ \text{e}.& {60}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\tan (2x+{60}^{\circ })& =\cot ({90}^{\circ }-3x)\\ \tan (2x+{60}^{\circ })& =\tan 3x\\ 2x+{60}^{\circ }& =3x\\ 2x-3x& =-{60}^{\circ }\\ -x& =-{60}^{\circ }\\ x& ={60}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 56.& \text{Jika nilai}\cot A+\cos A=x\\ & \text{dan}\cot A-\cos A=y,\\ & \text{maka nilai}(x^2-y^2)=....\\ & \begin{array}{l}\\ \text{a}.& 4\sqrt{xy}\\ \text{b}.& 2\sqrt{xy}\\ \text{c}.& xy\\ \text{d}.& 2\\ \text{e}.& 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}xy& =(\cot A+\cos A)(\cot A-\cos A)\\ & ={\cot }^{2}A-{\cos }^{2}A\\ & ={\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}-{\cos }^{2}A}\\ & ={\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}-{\cos }^{2}A\times \frac{{\sin }^{2}A}{{\sin }^{2}A}}\\ & ={\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}(1-{\sin }^{2}A)}\\ & ={\displaystyle \frac{{\cos }^{4}A}{{\sin }^{2}A}}\\ \sqrt{xy}& ={\displaystyle \frac{\cos A}{\sin A}\times \cos A}\\ \text{Se}& \text{lanjutnya}\\ x^2-y^2& ={(\cot A+\cos A)}^2-{(\cot A-\cos A)}^2\\ (x+y)& (x-y)=(\cot A+\cos A+\cot A-\cos A)\\ & \times (\cot A+\cos A-(\cot A-\cos A))\\ x^2-y^2& =2\cot A\times 2\cos A\\ & =4\times {\displaystyle \frac{\cos A}{\sin A}\times \cos A}\\ & =4\sqrt{xy}\end{array}\end{array}$

$\begin{array}{l}\\ 57.& \text{Jika nilai}\cos A+\sin A=\sqrt{2}\cos A\\ & \text{maka nilai}(\cos A-\sin A)=....\\ & \begin{array}{l}\\ \text{a}.& -\sqrt{2}\cos A\\ \text{b}.& -\sqrt{2}\sin A\\ \text{c}.& \sqrt{2}\sin A\\ \text{d}.& {\displaystyle \frac{1}{\sqrt{2}}\sec A}\\ \text{e}.& {\displaystyle \frac{1}{\sqrt{2}}\csc A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\cos A+\sin A& =\sqrt{2}\cos A\\ {(\cos A+\sin A)}^2& ={(\sqrt{2}\cos A)}^2\\ 1+2\sin A\cos A& =2{\cos }^{2}A\\ 2\sin A\cos A& =2{\cos }^{2}A-1\\ \text{maka}& \\ {(\cos A-\sin A)}^2& =1-2\sin A\cos A\\ & =1-(2{\cos }^{2}A-1)\\ & =2-2{\cos }^{2}A\\ & =2(1-{\cos }^{2}A)\\ & =2{\sin }^{2}A\\ \cos A-\sin A& =\sqrt{2{\sin }^{2}A}\\ & =\sin A\sqrt{2}\\ & =\sqrt{2}.\sin A\end{array}\end{array}$.

$\begin{array}{l}\\ 58.& \text{Nilai}\cos \gamma (\csc \gamma +\tan \gamma )\\ & \text{adalah ekivalen dengan}....\\ & \begin{array}{l}\\ \text{a}.& \cot \gamma +\sin \gamma \\ \text{b}.& \tan \gamma +\cos \gamma \\ \text{c}.& \cot \gamma -\sin \gamma \\ \text{d}.& \tan \gamma -\cos \gamma \\ \text{e}.& \cot \gamma -\tan \gamma \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\cos \gamma (\csc \gamma +\tan \gamma )& =\cos \gamma \left({\displaystyle \frac{1}{\sin \gamma }+{\displaystyle \frac{\sin \gamma }{\cos \gamma }}}\right)\\ & =\cos \gamma \left({\displaystyle \frac{\cos \gamma +{\sin }^2\gamma }{\sin \gamma \cos \gamma }}\right)\\ & ={\displaystyle \frac{\cos \gamma +{\sin }^2\gamma }{\sin \gamma }}\\ & ={\displaystyle \frac{\cos \gamma }{\sin \gamma }+{\displaystyle \frac{{\sin }^2\gamma }{\sin \gamma }}}\\ & =\cot \gamma +\sin \gamma \end{array}\end{array}$.

$\begin{array}{l}\\ 59.& \text{Jika diketahui}\sin \theta \cos \theta ={\displaystyle \frac{3}{8},}\\ & \text{maka nilai}{\displaystyle \frac{1}{\sin \theta }-{\displaystyle \frac{1}{\cos \theta }\text{adalah}=....}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{4}}\\ \text{b}.& {\displaystyle \frac{1}{2}}\\ \text{c}.& {\displaystyle \frac{3}{4}}\\ \text{d}.& {\displaystyle \frac{4}{3}}\\ \text{e}.& {\displaystyle 4}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{\sin \theta }-{\displaystyle \frac{1}{\cos \theta }}}\\ & =\sqrt{{\left({\displaystyle \frac{1}{\sin \theta }-{\displaystyle \frac{1}{\cos \theta }}}\right)}^2}\\ & =\sqrt{{\displaystyle \frac{1}{{\sin }^2\theta }-{\displaystyle \frac{2}{\sin \theta \cos \theta }+{\displaystyle \frac{1}{{\cos }^2\theta }}}}}\\ & =\sqrt{{\displaystyle \frac{{\cos }^2\theta +{\sin }^2\theta }{{\sin }^2\theta {\cos }^2\theta }-{\displaystyle \frac{2}{\sin \theta \cos \theta }}}}\\ & =\sqrt{{\displaystyle \frac{1}{{\left({\displaystyle \frac{3}{8}}\right)}^2}-{\displaystyle \frac{2}{\left({\displaystyle \frac{3}{8}}\right)}}}}=\sqrt{{\displaystyle \frac{1}{{\displaystyle \frac{9}{64}}}-{\displaystyle \frac{2}{{\displaystyle \frac{3}{8}}}}}}=\sqrt{{\displaystyle \frac{64}{9}-{\displaystyle \frac{16}{3}}}}\\ & =\sqrt{{\displaystyle \frac{64}{9}-{\displaystyle \frac{48}{9}}}}=\sqrt{{\displaystyle \frac{16}{9}}}={\displaystyle \frac{4}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 60.& \text{Bentuk}{\displaystyle \frac{(\sin \alpha +\sin \beta )(\sin \alpha -\sin \beta )}{{(\cos \alpha \cos \beta )}^2}}\\ & \text{ekivalen dengan}....\\ & \begin{array}{l}\\ \text{a}.& \tan \alpha -\tan \beta \\ \text{b}.& \tan \alpha +\tan \beta \\ \text{c}.& {\tan }^2\alpha -{\tan }^2\beta \\ \text{d}.& {\tan }^2\alpha +{\tan }^2\beta \\ \text{e}.& {\tan }^3\alpha -{\tan }^3\beta \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {\displaystyle \frac{(\sin \alpha +\sin \beta )(\sin \alpha -\sin \beta )}{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha -{\sin }^2\beta }{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha -{\sin }^2\alpha {\sin }^2\beta -{\sin }^2\beta +{\sin }^2\alpha {\sin }^2\beta }{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha (1-{\sin }^2\beta )-{\sin }^2\beta (1-{\sin }^2\alpha )}{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha {\cos }^2\beta -{\sin }^2\beta {\cos }^2\alpha }{{(\cos \alpha \cos \beta )}^2}}\\ & ={\displaystyle \frac{{\sin }^2\alpha {\cos }^2\beta }{{(\cos \alpha \cos \beta )}^2}-{\displaystyle \frac{{\sin }^2\beta {\cos }^2\alpha }{{(\cos \alpha \cos \beta )}^2}}}\\ & ={\displaystyle \frac{{\sin }^2\alpha }{{\cos }^2\alpha }-{\displaystyle \frac{{\sin }^2\beta }{{\cos }^2\beta }}}\\ & ={\tan }^2\alpha -{\tan }^2\beta \end{array}\end{array}$.

DAFTAR PUSTAKA

1. Sembiring, S., Zulkifli, M., Marsito, dan Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU.
2. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA