Contoh Soal 4 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 16.& \text{Jika}{\tan }^{2}x+\sec x=5\text{untuk}\\ & 0\le x\le {\displaystyle \frac{\pi }{2}\text{maka nilai}\cos x=....}\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& {\displaystyle \frac{1}{2}}\\ \text{c}.& {\displaystyle \frac{1}{3}}\\ \text{d}.& {\displaystyle \frac{1}{\sqrt{2}}}\\ \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Ingat bahwa}& 0\le x\le {\displaystyle \frac{\pi }{2}}\\ \text{berarti sudu}& \text{t}x\text{berada di kuadran I}\\ \text{sehingga ak}& \text{an menyebabkan nilai}\\ & \cos x=+\\ \text{Selanjutnya}& \\ {\tan }^{2}x+\sec x& =5\\ {\sec }^{2}x-1+\sec x& =5\\ {\sec }^{2}x+\sec x-6& =0\\ (\sec x+3)(\sec x-2)& =0\\ \sec x=-3\text{atau}& \sec x=2\\ \text{untuk}\sec x& =-3(\text{tidak memenuhi})\\ \text{untuk}\sec x& =2(\text{memenuhi})\\ \text{Selanjutnya}& \text{lagi}\\ \sec x& =2\\ {\displaystyle \frac{1}{\cos x}}& =2\\ \cos x& ={\displaystyle \frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Nilai}\\ & {(\sin A+\cos A)}^2+{(\sin A-\cos A)}^2=....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& {\displaystyle 2}\\ \text{c}.& {\displaystyle 3}\\ \text{d}.& {\displaystyle 3\cos A}\\ e.& {\displaystyle 4\sin A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {(\sin A+\cos A)}^2+{(\sin A-\cos A)}^2\\ & ={\sin }^{2}A+2\sin A\cos A+{\cos }^{2}A\\ & +{\sin }^{2}A-2\sin A\cos A+{\cos }^{2}A\\ & =1+1\\ & =2\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Nilai}\sqrt{{\displaystyle \frac{1+\sin A}{1-\sin A}}}=....\\ & \begin{array}{l}\\ \text{a}.& \sec A+\tan A\\ \text{b}.& {\sec }^{2}A+{\tan }^{2}A\\ \text{c}.& {\sec }^{2}A-{\tan }^{2}A\\ \text{d}.& {\tan }^{2}A-{\sec }^{2}A\\ e.& \sec A\times \tan A\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \sqrt{{\displaystyle \frac{1+\sin A}{1-\sin A}}}\\ & =\sqrt{{\displaystyle \frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}}}\\ & =\sqrt{{\displaystyle \frac{{(1+\sin A)}^2}{1-{\sin }^{2}A}}}\\ & =\sqrt{{\displaystyle \frac{{(1+\sin A)}^2}{{\cos }^{2}A}}}\\ & ={\displaystyle \frac{1+\sin A}{\cos A}}\\ & ={\displaystyle \frac{1}{\cos A}+\frac{\sin A}{\cos A}}\\ & =\sec A+\tan A\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Jika}{0}^{\circ }\le \theta ⩽{90}^{\circ },\text{maka nilai}\\ & \left({\displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta }}\right)=....\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& {\displaystyle 0}\\ \text{c}.& {\displaystyle \frac{1}{4}}\\ \text{d}.& {\displaystyle \frac{1}{2}}\\ e.& {\displaystyle 1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \left({\displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta }}\right)\\ & =\left({\displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }\times \frac{4+5\cos \theta }{4+5\cos \theta }}\right)\\ & -\left({\displaystyle \frac{3+5\sin \theta }{4+5\cos \theta }\times \frac{3-5\sin \theta }{3-5\sin \theta }}\right)\\ & ={\displaystyle \frac{25{\cos }^2-16-(9-25{\sin }^2\theta )}{(3-5\sin \theta )(4+5\cos \theta )}}\\ & ={\displaystyle \frac{25({\sin }^2\theta +{\cos }^2\theta )-25}{(3-5\sin \theta )(4+5\cos \theta )}}\\ & ={\displaystyle \frac{0}{(3-5\sin \theta )(4+5\cos \theta )}}\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Nilai}\\ & (1+\cot \theta -\csc \theta )(1+\tan \theta +\sec \theta )=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& (1+\cot \theta -\csc \theta )(1+\tan \theta +\sec \theta )\\ & =(1+{\displaystyle \frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }})(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta })\\ & =\left({\displaystyle \frac{\sin \theta +\cos \theta -1}{\sin \theta }}\right)\left({\displaystyle \frac{\cos \theta +\sin \theta +1}{\cos \theta }}\right)\\ & ={\displaystyle \frac{{(\sin \theta +\cos \theta )}^2-1}{\sin \theta \cos \theta }}\\ & ={\displaystyle \frac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }}\\ & =2\end{array}\end{array}$