Contoh Soal 12 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{l}\\ 56.& \text{Diketahui fungsi}f(x)={\displaystyle \frac{1}{2}\sin 2x\text{dengan}}\\ & 0^{\circ }<x<{360}^{\circ }.\text{Kurva akan cekung}\\ & \text{ke atas pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0^{\circ }<x<{90}^{\circ }\\ \text{b}.& 0^{\circ }<x<{90}^{\circ }\text{atau}{180}^{\circ }<x<{270}^{\circ }\\ \text{c}.& {45}^{\circ }<x<{225}^{\circ }\\ \text{d}.& {90}^{\circ }<x<{180}^{\circ }\text{atau}{270}^{\circ }<x<{360}^{\circ }\\ \text{e}.& {180}^{\circ }<x<{225}^{\circ }\text{atau}{225}^{\circ }<x<{360}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)={\displaystyle \frac{1}{2}\sin 2x}\\ & f^{\prime }(x)=\cos 2x\Rightarrow f^{″}(x)=-2\sin 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -2\sin 2x=0\iff \sin 2x=0\\ & \iff \sin 2x=\sin 0^{\circ }\\ & \iff 2x=0^{\circ }+k{.360}^{\circ }\text{atau}2x={180}^{\circ }+k{.360}^{\circ }\\ & \iff x=0^{\circ }+k{.180}^{\circ }\text{atau}x={90}^{\circ }+k{.180}^{\circ }\\ & \iff x=0^{\circ },x={90}^{\circ },x={180}^{\circ }\text{dan}x={270}^{\circ }\\ & \text{serta}x={360}^{\circ }\\ & \bullet \text{Selang}{0}^{\circ }<x<{90}^{\circ },\text{misal}x={45}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({45}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{90}^{\circ }<x<{180}^{\circ },\text{misal}x={135}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({135}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{180}^{\circ }<x<{270}^{\circ },\text{misal}x={225}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({225}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{270}^{\circ }<x<{360}^{\circ },\text{misal}x={315}^{\circ }\\ & \Rightarrow \Rightarrow f^{″}=-2\sin 2\left({315}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\end{array}\end{array}$

$\begin{array}{l}\\ 57.& \text{Diketahui fungsi}f(x)={\cos }^{2}x-{\sin }^{2}x\text{dengan}\\ & 0<x<2\pi .\text{Kurva akan cekung ke bawah}\\ & \text{pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{\pi }{2}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\text{atau}{\displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4}}}\\ \text{c}.& {\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}\text{atau}{\displaystyle \frac{7\pi }{4}<x<2\pi }}\\ \text{d}.& {\displaystyle \frac{7\pi }{4}<x<2\pi }\\ \text{e}.& {\displaystyle \frac{5\pi }{4}<x<2\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& f(x)={\cos }^{2}x-{\sin }^{2}x=\cos 2x\\ & f^{\prime }(x)=-2\sin 2x\Rightarrow f^{″}(x)=-4\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -4\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x={\displaystyle \frac{\pi }{4}+k.\pi }}\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4}\text{dan}x=\frac{7\pi }{4}}\\ & \text{Ingat bahwa domain}0<x<2\pi \text{saja}\\ & \bullet \text{Selang}0<x<{\displaystyle \frac{\pi }{4},\text{misal}x={30}^{\circ }=\frac{\pi }{6}}\\ & \Rightarrow f^{″}({30}^{\circ })=-4\cos 2\left({30}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{3\pi }{4},\text{misal}x={120}^{\circ }={\displaystyle \frac{2\pi }{3}}}}\\ & \Rightarrow f^{″}({120}^{\circ })=-4\cos 2\left({90}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\text{misal}x={210}^{\circ }=\frac{7\pi }{6}}\\ & \Rightarrow f^{″}({210}^{\circ })=-4\cos 2\left({210}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\text{misal}x={300}^{\circ }={\displaystyle \frac{5\pi }{3}}}\\ & \Rightarrow f^{″}({300}^{\circ })=-4\cos 2\left({300}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{7\pi }{4}<x<2\pi ,\text{misal}x={330}^{\circ }=\frac{11\pi }{6}}\\ & \Rightarrow f^{″}=-4\cos 2\left({330}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\end{array}\end{array}$

$\begin{array}{l}\\ 58.& \text{Diketahui fungsi}f(x)={\sin }^{2}x\text{dengan}\\ & 0<x<2\pi .\text{Kurva fungsi tersebut akan}\\ & \text{cekung ke bawah pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\text{atau}\frac{5\pi }{4}<x<\frac{7\pi }{4}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\text{atau}\frac{7\pi }{4}<x<2\pi }\\ \text{c}.& {\displaystyle 0<x<\frac{\pi }{2}\text{atau}\frac{3\pi }{4}<x<\frac{5\pi }{4}}\\ \text{d}.& {\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}}\\ \text{e}.& {\displaystyle 0<x<\frac{\pi }{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& f(x)={\sin }^{2}x\\ & f^{\prime }(x)=2\sin x\cos x=\sin 2x\\ & f^{″}(x)=2\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & 2\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x={\displaystyle \frac{\pi }{4}+k.\pi }}\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x=\frac{5\pi }{4}\text{dan}x=\frac{7\pi }{4}}\\ & \text{Ingat bahwa domain}0<x<2\pi \text{saja}\\ & \bullet \text{Selang}0<x<{\displaystyle \frac{\pi }{4},\text{misal}x={30}^{\circ }=\frac{\pi }{6}}\\ & \Rightarrow f^{″}({30}^{\circ })=2\cos 2\left({30}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{3\pi }{4},\text{misal}x={120}^{\circ }={\displaystyle \frac{2\pi }{3}}}}\\ & \Rightarrow f^{″}({120}^{\circ })=2\cos 2\left({90}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\text{misal}x={210}^{\circ }=\frac{7\pi }{6}}\\ & \Rightarrow f^{″}({210}^{\circ })=2\cos 2\left({210}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\\ & \bullet \text{Selang}{\displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\text{misal}x={300}^{\circ }={\displaystyle \frac{5\pi }{3}}}\\ & \Rightarrow f^{″}({300}^{\circ })=2\cos 2\left({300}^{\circ }\right)=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{7\pi }{4}<x<2\pi ,\text{misal}x={330}^{\circ }=\frac{11\pi }{6}}\\ & \Rightarrow f^{″}=2\cos 2\left({330}^{\circ }\right)=2>0\\ & \text{pada selang ini kurva cekung ke atas}\end{array}\end{array}$

$\begin{array}{l}\\ 59.& \text{Diketahui fungsi}f(x)=2\sin x-2\cos x\text{dengan}\\ & 0<x<2\pi .\text{Kurva akan cekung ke atas}\\ & \text{pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{3\pi }{4}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{5\pi }{4}}}\\ \text{c}.& {\displaystyle \frac{3\pi }{4}<x<2\pi }\\ \text{d}.& 0<x<\frac{\pi }{4}\text{atau}{\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}}\\ \text{e}.& 0<x<\frac{\pi }{4}\text{atau}{\displaystyle \frac{5\pi }{4}<x<2\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& f(x)=2\sin x-2\cos x\\ & f^{\prime }(x)=2\cos x+2\sin x\\ & f^{″}(x)=-2\sin x+2\cos x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -2\sin x+2\cos x=0\iff \sin x=\cos x\\ & \iff \tan x=1\\ & \iff x={\displaystyle \frac{\pi }{4}+k.\pi }\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{5\pi }{4}}\\ & \text{Ingat bahwa domain}0<x<2\pi \text{saja}\\ & \text{Sebagai gambaran saja}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{3\pi }{4},\text{misal}x={90}^{\circ }={\displaystyle \frac{\pi }{2}}}}\\ & \Rightarrow f^{″}({90}^{\circ })=-2\sin {90}^{\circ }+2\cos {90}^{\circ }=-2<0\\ & \text{pada selang ini kurva cekung ke bawah}\end{array}\end{array}$

$\begin{array}{l}\\ 60.& \text{Diketahui fungsi}f(x)=\sin (3x+{\displaystyle \frac{\pi }{2}})\\ & \text{dengan}0<x<2\pi .\text{Kurva fungsi tersebut}\\ & \text{akan cekung ke atas pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0<x<\frac{\pi }{6}\text{atau}\frac{\pi }{2}<x<\frac{5\pi }{6}}\\ \text{b}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\text{atau}\frac{5\pi }{6}<x<\pi }\\ \text{c}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\text{atau}\frac{3\pi }{4}<x<\frac{5\pi }{6}}\\ \text{d}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\text{atau}\frac{3\pi }{4}<x<\frac{5\pi }{6}}\\ \text{e}.& {\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\text{atau}\frac{5\pi }{6}<x<\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& f(x)=\sin (3x+{\displaystyle \frac{\pi }{2}})\\ & f^{\prime }(x)=3\cos (3x+{\displaystyle \frac{\pi }{2}})\\ & f^{″}(x)=-9\sin (3x+{\displaystyle \frac{\pi }{2}})\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -9\sin (3x+{\displaystyle \frac{\pi }{2}})=0\iff \sin (3x+{\displaystyle \frac{\pi }{2}})=0\\ & \iff \sin (3x+{\displaystyle \frac{\pi }{2}})=\sin 0\\ & \iff (3x+{\displaystyle \frac{\pi }{2}})=0+k.2\pi \iff (3x+{\displaystyle \frac{\pi }{2}})={\displaystyle \pi +k.2\pi }\\ & \iff 3x=-{\displaystyle \frac{\pi }{2}+k.2\pi \iff 3x={\displaystyle \frac{\pi }{2}+k.2\pi }}\\ & \iff x=-{\displaystyle \frac{\pi }{6}+k.\frac{2\pi }{3}\iff x={\displaystyle \frac{\pi }{6}+k.{\displaystyle \frac{2\pi }{3}}}}\\ & \iff x={\displaystyle \frac{\pi }{6},x={\displaystyle \frac{\pi }{2},x=\frac{5\pi }{6},x=\frac{7\pi }{6},}}\\ & \text{dan}x=\frac{3\pi }{2},\text{serta}x=\frac{11\pi }{6}\\ & \text{Ingat bahwa domain}0\le x\le 2\pi \text{saja}\\ & \text{Sebagai GAMBARAN saja, diberikan 2 nilai selang}\\ & \bullet \text{Selang}0<x<{\displaystyle \frac{\pi }{6},\text{misal}x={15}^{\circ }=\frac{\pi }{12}}\\ & \Rightarrow f^{″}({15}^{\circ })=-9\sin (3\left({\displaystyle \frac{\pi }{12}}\right)+{\displaystyle \frac{\pi }{2}})=-{\displaystyle \frac{9}{2}\sqrt{2}<0}\\ & \text{pada selang ini kurva cekung ke bawah}\\ & \bullet \text{Selang}{\displaystyle \frac{\pi }{6}<x<{\displaystyle \frac{\pi }{2},\text{misal}x={60}^{\circ }={\displaystyle \frac{\pi }{3}}}}\\ & \Rightarrow f^{″}({60}^{\circ })=-9\sin (3\left({\displaystyle \frac{\pi }{3}}\right)+{\displaystyle \frac{\pi }{2}})=9>0\\ & \text{pada selang ini kurva cekung ke atas}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan MAtematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
2. Tasari, Aksin, N., Miyanto, Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten. PT. INTAN PARIWARA.

Contoh Soal 11 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{l}\\ 51.& \text{Diketahui}f(x)={\cos }^{2}2x.\text{Jika}\\ & f^{″}(x)=a{\sin }^{2}bx+c{\cos }^{2}dx,\text{nilai untuk}\\ & {\displaystyle \frac{a-b}{c-d}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{3}}\\ \text{b}.& {\displaystyle \frac{2}{3}}\\ \text{c}.& -{\displaystyle \frac{3}{5}}\\ \text{d}.& -{\displaystyle \frac{6}{5}}\\ \text{e}.& -{\displaystyle \frac{9}{5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& f(x)={\cos }^{2}2x\\ & f^{\prime }(x)=2\cos 2x(-\sin 2x)(2)\\ & =-4\sin 2x\cos 2x\\ & f^{″}(x)=-4\cos 2x.(2).\cos 2x-4\sin 2x.(-\sin 2x)(2)\\ & =8{\sin }^{2}2x-8{\cos }^{2}2x\\ & \text{Bandingkan dengan}\\ & f^{″}(x)=a{\sin }^{2}bx+c{\cos }^{2}dx\\ & \text{maka},a=8,b=2,c=-8,d=2\\ & \text{Jadi},{\displaystyle \frac{a-b}{c-d}=\frac{8-2}{-8-2}=-\frac{3}{5}}\end{array}\end{array}$

$\begin{array}{l}\\ 52.& \text{Diketahui}f(x)={\displaystyle \frac{\cos x}{\sin x+\cos x}.\text{Jika}}\\ & f^{″}(x)={\displaystyle \frac{m\cos 2x}{{(\sin 2x+n)}^2},\text{nilai dari}}\\ & m.n=....\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 4\\ \text{c}.& 5\\ \text{d}.& 8\\ \text{e}.& 10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& f(x)={\displaystyle \frac{\cos x}{\sin x+\cos x}}\\ & f^{\prime }(x)={\displaystyle \frac{-\sin x(\sin x+\cos x)-\cos x(\cos x-\sin x)}{(\sin x+\cos x{)}^2}}\\ & ={\displaystyle \frac{-{\sin }^{2}x-{\cos }^{2}x+0}{{\sin }^2+2\sin x\cos x+{\cos }^{2}x}}\\ & ={\displaystyle \frac{-1}{1+\sin 2x}}\\ & f^{″}(x)={\displaystyle \frac{0-((-1).2\cos 2x)}{{(\sin 2x+1)}^2}=\frac{2\cos 2x}{{(\sin 2x+1)}^2}}\\ & \text{Bandingkan dengan yang diketahui}\\ & f^{″}(x)={\displaystyle \frac{m\cos 2x}{{(\sin 2x+n)}^2}}\\ & \{\begin{array}{ll}m& =2\\ n& =2\end{array}\\ & \text{Jadi},m.n=2.1=2\end{array}\end{array}$

$\begin{array}{l}\\ 53.& \text{Salah satu titik belok dari fungsi}\\ & f(x)=\sin 2x\text{dengan}0\le x\le 2\pi \\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{4},0}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{2},0}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{4},1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{2},1}\right)\\ \text{e}.& (\pi ,1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& f(x)=\sin 2x\\ & f^{\prime }(x)=2\cos 2x\Rightarrow f^{″}(x)=-4\sin 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & -4\sin 2x=0\iff \sin 2x=0\\ & \iff \sin 2x=\sin 0\\ & \iff 2x=0+k.2\pi \text{atau}2x=\pi +k.2\pi \\ & \iff x=0+k.\pi \text{atau}x=\frac{\pi }{2}+k.\pi \\ & \iff x=0,x={\displaystyle \frac{\pi }{2},x=\pi ,x={\displaystyle \frac{3\pi }{2}\text{atau}x=2\pi }}\\ & \bullet f\left({\displaystyle \frac{\pi }{2}}\right)=\sin 2\left({\displaystyle \frac{\pi }{2}}\right)=0\Rightarrow \left({\displaystyle \frac{\pi }{2},0}\right)\\ & \bullet f\left({\displaystyle \pi }\right)=\sin 2\left({\displaystyle \pi }\right)=0\Rightarrow \left({\displaystyle \pi ,0}\right)\\ & \bullet f\left({\displaystyle \frac{3\pi }{2}}\right)=\sin 2\left({\displaystyle \frac{3\pi }{2}}\right)=0\Rightarrow \left({\displaystyle \frac{3\pi }{2},0}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 54.& \text{Diketahui fungsi}f(x)=-3\cos 2x+1\text{dengan}\\ & 0<x<2\pi .\text{Salah satu koordinat titik belok}\\ & \text{dari fungsi}f(x)\text{tersebut}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{2},2}\right)\\ \text{b}.& \left({\displaystyle \frac{2\pi }{3},\frac{5}{2}}\right)\\ \text{c}.& \left({\displaystyle \frac{3\pi }{2},4}\right)\\ \text{d}.& \left({\displaystyle \frac{5\pi }{4},1}\right)\\ \text{e}.& \left({\displaystyle \frac{5\pi }{3},\frac{5}{2}}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)=-3\cos 2x+1\text{untuk}0<x<2\pi \\ & f^{\prime }(x)=6\sin 2x\Rightarrow f^{″}(x)=12\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & 12\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x=\pm \frac{\pi }{4}+k.\pi }\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x={\displaystyle \frac{5\pi }{4}\text{atau}x=\frac{7\pi }{4}}}\\ & \bullet f\left({\displaystyle \frac{\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{\pi }{4},1}\right)\\ & \bullet f\left({\displaystyle \frac{3\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{3\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{3\pi }{4},1}\right)\\ & \bullet f\left({\displaystyle \frac{5\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{5\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{5\pi }{4},1}\right)\\ & \bullet f\left({\displaystyle \frac{7\pi }{4}}\right)=-3\cos 2\left({\displaystyle \frac{7\pi }{4}}\right)+1=1\Rightarrow \left({\displaystyle \frac{7\pi }{4},1}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 55.& \text{Diketahui fungsi}f(x)={\sin }^{2}x+2\text{dengan}\\ & 0<x<2\pi .\text{Salah satu koordinat titik belok}\\ & \text{dari fungsi}f(x)\text{tersebut}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{4},\frac{5}{2}}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{3},\frac{11}{4}}\right)\\ \text{c}.& \left({\displaystyle \pi ,2}\right)\\ \text{d}.& \left({\displaystyle \frac{4\pi }{3},\frac{11}{4}}\right)\\ \text{e}.& \left({\displaystyle \frac{11\pi }{6},\frac{9}{4}}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& f(x)={\sin }^{2}x+2\text{untuk}0<x<2\pi \\ & f^{\prime }(x)=2\sin x\cos x\Rightarrow f^{\prime }(x)=\sin 2x\\ & f^{″}(x)=2\cos 2x\\ & \text{Syarat belok}{f}^{″}(x)=0\\ & 2\cos 2x=0\iff \cos 2x=0\\ & \iff \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi \iff x=\pm \frac{\pi }{4}+k.\pi }\\ & \iff x={\displaystyle \frac{\pi }{4},x=\frac{3\pi }{4},x={\displaystyle \frac{5\pi }{4}\text{atau}x=\frac{7\pi }{4}}}\\ & \bullet f\left({\displaystyle \frac{\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{\pi }{4},\frac{5}{2}}\right)\\ & \bullet f\left({\displaystyle \frac{3\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{3\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{3\pi }{4},\frac{5}{2}}\right)\\ & \bullet f\left({\displaystyle \frac{5\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{5\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{5\pi }{4},\frac{5}{2}}\right)\\ & \bullet f\left({\displaystyle \frac{7\pi }{4}}\right)={\sin }^2\left({\displaystyle \frac{7\pi }{4}}\right)+2=\frac{5}{2}\Rightarrow \left({\displaystyle \frac{7\pi }{4},\frac{5}{2}}\right)\end{array}\end{array}$

Contoh Soal 10 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{l}\\ 46.& \text{Turunan kedua dari}f(x)=x^3-\sin 3x\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 6x^2+9\sin 3x\\ \text{b}.& 3x^2+6\sin 3x\\ \text{c}.& 3x-9\sin 3x\\ \text{d}.& 6x+9\sin 3x\\ \text{e}.& 9x-6\sin 3x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}f(x)& =x^3-\sin 3x\\ f^{\prime }(x)& =3x^2-3\cos 3x\\ f^{″}(x)& =6x+9\sin 3x\end{array}\end{array}$

$\begin{array}{l}\\ 47.& \text{Diketahui fungsi}g(x)={\displaystyle \frac{1-\cos x}{\sin x}.\text{Nilai}}\\ & \text{turunan kedua saat}x={\displaystyle \frac{\pi }{4}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{2}+4\\ \text{b}.& 2\sqrt{2}-3\\ \text{c}.& 2\sqrt{2}+3\\ \text{d}.& 3\sqrt{2}-4\\ \text{e}.& 3\sqrt{2}+4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}g(x)& ={\displaystyle \frac{1-\cos x}{\sin x}}\\ g^{\prime }(x)& ={\displaystyle \frac{\sin x(\sin x)-\cos x(1-\cos x)}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{{\sin }^{2}x-\cos x+{\cos }^{2}x}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{1-\cos x}{{\sin }^{2}x}}\\ g^{″}(x)& ={\displaystyle \frac{\sin x({\sin }^{2}x)-2\sin x\cos x(1-\cos x)}{{\sin }^{4}x}}\\ & ={\displaystyle \frac{\sin x({\sin }^{2}x)-\sin 2x(1-\cos x)}{{\sin }^{4}x}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{\pi }{4}({\sin }^2{\displaystyle \frac{\pi }{4})-\sin 2{\displaystyle \frac{\pi }{4}(1-\cos {\displaystyle \frac{\pi }{4})}}}}}{{\sin }^4{\displaystyle \frac{\pi }{4}}}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{1}{\sqrt{2}}}\right){\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2-1.(1-\left({\displaystyle \frac{1}{\sqrt{2}}}\right))}{{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^4}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}{\displaystyle \frac{1}{\sqrt{2}}-1+{\displaystyle \frac{1}{\sqrt{2}}}}}}{{\displaystyle \frac{1}{4}}}\times {\displaystyle \frac{4}{4}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2}{\sqrt{2}}-4+\frac{4}{\sqrt{2}}}}{1}}\\ & ={\displaystyle \frac{6}{\sqrt{2}}-4=3\sqrt{2}-4}\end{array}\end{array}$

$\begin{array}{l}\\ 48.& \text{Turunan kedua fungsi}f(x)={\sin }^{2}x-{\cos }^{2}x\\ & \text{adalah}{f}^{″}(x)=....\\ & \begin{array}{l}\\ \text{a}.& 6\sin 2x\\ \text{b}.& 4\cos 2x\\ \text{c}.& 2\cos 2x\\ \text{d}.& -2\cos 2x\\ \text{e}.& -4\cos 2x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(x)& ={\sin }^{2}x-{\cos }^{2}x\\ f^{\prime }(x)& =2\sin x\cos x-2\cos x(-\sin x)\\ & =2\sin x\cos x+2\sin x\cos x\\ & =2(2\sin x\cos x)=2\sin 2x\\ f^{″}(x)& =2.2\cos 2x=4\cos 2x\end{array}\end{array}$

$\begin{array}{l}\\ 49.& \text{Diketahui}f(x)=\sqrt{\sin x}.\text{Jika}{f}^{″}(x)\\ & \text{adalah turunan keduafungsi}f,\text{maka}\\ & \text{nilai dari}{f}^{″}\left({\displaystyle \frac{\pi }{2}}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}}\\ \text{b}.& -{\displaystyle \frac{1}{4}}\\ \text{c}.& 0\\ \text{d}.& {\displaystyle \frac{1}{4}}\\ \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}f(x)& =\sqrt{\sin x}={\sin }^{\frac{1}{2}}x\\ f^{\prime }(x)& ={\displaystyle \frac{1}{2}{\sin }^{-\frac{1}{2}}x.\cos x={\displaystyle \frac{\cos x}{2{\sin }^{\frac{1}{2}}x}}}\\ f^{″}(x)& ={\displaystyle \frac{-\sin x(2{\sin }^{\frac{1}{2}}x)-\cos x\left(2.{\displaystyle \frac{1}{2}{\sin }^{-\frac{1}{2}}x.\cos x}\right)}{4\sin x}}\\ & ={\displaystyle \frac{-2\sin x\sqrt{\sin x}-{\displaystyle \frac{{\cos }^{2}x}{\sqrt{\sin x}}}}{4\sin x}}\\ f^{″}\left({\displaystyle \frac{\pi }{2}}\right)& ={\displaystyle \frac{-2\sin {\displaystyle \frac{\pi }{2}.\sqrt{\sin {\displaystyle \frac{\pi }{2}}}-{\displaystyle \frac{{\cos }^2{\displaystyle \frac{\pi }{2}}}{\sin {\displaystyle \frac{\pi }{2}}}}}}{4\sin {\displaystyle \frac{\pi }{2}}}}\\ & ={\displaystyle \frac{-2.1.1-0}{4.1}=-\frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 50.& \text{Jika}f(x)={\tan }^2(3x-2)\text{maka}{f}^{″}(x)=....\\ & \begin{array}{l}\\ \text{a}.& 36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & -18{\sec }^4(3x-2)\\ \text{b}.& 36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +18{\sec }^2(3x-2)\\ \text{c}.& 36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +18{\sec }^4(3x-2)\\ \text{d}.& 18{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +36{\sec }^4(3x-2)\\ \text{e}.& 18{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +18{\sec }^4(3x-2)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(x)& ={\tan }^2(3x-2)\\ f^{\prime }(x)& =2\tan (3x-2){\sec }^2(3x-2)(3)\\ & =6\tan (3x-2){\sec }^2(3x-2)\\ f^{″}(x)& =6{\sec }^2(3x-2).(3){\sec }^2(3x-2)\\ & +6\tan (3x-2).2\sec (3x-2).\sec (3x-2)\tan (3x-2)(3)\\ & =18{\sec }^4(3x-2)\\ & +36{\tan }^2(3x-2){\sec }^2(3x-2)\end{array}\end{array}$

Contoh Soal 9 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{l}\\ 41.& \text{Sebuah mesin diprogram untuk dapat}\\ & \text{begerak tiap waktu mengikuti posisi}\\ & x=2\cos 3t\text{dan}y=2\cos 2t\text{di mana}\\ & x,y\text{dalam}cm,\text{dan}t\text{dalam detik}\\ & \text{Jika kecepatakan dirumuskan dengan}\\ & v=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2},\text{maka nilai}v\\ & \text{saat}t=30detik\text{adalah}...cm/detik\\ & \begin{array}{l}\\ \text{a}.& 4\sqrt{3}\\ \text{b}.& 2\sqrt{11}\\ \text{c}.& 2\sqrt{10}\\ \text{d}.& 6\\ \text{e}.& 4\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Diketahui Kecepatan gerak mesin}\\ & \{\begin{array}{ll}x=2\cos 3x& \Rightarrow {\displaystyle \frac{dx}{dt}=-6\sin 3t}\\ y=2\cos 2x& \Rightarrow {\displaystyle \frac{dy}{dt}=-4\sin 2t}\end{array}\\ & \text{Maka kecepatan mesin saat}t=30\\ & v=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}\\ & v=\sqrt{{(-6\sin 3t)}^2+{(-4\sin 2t)}^2}\\ & =\sqrt{{(-6\sin 3(30))}^2+{(-4\sin 2(30))}^2}\\ & =\sqrt{{(-6(1))}^2+{(-4\left({\displaystyle \frac{1}{2}\sqrt{3}}\right))}^2}\\ & =\sqrt{36+12}=\sqrt{48}=\sqrt{16.3}\\ & =4\sqrt{3}\end{array}\end{array}$

$\begin{array}{l}\\ 42.& \text{Sebuah benda duhubungkan dengan}\\ & \text{pegas dan bergerak sepanjang sumbu}\\ & \text{X dengan formula persamaan}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \text{Jarak terjauh dari titik}O\text{yang dapat}\\ & \text{dicapai oleh benda tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 4\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui gerak benda yang bergerak}\\ & \text{mengikuti formula}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \text{Jarak terjauh dicapai saat}{x}^{\prime }={\displaystyle \frac{dx}{dt}=0}\\ & x^{\prime }=2\cos 2t-2\sqrt{3}\sin 2t=0\\ & \iff 2\cos 2t=2\sqrt{3}\sin 2t\\ & \iff {\displaystyle \frac{\sin 2t}{\cos 2t}={\displaystyle \frac{1}{3}\sqrt{3}}}\\ & \iff \tan 2t=\tan {30}^{\circ }\\ & \iff 2t={30}^{\circ }+k{.180}^{\circ }\\ & \iff t={15}^{\circ }+k{.90}^{\circ }\{\begin{array}{ll}k=0,& t={15}^{\circ }\\ k=1,& t={105}^{\circ }\\ k=2,& t={195}^{\circ }\\ k=3,& t={285}^{\circ }\\ k=4,& t={375}^{\circ }\\ & \text{dst}\end{array}\\ & \text{Ambil}t={15}^{\circ },\text{maka nilai}\\ & x-\text{nya adalah}:\\ & x=\sin 2t+\sqrt{3}\cos 2t\\ & \iff x=\sin 2({15}^{\circ })+\sqrt{3}\cos 2({15}^{\circ })\\ & \iff x={\displaystyle \frac{1}{2}+\sqrt{3}\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}\\ & \iff x={\displaystyle \frac{1}{2}+\frac{3}{2}=2}\end{array}\end{array}$

$\begin{array}{l}\\ 43.& \text{Pada kurva}y=\sin x\text{dibuat}\\ & \text{garis singgung melalui titik}\left({\displaystyle \frac{2\pi }{3},k}\right)\\ & \text{garis singgung tersebut memotong}\\ & \text{sumbu-X di A dan sumbu-Y di B}.\\ & \text{Luas}\mathrm{△}AOB\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{{(3\pi +2\sqrt{3})}^2}{36}}\\ \text{b}.& {\displaystyle \frac{{(3\pi +3\sqrt{3})}^2}{36}}\\ \text{c}.& {\displaystyle \frac{{(3\pi +2\sqrt{3})}^2}{16}}\\ \text{d}.& {\displaystyle \frac{{(3\pi +2\sqrt{3})}^2}{18}}\\ \text{e}.& {\displaystyle \frac{{(3\pi +3\sqrt{3})}^2}{18}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Perhatikan ilustrasi berikut}\end{array}$

$.\begin{array}{rl}& \text{Misalkan koordinat titik}P\left({\displaystyle \frac{2\pi }{3},k}\right)\\ & \text{maka},\\ & x_p={\displaystyle \frac{2\pi }{3},y_p=k=\sin {\displaystyle \frac{2\pi }{3}={\displaystyle \frac{1}{2}\sqrt{3}}}}\\ & \text{Persamaan garis singgung di titik P}:\\ & y=m_{x_p}(x-x_p)+y_p\\ & \{\begin{array}{ll}\left({\displaystyle \frac{2\pi }{3},k}\right)& =\left({\displaystyle \frac{2\pi }{3},\frac{1}{2}\sqrt{3}}\right)\\ m_{x_p}={\displaystyle \frac{dy}{dx}}& =y^{\prime }=\cos x\\ m_{x_p}& =\cos \left({\displaystyle \frac{2\pi }{3}}\right)=-\frac{1}{2}\end{array}\\ & \text{Sehingga persamaan garis singgungnya}\\ & y=(-{\displaystyle \frac{1}{2}})(x-{\displaystyle \frac{2\pi }{3}})+{\displaystyle \frac{1}{2}\sqrt{3}}\\ & \iff 2y=-x+{\displaystyle \frac{2\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \bullet \text{memotong sumbu-X, maka}{y}_A=0\\ & 2y_A=-x_A+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & 0=-x_A+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & x_A={\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & \bullet \text{memotong sumbu-Y, maka}{x}_B=0\\ & 2y_B=-x_B+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & 2y_B=0+{\displaystyle \frac{2\pi }{3}+\sqrt{3}}\\ & y_B={\displaystyle \frac{\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \bullet \text{Luas}\mathrm{△}AOB=\left[AOB\right]={\displaystyle \frac{x_A.y_B}{2}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{2\pi }{3}+\sqrt{3}}\right).\left({\displaystyle \frac{\pi }{3}+{\displaystyle \frac{1}{2}\sqrt{3}}}\right)}{2}}\\ & ={\displaystyle \frac{1}{6}(2\pi +3\sqrt{3}).{\displaystyle \frac{1}{6}(2\pi +3\sqrt{3})}}\\ & ={\displaystyle \frac{1}{36}{(2\pi +3\sqrt{3})}^2}\end{array}$

$\begin{array}{l}\\ 44.& \text{Sebuah wadah penampung air hujan}\\ & \text{memiliki ukuran sisi samping 3 m dan}\\ & \text{sisi horisontal juga 3 m. Sisi samping}\\ & \text{membentuk sudut}\theta (0\le \theta \le {\displaystyle \frac{\pi }{2}})\\ & \text{dengan garis vertikal (lihat gambar)}\\ & \text{Nilai}\theta \text{supaya wadah dapat menampung}\\ & \text{air hujan maksimum adalah}....\end{array}$

$\begin{array}{ll}.& \\ & \begin{array}{ll}\text{a}.& {\displaystyle \frac{\pi }{3}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}}\\ \text{c}.& {\displaystyle \frac{\pi }{5}}\\ \text{d}.& {\displaystyle \frac{\pi }{6}}\\ \text{e}.& {\displaystyle \frac{\pi }{8}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Supaya memuat dapat maksimum}\\ & \text{maka luas penampang haruslah}\\ & \text{MAKSIMUM, yaitu}\end{array}\end{array}$

gambar 1

gambar 2

$.\begin{array}{rl}& \text{Luas penampang}=\text{Luas Trapesium}\\ & \text{dengan}\{\begin{array}{ll}t& =3\sin \theta \\ n& =3\cos \theta \end{array}\\ & \text{Luas Penampang}={\displaystyle \frac{1}{2}(\sum \text{sisi sejajar})\times t}\\ & \iff L={\displaystyle \frac{1}{2}(6+2n)\times t}\\ & \iff L=(3+n)\times t\\ & \iff L=(3+3\cos \theta )\times 3\sin \theta \\ & \iff L=9\sin \theta +9\sin \theta \cos \theta \\ & \iff L=9\sin \theta +{\displaystyle \frac{9}{2}\sin 2\theta }\\ & \text{Suapa luas penampang}\text{MAKSIMUM}\\ & \text{maka}{L}^{\prime }={\displaystyle \frac{dL}{d\theta }=0}\\ & \iff L^{\prime }=9\cos \theta +9\cos 2\theta =0\\ & \iff 9\cos \theta +9\cos 2\theta =0\\ & \iff 9\cos \theta +9(2{\cos }^2\theta -1)=0\\ & \iff 2{\cos }^2\theta +\cos \theta -1=0\\ & \iff (\cos \theta +1)(2\cos \theta -1)=0\\ & \iff \cos \theta =-1\text{atau}2\cos \theta =1\\ & \iff \cos \theta =-1\text{atau}\cos \theta ={\displaystyle \frac{1}{2}}\\ & \iff \cos \theta =\cos \pi \text{atau}\cos \theta =\cos {\displaystyle \frac{\pi }{3}}\\ & \iff \theta =\pi \text{atau}\theta ={\displaystyle \frac{\pi }{3}}\end{array}$

$\begin{array}{l}\\ 45.& \text{Seseorang melempar bola dari atap}\\ & \text{sebuah rumah. Ketinggian bola saat}\\ & t(detik)\text{dinyatakan dengan persamaan}\\ & h(t)=5+{\cos }^2\pi t.\text{Kecepatan bola}\\ & \text{ditentukan dengan formula}v={\displaystyle \frac{dh}{dt}}\\ & \text{Besar kecepatan bola saat}t=0,25\\ & \text{detik adalah}....\\ & \begin{array}{ll}\text{a}.& 0\\ \text{b}.& \pi \\ \text{c}.& 2\pi \\ \text{d}.& 3\pi \\ \text{e}.& 4\pi \end{array}\\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui}h(t)=5+{\cos }^2\pi t.\text{maka}\\ & v={\displaystyle \frac{dh}{dt}=2\cos \pi t(-\sin \pi t).(\pi )}\\ & \iff v=-\pi \sin 2\pi t\\ & \text{Saat}t=0,25={\displaystyle \frac{1}{4},\text{maka}}\\ & \text{besar kecepatannya adalah}:\\ & \iff v=-\pi \sin 2\pi \left({\displaystyle \frac{1}{4}}\right)\\ & \iff =-\pi \sin {\displaystyle \frac{\pi }{2}}\\ & \iff =-\pi \\ & \text{Tanda negatif menunjukkan}\\ & \text{arah kecepatan ke bawah}\\ & \text{Karena kecepatan merupakan salah}\\ & \text{satu besaran}\text{VEKTOR}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., & Akhmad G. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA
2. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan MAtematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
3. Sembiring, S., Zulkifli, M., Marsito, & Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU

Contoh Soal 8 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{l}\\ 36.& \text{Titik stasioner fungsi}f(x)=\cos 3x\\ & \text{pada}0\le x\le \pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& (0,1),\left({\displaystyle \frac{\pi }{4},1}\right),\left({\displaystyle \frac{\pi }{3},1}\right),\text{dan}\left({\displaystyle \frac{\pi }{2},-1}\right)\\ \text{b}.& (0,1),\left({\displaystyle \frac{\pi }{3},1}\right),\left({\displaystyle \frac{\pi }{2},-1}\right),\text{dan}\left({\displaystyle \pi ,-1}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{6},-1}\right),\left({\displaystyle \frac{\pi }{3},1}\right),\left({\displaystyle \frac{\pi }{2},-1}\right),\text{dan}\left({\displaystyle \frac{2\pi }{3},1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{6},1}\right),\left({\displaystyle \frac{\pi }{3},-1}\right),\left({\displaystyle \frac{\pi }{2},1}\right),\text{dan}\left({\displaystyle \frac{2\pi }{3},-1}\right)\\ \text{e}.& (0,1),\left({\displaystyle \frac{\pi }{3},-1}\right),\left({\displaystyle \frac{2\pi }{3},1}\right),\text{dan}\left({\displaystyle \pi ,-1}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\cos 3x\Rightarrow f^{\prime }(x)=-3\sin 3x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & -\sin 3x=0\iff \sin 3x=0\iff \sin 3x=\sin 0\\ & \iff 3x=0+k.2\pi \text{atau}3x=\pi +k.2\pi \\ & \iff x=k.{\displaystyle \frac{2\pi }{3}\text{atau}x={\displaystyle \frac{\pi }{3}+k.\frac{2\pi }{3}}}\\ & \iff k=0\Rightarrow x=0\text{atau}x={\displaystyle \frac{\pi }{3}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{2\pi }{3}\text{atau}x={\displaystyle \pi }}\\ & \text{Sekarang kita tentukan nilai dan titiknya}\\ & x=0\Rightarrow f(0)=\cos 3(0)=1\to (0,1)\\ & x={\displaystyle \frac{\pi }{3}\Rightarrow f\left({\displaystyle \frac{\pi }{3}}\right)=\cos 3\left({\displaystyle \frac{\pi }{3}}\right)=\cos \pi }\\ & =-1\to \left({\displaystyle \frac{\pi }{3},-1}\right)\\ & \text{dan seterusnya}\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \text{Titik stasioner fungsi}f(x)=\sin (2x-{\displaystyle \frac{\pi }{6}})\\ & \text{pada}0\le x\le \pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& (0,1)\text{dan}\left({\displaystyle \frac{\pi }{6},-1}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{6},1}\right)\text{dan}\left({\displaystyle \frac{\pi }{3},-1}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{4},-1}\right)\text{dan}\left({\displaystyle \frac{\pi }{2},-1}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{3},1}\right)\text{dan}\left({\displaystyle \frac{5\pi }{6},-1}\right)\\ \text{e}.& \left({\displaystyle \frac{\pi }{2},-1}\right)\text{dan}\left({\displaystyle \pi ,1}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin (2x-{\displaystyle \frac{\pi }{6}})\Rightarrow f^{\prime }(x)=2\cos (2x-{\displaystyle \frac{\pi }{6}})\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & 2\cos (2x-{\displaystyle \frac{\pi }{6}})=0\iff \cos (2x-{\displaystyle \frac{\pi }{6}})=0\\ & \cos (2x-{\displaystyle \frac{\pi }{6}})=\cos {\displaystyle \frac{\pi }{2}}\\ & \iff (2x-{\displaystyle \frac{\pi }{6}})=\pm {\displaystyle \frac{\pi }{2}+k.2\pi }\\ & \iff x={\displaystyle \frac{\pi }{12}\pm \frac{\pi }{4}+k.\pi \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{3}+k.\pi }\\ x& =-{\displaystyle \frac{\pi }{6}+k.\pi }\end{array}}\\ & \iff k=0\Rightarrow \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{3}}\\ x& =-{\displaystyle \frac{\pi }{6}(\text{tm})}\end{array}\\ & \iff k=1\Rightarrow \{\begin{array}{ll}x& ={\displaystyle \frac{4\pi }{3}\text{tm}}\\ x& ={\displaystyle \frac{5\pi }{6}}\end{array}\\ & \text{Sekarang kita tentukan nilai dan titiknya}\\ & x={\displaystyle \frac{\pi }{3}\Rightarrow f\left({\displaystyle \frac{\pi }{3}}\right)=\sin \left(2.{\displaystyle \frac{\pi }{3}-{\displaystyle \frac{\pi }{6}}}\right)=\sin \frac{\pi }{2}=1}\\ & =1\to \left({\displaystyle \frac{\pi }{3},1}\right)\\ & x={\displaystyle \frac{5\pi }{6}\Rightarrow f\left({\displaystyle \frac{5\pi }{6}}\right)=\sin \left(2.{\displaystyle \frac{5\pi }{6}-{\displaystyle \frac{\pi }{6}}}\right)}\\ & =\sin \frac{3\pi }{2}=-1\to \left({\displaystyle \frac{5\pi }{6},-1}\right)\end{array}\end{array}$

$\begin{array}{l}\\ 38.& \text{Nilai}x\text{pada titik stasioner}\\ & \text{fungsi}f(x)=x+\sin x\text{untuk}\\ & 0^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {90}^{\circ }\\ \text{b}.& {135}^{\circ }\\ \text{c}.& {150}^{\circ }\\ \text{d}.& {180}^{\circ }\\ \text{e}.& 360\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=x+\sin x\Rightarrow f^{\prime }(x)=1+\cos x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & 1+\cos =0\iff \cos x=-1\\ & \iff \cos x=\cos {180}^{\circ }\\ & \iff x=\pm {180}^{\circ }+k{.360}^{\circ }\\ & \iff k=0\Rightarrow x=\{\begin{array}{ll}{180}^{\circ }& \text{mungkin}\\ -{180}^{\circ }& \text{tidak mungkin}\end{array}\\ & \iff k=1\Rightarrow x=\{\begin{array}{ll}{540}^{\circ }& \text{tidak mungkin}\\ {180}^{\circ }& \text{mungkin}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Nilai}y\text{pada titik stasioner}\\ & \text{fungsi}f(x)=4\cos x+\cos 2x\\ & \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -5\text{dan}3\\ \text{b}.& -4\text{dan}2\\ \text{c}.& -3\text{dan}5\\ \text{d}.& -2\text{dan}4\\ \text{e}.& 3\text{dan}5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=4\cos x+\cos 2x\\ & \Rightarrow f^{\prime }(x)=-4\sin x-2\sin 2x\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & -4\sin x-2\sin 2x=0\\ & \iff -4\sin x-4\sin x\cos x=0\\ & \iff -4\sin x(1+\cos x)=0\\ & \iff \sin x(1+\cos x)=0\\ & \iff \sin x=0\text{atau}1+\cos x=0\\ & \iff \sin x=0\text{atau}\cos x=-1\\ & \iff \sin x=\sin 0^{\circ }\text{atau}\cos x=\cos {180}^{\circ }\\ & \iff x=\{\begin{array}{l}{0}^{\circ }+k{.360}^{\circ }\\ {180}^{\circ }+k{.360}^{\circ }\end{array}\text{atau}x=\{\begin{array}{l}{180}^{\circ }+k{.360}^{\circ }\\ -{180}^{\circ }+k{.360}^{\circ }\end{array}\\ & \iff k=0\Rightarrow x=0^{\circ }\text{atau}{180}^{\circ }\\ & \text{Nilai}y-\text{nya}\\ & x=0^{\circ }\Rightarrow f(0^{\circ })\\ & =4\cos 0^{\circ }+\cos 2(0^{\circ })=4+1=5\\ & x={180}^{\circ }\Rightarrow f({180}^{\circ })\\ & =4\cos {180}^{\circ }+\cos 2({180}^{\circ })=-4+1=-3\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \text{Nilai stasioner fungsi}\\ & f(x)={\displaystyle \frac{\sin x}{2-\cos x}}\\ & \text{untuk}0\le x\le 2\pi \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left({\displaystyle \frac{\pi }{2},\frac{1}{2}}\right)\text{dan}\left({\displaystyle \frac{\pi }{2},-\frac{1}{2}}\right)\\ \text{b}.& \left({\displaystyle \frac{\pi }{3},\frac{1}{2}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{\pi }{3},-\frac{1}{2}\sqrt{3}}\right)\\ \text{c}.& \left({\displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{2\pi }{3},-\frac{1}{3}\sqrt{3}}\right)\\ \text{d}.& \left({\displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{5\pi }{3},-\frac{1}{3}\sqrt{3}}\right)\\ \text{e}.& \left({\displaystyle \frac{\pi }{4},\frac{1}{4}\sqrt{3}}\right)\text{dan}\left({\displaystyle \frac{3\pi }{4},-\frac{1}{4}\sqrt{3}}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)={\displaystyle \frac{\sin x}{2-\cos x}\Rightarrow f^{\prime }(x)={\displaystyle \frac{2\cos x-1}{(2-\cos x{)}^2}}}\\ & \text{Stasioner fungsi}f\text{saat}{f}^{\prime }(x)=0\text{maka},\\ & {\displaystyle \frac{2\cos x-1}{(2-\cos x{)}^2}=0\iff 2\cos x-1=0}\\ & \iff \cos x={\displaystyle \frac{1}{2}\iff \cos x=\cos {\displaystyle \frac{\pi }{3}}}\\ & \iff x=\pm {\displaystyle \frac{\pi }{3}+k.2\pi }\\ & \iff k=0\Rightarrow x=\pm {\displaystyle \frac{\pi }{3}\iff x=\{\begin{array}{ll}{\displaystyle \frac{\pi }{3}}& \text{memenuhi}\\ -{\displaystyle \frac{\pi }{3}}& \text{tidak memenuhi}\end{array}}\\ & \iff k=1\Rightarrow x=\pm {\displaystyle \frac{\pi }{3}+2\pi \iff x=\{\begin{array}{ll}{\displaystyle \frac{7\pi }{3}}& \text{tidak memenuhi}\\ {\displaystyle \frac{5\pi }{3}}& \text{memenuhi}\end{array}}\\ & \text{Titiknya adalah}\\ & x={\displaystyle \frac{\pi }{3}\Rightarrow f\left({\displaystyle \frac{\pi }{3}}\right)}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{\pi }{3}}}{2-\cos {\displaystyle \frac{\pi }{3}}}={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{3}}}{2-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{3}\sqrt{3}}}}\\ & \left({\displaystyle \frac{\pi }{3},{\displaystyle \frac{1}{3}\sqrt{3}}}\right)\\ & x={\displaystyle \frac{5\pi }{3}\Rightarrow f\left({\displaystyle \frac{5\pi }{3}}\right)}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{5\pi }{3}}}{2-\cos {\displaystyle \frac{5\pi }{3}}}={\displaystyle \frac{-{\displaystyle \frac{1}{2}\sqrt{3}}}{2-{\displaystyle \frac{1}{2}}}=-{\displaystyle \frac{1}{3}\sqrt{3}}}}\\ & \left({\displaystyle \frac{5\pi }{3},-{\displaystyle \frac{1}{3}\sqrt{3}}}\right)\end{array}\end{array}$

Contoh Soal 7 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{l}\\ 31.& \text{Fungsi}f(x)=\sin x-\cos x\text{dengan}\\ & 0<x<2\pi \text{naik pada interval}....\\ & \begin{array}{l}\\ \text{a}.& 0<x<{\displaystyle \frac{\pi }{4}}\\ \text{b}.& {\displaystyle \frac{\pi }{4}<x<2\pi }\\ \text{c}.& {\displaystyle \frac{3\pi }{4}<x<{\displaystyle \frac{7\pi }{4}}}\\ \text{d}.& 0<x<{\displaystyle \frac{3\pi }{4}\text{atau}{\displaystyle \frac{7\pi }{4}<x<2\pi }}\\ \text{e}.& 0<x<{\displaystyle \frac{\pi }{4}\text{atau}{\displaystyle \frac{3\pi }{4}<x<2\pi }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)=\sin x-\cos x\\ & \text{Fungsi}f\text{naik, jika}{f}^{\prime }(x)>0\\ & \text{Selanjutnya}\\ & f^{\prime }(x)=\cos x+\cos x=0\\ & \sin x=-\cos x\iff {\displaystyle \frac{\sin x}{\cos x}=-1}\\ & \iff \tan x=-1\\ & \iff \tan x=\tan {\displaystyle \frac{3\pi }{4}}\\ & \iff x={\displaystyle \frac{3\pi }{4}\pm k.\pi }\\ & \iff k=0\Rightarrow x={\displaystyle \frac{3\pi }{4}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{3\pi }{4}\pm \pi =\frac{7\pi }{4}}\\ & \iff k=2\Rightarrow x={\displaystyle \frac{3\pi }{4}\pm 2\pi =\text{tm}}\\ & \begin{array}{c}\\ & & & & & & & \\ & ++& & --& & & ++& \\ 0& & {\displaystyle \frac{3\pi }{4}}& & & {\displaystyle \frac{7\pi }{4}}& & 2\pi \end{array}\\ & \text{ambil titik uji}x={\displaystyle \frac{1}{2}\pi }\\ & \text{untuk}x={\displaystyle \frac{1}{2}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{1}{2}\pi }\right)}\\ & =\cos {\displaystyle \frac{1}{2}\pi +\sin {\displaystyle \frac{1}{2}\pi =0+1=1(\text{positif})}}\\ & \text{untuk}x={\displaystyle \frac{3}{2}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{3}{2}\pi }\right)}\\ & =\cos {\displaystyle \frac{3}{2}\pi +\sin {\displaystyle \frac{3}{2}\pi =0-1=-1(\text{negatif})}}\\ & \text{untuk}x={\displaystyle \frac{11}{6}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{11}{6}\pi }\right)}\\ & =\cos {\displaystyle \frac{11}{6}\pi +\sin {\displaystyle \frac{11}{6}\pi ={\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}(\text{positif})}}}\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Fungsi}f(x)={\sin }^{2}x\text{dengan}\\ & 0<x<2\pi \text{naik pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{2}<x<\pi \text{atau}{\displaystyle \frac{3\pi }{2}<x<2\pi }}\\ \text{b}.& {\displaystyle \frac{2\pi }{3}<x<\pi }\\ \text{c}.& 0<x<{\displaystyle \frac{\pi }{2}\text{atau}\pi <x<{\displaystyle \frac{3\pi }{2}}}\\ \text{d}.& {\displaystyle \frac{4\pi }{3}<x<2\pi }\\ \text{e}.& {\displaystyle \frac{\pi }{3}<x<\pi \text{atau}{\displaystyle \frac{4\pi }{3}<x<2\pi }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)={\sin }^{2}x\\ & \text{Fungsi}f\text{naik, jika}{f}^{\prime }(x)>0\\ & \text{Selanjutnya}\\ & f^{\prime }(x)=2\sin x\cos x=\sin 2x=0\\ & \iff \sin 2x=0\\ & \iff \sin 2x=\sin 0\\ & \iff 2x=\pm k.2\pi \text{atau}2x=\pi \pm k.2\pi \\ & \iff x=\pm k.\pi \text{atau}x=\frac{\pi }{2}\pm k.\pi \\ & \iff k=0\Rightarrow x=0\text{atau}x={\displaystyle \frac{\pi }{2}}\\ & \iff k=1\Rightarrow x=\pi \text{atau}x={\displaystyle \frac{\pi }{2}+\pi =\frac{3\pi }{2}}\\ & \iff k=2\Rightarrow x=2\pi \text{atau}x={\displaystyle \frac{\pi }{2}+2\pi ={\displaystyle \frac{5}{2}\pi (\text{tm})}}\\ & \begin{array}{c}\\ & & & & & & & & \\ & ++& & --& & ++& & --& \\ 0& & {\displaystyle \frac{\pi }{2}}& & \pi & & {\displaystyle \frac{3\pi }{2}}& & 2\pi \end{array}\\ & \text{ambil titik uji}x={\displaystyle \frac{1}{6}\pi }\\ & \text{untuk}x={\displaystyle \frac{1}{6}\pi \Rightarrow f^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)}\\ & =\sin 2\left({\displaystyle \frac{1}{6}\pi }\right)=\sin {\displaystyle \frac{1}{3}\pi =\frac{1}{2}(\text{positif})}\\ & \text{untuk}x={\displaystyle \frac{3}{4}\pi \Rightarrow f\left({\displaystyle \frac{3}{4}\pi }\right)}\\ & =\sin 2\left({\displaystyle \frac{3}{4}\pi }\right)=-1(\text{negatif})\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Fungsi}f(x)={\cos }^{2}2x\text{untuk}\\ & 0^{\circ }<x<{360}^{\circ }\text{turun pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {45}^{\circ }<x<{90}^{\circ }\\ \text{b}.& {135}^{\circ }<x<{180}^{\circ }\\ \text{c}.& {225}^{\circ }<x<{270}^{\circ }\\ \text{d}.& {270}^{\circ }<x<{300}^{\circ }\\ \text{e}.& {315}^{\circ }<x<{360}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)={\cos }^{2}2x\\ & \text{Fungsi}f\text{turun, jika}{f}^{\prime }(x)<0\\ & f^{\prime }(x)=2\cos 2x(-\sin 2x)(2)=-2\sin 4x\\ & \text{Selanjutnya}\\ & \iff -2\sin 4x=0\iff \sin 4x=0\iff \sin 4x=\sin 0^{\circ }\\ & \iff \{\begin{array}{ll}4x=0^{\circ }+k{.360}^{\circ }& \Rightarrow x=k{.90}^{\circ }\\ 4x={180}^{\circ }+k{.360}^{\circ }& \Rightarrow x={45}^{\circ }+k{.90}^{\circ }\end{array}\\ & \iff k=0\Rightarrow x=0^{\circ }\text{atau}x={45}^{\circ }\\ & \iff k=1\Rightarrow x={90}^{\circ }\text{atau}x={135}^{\circ }\\ & \iff k=2\Rightarrow x={180}^{\circ }\text{atau}x={225}^{\circ }\\ & \iff k=3\Rightarrow x={270}^{\circ }\text{atau}x={315}^{\circ }\\ & \iff k=4\Rightarrow x={360}^{\circ }\text{atau}x={405}^{\circ }(\text{tm})\\ & \text{Gunakan titik uji pada}x={30}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({30}^{\circ })=-2\sin 4({30}^{\circ })=-\sqrt{3}(\text{negatif})\\ & \text{Gunakan titik uji pada}x={60}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({60}^{\circ })=-2\sin 4({60}^{\circ })=\sqrt{3}(\text{positif})\\ & \text{Gunakan titik uji pada}x={120}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({120}^{\circ })=-2\sin 4({120}^{\circ })=-\sqrt{3}(\text{negatif})\\ & \text{Gunakan titik uji pada}x={150}^{\circ }\\ & \bullet \text{untuk}{f}^{\prime }({150}^{\circ })=-2\sin 4({150}^{\circ })=\sqrt{3}(\text{positif})\\ & \text{dan seterusnya}...\\ & \begin{array}{c}\\ & & & & & & & & \\ & --& & ++& & --& & ++\\ 0& & {\displaystyle {45}^{\circ }}& & {90}^{\circ }& & {135}^{\circ }& & {180}^{\circ }\\ & --& & ++& & --& & ++\\ {180}^{\circ }& & {\displaystyle {225}^{\circ }}& & {270}^{\circ }& & {315}^{\circ }& & {360}^{\circ }\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 34.& (\mathbf{\text{SBMPTN 2015}})\\ & \text{Fungsi}f(x)=2\sqrt{{\sin }^{2}x+{\displaystyle \frac{x\sqrt{3}}{2}}}\\ & \text{pada}0<x<\pi \text{turun pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5\pi }{12}<x<\frac{11\pi }{12}}\\ \text{b}.& {\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}}\\ \text{c}.& {\displaystyle \frac{2\pi }{3}<x<\frac{5\pi }{6}}\\ \text{d}.& {\displaystyle \frac{3\pi }{4}<x<\pi }\\ \text{e}.& {\displaystyle \frac{3\pi }{4}<x<\frac{3\pi }{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)=2\sqrt{{\sin }^{2}x+{\displaystyle \frac{x\sqrt{3}}{2}}}\\ & \text{Fungsi}f\text{turun, jika}{f}^{\prime }(x)<0\\ & f^{\prime }(x)={\displaystyle \frac{\sin 2x+{\displaystyle \frac{1}{2}\sqrt{3}}}{\sqrt{{\sin }^{2}x+{\displaystyle \frac{x\sqrt{3}}{2}}}}=0}\\ & \sin 2x+{\displaystyle \frac{1}{2}\sqrt{3}=0\iff \sin 2x=-{\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \iff \sin 2x=\sin {\displaystyle \frac{4\pi }{3}}\\ & \iff 2x={\displaystyle \frac{4\pi }{3}+k.2\pi \text{atau}2x=\pi -\frac{4\pi }{3}+k.2\pi }\\ & \iff x={\displaystyle \frac{2\pi }{3}+k.\pi \text{atau}x=-\frac{\pi }{6}+k.\pi }\\ & \iff k=0\Rightarrow x={\displaystyle \frac{2\pi }{3}\text{atau}x=-{\displaystyle \frac{\pi }{6}(\text{tm})}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{5\pi }{3}\text{atau}x={\displaystyle \frac{5\pi }{6}}}\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{\pi }{2}={90}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\left({\displaystyle \frac{\pi }{2}}\right)={\displaystyle \frac{\sin 2\left({\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{1}{2}\sqrt{3}}}{\sqrt{{\sin }^2\left({\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{\left({\displaystyle \frac{\pi }{2}}\right)\sqrt{3}}{2}}}}=+}\\ & (\text{positif})\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{3\pi }{4}={135}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\left({\displaystyle \frac{3\pi }{4}}\right)={\displaystyle \frac{\sin 2\left({\displaystyle \frac{3\pi }{4}}\right)+{\displaystyle \frac{1}{2}\sqrt{3}}}{\sqrt{{\sin }^2\left({\displaystyle \frac{3\pi }{4}}\right)+{\displaystyle \frac{\left({\displaystyle \frac{3\pi }{4}}\right)\sqrt{3}}{2}}}}=-}\\ & (\text{negatif})\\ & \begin{array}{c}\\ & & & & \\ & ++& & --\\ 0& & {\displaystyle {\displaystyle \frac{2\pi }{3}}}& & {\displaystyle \frac{5\pi }{6}}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Fungsi}f(x)=\sqrt{{\sin }^{2}x+{\displaystyle \frac{x}{2}}}\\ & \text{dengan}x>0\text{turun pada interval}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5\pi }{12}<x\le \frac{13\pi }{12}}\\ \text{b}.& {\displaystyle \frac{7\pi }{12}<x<\frac{11\pi }{12}}\\ \text{c}.& {\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}}\\ \text{d}.& {\displaystyle \frac{7\pi }{6}<x\le {\displaystyle \frac{13\pi }{6}}}\\ \text{e}.& {\displaystyle \frac{7\pi }{6}<x\le \frac{11\pi }{6}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Diketahui}f(x)=\sqrt{{\sin }^{2}x+{\displaystyle \frac{x}{2}}}\\ & \text{Fungsi}f\text{turun, jika}{f}^{\prime }(x)<0\\ & f^{\prime }(x)={\displaystyle \frac{\sin 2x+{\displaystyle \frac{1}{2}}}{2\sqrt{{\sin }^{2}x+{\displaystyle \frac{x}{2}}}}=0}\\ & \sin 2x+{\displaystyle \frac{1}{2}=0\iff \sin 2x=-{\displaystyle \frac{1}{2}}}\\ & \iff \sin 2x=\sin {\displaystyle \frac{7\pi }{6}}\\ & \iff 2x={\displaystyle \frac{7\pi }{6}+k.2\pi \text{atau}2x=\pi -\frac{7\pi }{6}+k.2\pi }\\ & \iff x={\displaystyle \frac{7\pi }{12}+k.\pi \text{atau}x=-\frac{\pi }{12}+k.\pi }\\ & \iff k=0\Rightarrow x={\displaystyle \frac{7\pi }{12}\text{atau}x=-{\displaystyle \frac{\pi }{12}(\text{tm})}}\\ & \iff k=1\Rightarrow x={\displaystyle \frac{19\pi }{12}\text{atau}x={\displaystyle \frac{11\pi }{12}}}\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{\pi }{2}={90}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\left({\displaystyle \frac{\pi }{2}}\right)={\displaystyle \frac{\sin 2\left({\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{1}{2}}}{\sqrt{{\sin }^2\left({\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{\left({\displaystyle \frac{\pi }{2}}\right)}{2}}}}=+}\\ & (\text{positif})\\ & \text{Gunakan titik uji pada}x={\displaystyle \frac{3\pi }{4}={135}^{\circ }}\\ & \bullet \text{untuk}{f}^{\prime }\left({\displaystyle \frac{3\pi }{4}}\right)={\displaystyle \frac{\sin 2\left({\displaystyle \frac{3\pi }{4}}\right)+{\displaystyle \frac{1}{2}}}{\sqrt{{\sin }^2\left({\displaystyle \frac{3\pi }{4}}\right)+{\displaystyle \frac{\left({\displaystyle \frac{3\pi }{4}}\right)}{2}}}}=-}\\ & (\text{negatif})\\ & \begin{array}{c}\\ & & & & \\ & ++& & --\\ 0& & {\displaystyle {\displaystyle \frac{7\pi }{12}}}& & {\displaystyle \frac{11\pi }{12}}\end{array}\end{array}\end{array}$

Contoh Soal 6 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{l}\\ 25.& \text{Persamaan garis singgung pada kurva}\\ & y=3\sin x\text{pada titik yang berabsis}{\displaystyle \frac{\pi }{3}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y={\displaystyle \frac{2}{3}(x-\frac{\pi }{3})-\frac{2\sqrt{2}}{3}}\\ \text{b}.& y={\displaystyle \frac{2}{3}(x-\frac{\pi }{3})+\frac{2\sqrt{2}}{3}}\\ \text{c}.& y={\displaystyle \frac{3}{2}(x-\frac{\pi }{3})-\frac{3\sqrt{3}}{2}}\\ \text{d}.& y={\displaystyle \frac{3}{2}(x-\frac{\pi }{3})+\frac{3\sqrt{3}}{2}}\\ \text{e}.& y={\displaystyle \frac{3}{2}(x-\frac{\pi }{3})-\frac{3\sqrt{2}}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& y=3\sin x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & y_0=3\sin \left({\displaystyle \frac{\pi }{3}}\right)=3\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)=\frac{3\sqrt{3}}{2}\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=y^{\prime }=3\cos x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & m=3\cos \left({\displaystyle \frac{\pi }{3}}\right)=3\left({\displaystyle \frac{1}{2}}\right)=\frac{3}{2}\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & y={\displaystyle \frac{3}{2}(x-{\displaystyle \frac{\pi }{3}})+{\displaystyle \frac{3\sqrt{3}}{2}}}\end{array}\end{array}$

$\begin{array}{l}\\ 26.& \text{Kurva}y=\sin x+\cos x\text{untuk}\\ & 0<x<\pi \text{memotong sumbu X}\\ & \text{di titik A. Persamaan garis}\\ & \text{singgung di titik A adalah}....\\ & \begin{array}{l}\\ \text{a}.& y=-\sqrt{2}(x-{\displaystyle \frac{\pi }{4}})\\ \text{b}.& y=-\sqrt{2}(x-{\displaystyle \frac{\pi }{2}})\\ \text{c}.& y=-\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})\\ \text{d}.& y=\sqrt{2}(x-{\displaystyle \frac{\pi }{4}})\\ \text{e}.& y=\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Kurva memotong sumbu X}\\ & \text{di titik A, berarti}y=0\\ & \sin x+\cos x=0\\ & \sin x=-\cos x\\ & {\displaystyle \frac{\sin x}{\cos x}=-1\iff \tan x=-1}\\ & \tan x=\tan \left({\displaystyle \frac{3\pi }{4}}\right)\Rightarrow x={\displaystyle \frac{3\pi }{4}}\\ & \text{Jadi, titik A-nya}:\left({\displaystyle \frac{3\pi }{4},0}\right)\\ & \text{dan nilai gradien}m=y^{\prime },\text{yaitu}:\\ & m=\cos x-\sin x\\ & m=\cos \left({\displaystyle \frac{3\pi }{4}}\right)-\sin \left({\displaystyle \frac{3\pi }{4}}\right)\\ & m=-{\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}}\\ & \text{Persamaan garis singgung di A}:\\ & y=m(x-x_0)+y_0\\ & \iff y=-\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})+0\\ & \iff y=-\sqrt{2}(x-{\displaystyle \frac{3\pi }{4}})\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Persamaan garis singgung pada}\\ & \text{kurva}y={\sec }^{2}x\text{pada titik yang}\\ & \text{berabsis}{\displaystyle \frac{\pi }{3}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& y=8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})-4\\ \text{b}.& y=8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})+4\\ \text{c}.& y=-8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})-4\\ \text{d}.& y=-8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})+4\\ \text{e}.& y=4\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})-4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& y={\sec }^{2}x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & y_0={\sec }^2\left({\displaystyle \frac{\pi }{3}}\right)=(2{)}^2=4\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=y^{\prime }=2{\sec }^{2}x\tan x,\text{saat}{x}_0={\displaystyle \frac{\pi }{3}}\\ & m=2{\sec }^2\left({\displaystyle \frac{\pi }{3}}\right)\tan \left({\displaystyle \frac{\pi }{3}}\right)\\ & =2(4)\sqrt{3}=8\sqrt{3}\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & y=8\sqrt{3}(x-{\displaystyle \frac{\pi }{3}})+4\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Kurva berikut yang memiliki}\\ & \text{garis singgung dengan gradien}\\ & 4\sqrt{3}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y=2\sin x\text{pada titik}\left({\displaystyle \frac{\pi }{3},\sqrt{3}}\right)\\ \text{b}.& y=\cos 2x\text{pada titik}\left({\displaystyle \frac{\pi }{12},\frac{1}{2}}\right)\\ \text{c}.& y=\tan x\text{pada titik}\left({\displaystyle \pi ,0}\right)\\ \text{d}.& y=2\sec x\text{pada titik}\left({\displaystyle \frac{\pi }{3},2}\right)\\ \text{e}.& y=\cot x\text{pada titik}\left({\displaystyle \frac{\pi }{4},1}\right)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{|cll|}\hline \text{a}& y=2\sin x& m=2\cos {\displaystyle \frac{\pi }{3}}\\ & y^{\prime }=2\cos x& m=2.{\displaystyle \frac{1}{2}=1}\\ \text{b}& y=\cos 2x& m=-2\sin 2\left({\displaystyle \frac{\pi }{12}}\right)\\ & y^{\prime }=-2\sin 2x& m=-2.{\displaystyle \frac{1}{2}=-1}\\ \text{c}& y=\tan x& m={\sec }^2\left(\pi \right)\\ & y^{\prime }={\sec }^{2}x& m=(-1{)}^2=1\\ \text{d}& y=2\sec x& m=2\sec \left({\displaystyle \frac{\pi }{3}}\right)\tan \left({\displaystyle \frac{\pi }{3}}\right)\\ & y^{\prime }=2\sec x\tan x& m=2.2.\sqrt{3}=4\sqrt{3}\\ \text{e}& y=\cot x& m=-{\csc }^2\left({\displaystyle \frac{\pi }{4}}\right)\\ & y^{\prime }=-{\csc }^{2}x& m=-{\left(\sqrt{2}\right)}^2=-2\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 29.& \text{Persamaan garis singgung pada}\\ & \text{kurva}y=\sec x\text{di titik yang}\\ & \text{berabsis}{\displaystyle \frac{\pi }{4}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& y=\sqrt{3}x-{\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}}\\ \text{b}.& y=\sqrt{3}x+{\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}}\\ \text{c}.& y=\sqrt{2}x-{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}}\\ \text{d}.& y=\sqrt{2}x+{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}}\\ \text{e}.& y=\sqrt{2}x-{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& y=\sec x,\text{saat}{x}_0={\displaystyle \frac{\pi }{4}}\\ & y_0=\sec \left({\displaystyle \frac{\pi }{4}}\right)=\sqrt{2}\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=y^{\prime }=\sec x\tan x,\text{saat}{x}_0={\displaystyle \frac{\pi }{4}}\\ & m=\sec \left({\displaystyle \frac{\pi }{4}}\right)\tan \left({\displaystyle \frac{\pi }{4}}\right)=\sqrt{2}.1=\sqrt{2}\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & \iff y=\sqrt{2}(x-{\displaystyle \frac{\pi }{4}})+\sqrt{2}\\ & \iff y=\sqrt{2}x-{\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 30.& \text{Persamaan garis singgung pada}\\ & \text{kurva}y=\sin x+\cos x\text{di titik yang}\\ & \text{berabsis}{\displaystyle \frac{\pi }{2}\text{akan memotong sumbu}}\\ & \text{Y dengan ordinatnya berupa}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\pi }{2}+1}\\ \text{b}.& {\displaystyle \frac{\pi }{2}-1}\\ \text{c}.& 1-{\displaystyle \frac{\pi }{2}}\\ \text{d}.& 2+{\displaystyle \frac{\pi }{2}}\\ \text{e}.& 2-{\displaystyle \frac{\pi }{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& y=\sin x+\cos x,\text{saat}{x}_0={\displaystyle \frac{\pi }{2}}\\ & y_0=\sin \left({\displaystyle \frac{\pi }{2}}\right)+\cos \left({\displaystyle \frac{\pi }{2}}\right)=1+0=1\\ & \text{kita cari gradien}m\text{saat}{y}^{\prime },\text{yaitu}:\\ & m=\cos x-\sin x\\ & m=\cos \left({\displaystyle \frac{\pi }{2}}\right)-\sin \left({\displaystyle \frac{\pi }{2}}\right)\\ & m=0-1=-1\\ & \text{Persamaan garis singgungnya adalah}:\\ & y=m(x-x_0)+y_0\\ & \iff y=-1(x-{\displaystyle \frac{\pi }{2}})+1\\ & \iff y=-x+{\displaystyle \frac{\pi }{2}+1}\\ & \text{Ordinat garis singgungnya saat}\\ & \text{memotong sumbu-Y adalah}:x=0,\\ & \text{maka}\\ & y=-0+{\displaystyle \frac{\pi }{2}+1={\displaystyle \frac{\pi }{2}+1}}\end{array}\end{array}$