Contoh Soal 5 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 21.& \text{Nilai dari}{\displaystyle \frac{1}{{\sec }^{2}A}+\frac{1}{{\csc }^{2}A}=....}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ e.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{{\sec }^{2}A}+\frac{1}{{\csc }^{2}A}}\\ & ={\cos }^{2}A+{\sin }^2=1\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Nilai dari}{\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}=....}\\ & \begin{array}{l}\\ \text{a}.& \cot B\times \cot C\\ \text{b}.& {\displaystyle \tan B\times \tan C}\\ \text{c}.& {\displaystyle \sec B\times \csc C}\\ \text{d}.& {\displaystyle \tan B\times \cot C}\\ e.& {\displaystyle \tan B\times \csc C}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}}\\ & ={\displaystyle \frac{\tan B+\tan C}{{\displaystyle \frac{1}{\tan B}+\frac{1}{\tan C}}}}\\ & ={\displaystyle \frac{\tan B+\tan C}{\left({\displaystyle \frac{\tan B+\tan C}{\tan B\times \tan C}}\right)}}\\ & =\tan B\times \tan C\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Nilai dari}\\ & {\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=....}\\ & \begin{array}{l}\\ \text{a}.& 2\tan A\\ \text{b}.& 2\cot A\\ \text{c}.& {\displaystyle 2\sec A}\\ \text{d}.& {\displaystyle 2\csc A}\\ e.& {\displaystyle 2\tan A.\sec A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}}\\ & =\tan A\left({\displaystyle \frac{1}{{\displaystyle \frac{1}{\cos A}-1}}+\frac{1}{{\displaystyle \frac{1}{\cos A}+1}}}\right)\\ & ={\displaystyle \frac{\sin A}{\cos A}\left({\displaystyle \frac{\cos A}{1-\cos A}+\frac{\cos A}{1+\cos A}}\right)}\\ & ={\displaystyle \frac{\sin A}{1-\cos A}+\frac{\sin A}{1+\cos A}}\\ & ={\displaystyle \frac{\sin A(1+\cos A)+\sin A(1-\cos A)}{(1-\cos A)(1+\cos A)}}\\ & ={\displaystyle \frac{2\sin A}{1-{\cos }^2}}\\ & ={\displaystyle \frac{2\sin A}{{\sin }^{2}A}}\\ & ={\displaystyle \frac{2}{\sin A}}\\ & =2\csc A\end{array}\\ \\ & \text{Sebagai catatanya}\\ & \text{Anda bisa gunakan cara yang lain}\end{array}$

$\begin{array}{l}\\ 24.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & \sin (2x-{20}^{\circ })=-\cos (3x+{50}^{\circ })\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{30}^{\circ }\\ \text{b}.& -{25}^{\circ }\\ \text{c}.& {20}^{\circ }\\ \text{d}.& {25}^{\circ }\\ e.& {30}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sin (2x-{20}^{\circ })& =-\cos (3x+{50}^{\circ })\\ \sin ({20}^{\circ }-2x)& =\cos (3x+{50}^{\circ })\\ \sin A& =\cos B,\text{artinya}\\ A+B& ={90}^{\circ },\text{maka}\\ ({20}^{\circ }-2x)+(3x+{50}^{\circ })& ={90}^{\circ }\\ x+{70}^{\circ }& ={90}^{\circ }\\ x& ={90}^{\circ }-{70}^{\circ }\\ & ={20}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & \tan (2x+{60}^{\circ })=\cot ({90}^{\circ }-3x)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {20}^{\circ }\\ \text{b}.& {30}^{\circ }\\ \text{c}.& {40}^{\circ }\\ \text{d}.& {50}^{\circ }\\ \text{e}.& {60}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\tan (2x+{60}^{\circ })& =\cot ({90}^{\circ }-3x)\\ \tan (2x+{60}^{\circ })& =\tan 3x\\ 2x+{60}^{\circ }& =3x\\ 2x-3x& =-{60}^{\circ }\\ -x& =-{60}^{\circ }\\ x& ={60}^{\circ }\end{array}\end{array}$