Contoh Soal 12 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{ll}\\ 56.&\textrm{Diketahui fungsi}\: \: f(x)=\displaystyle \frac{1}{2}\sin 2x\: \: \textrm{dengan}\\ &0^{\circ}<x<360^{\circ} \: .\: \textrm{Kurva akan cekung}\\ &\textrm{ke atas pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0^{\circ}<x<90^{\circ}\\ \textrm{b}.&0^{\circ}<x<90^{\circ}\: \: \textrm{atau}\: \: 180^{\circ}<x<270^{\circ}\\ \textrm{c}.&45^{\circ}<x<225^{\circ}\\ \color{red}\textrm{d}.&90^{\circ}<x<180^{\circ}\: \: \textrm{atau}\: \: 270^{\circ}<x<360^{\circ}\\ \textrm{e}.&180^{\circ}<x<225^{\circ}\: \: \textrm{atau}\: \: 225^{\circ}<x<360^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&f(x)=\displaystyle \frac{1}{2}\sin 2x\\ &f'(x)=\cos 2x\Rightarrow f''(x)=-2\sin 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-2\sin 2x=0\Leftrightarrow \sin 2x=0\\ &\Leftrightarrow \sin 2x=\sin 0^{\circ}\\ &\Leftrightarrow 2x=0^{\circ}+k.360^{\circ}\: \: \textrm{atau}\: \: 2x=180^{\circ}+k.360^{\circ}\\ &\Leftrightarrow x=0^{\circ}+k.180^{\circ}\: \: \textrm{atau}\: \: x=90^{\circ}+k.180^{\circ}\\ &\Leftrightarrow \color{red}x=0^{\circ},\: x=90^{\circ} \: ,\: x=180^{\circ}\: \: \textrm{dan}\: \: x=270^{\circ}\\ &\qquad \color{red}\textrm{serta}\: \: x=360^{\circ}\\ &\bullet \color{red}\textrm{Selang}\: \: 0^{\circ}<x<90^{\circ},\: \: \color{black}\textrm{misal}\: \: x=45^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 45^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: 90^{\circ}<x<180^{\circ},\: \: \color{black}\textrm{misal}\: \: x=135^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 135^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: 180^{\circ}<x<270^{\circ},\: \: \color{black}\textrm{misal}\: \: x=225^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 225^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: 270^{\circ}<x<360^{\circ},\: \: \color{black}\textrm{misal}\: \: x=315^{\circ}\\ &\Rightarrow \Rightarrow f''=-2\sin 2\left ( 315^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 57.&\textrm{Diketahui fungsi}\: \: f(x)=\cos ^{2}x-\sin ^{2}x\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Kurva akan cekung ke bawah}\\ &\textrm{pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{\pi }{2}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4}\\ \color{red}\textrm{c}.&\displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi \\ \textrm{d}.&\displaystyle \frac{7\pi }{4}<x<2\pi \\ \textrm{e}.&\displaystyle \frac{5\pi }{4}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&f(x)=\color{red}\cos ^{2}x-\sin ^{2}x\color{blue}=\cos 2x\\ &f'(x)=-2\sin 2x\Rightarrow f''(x)=-4\cos 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-4\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi \: \: \Leftrightarrow \: \: x=\displaystyle \frac{\pi }{4}+k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi }{4} \: ,\: x=\frac{5\pi }{4}\: \: \color{black}\textrm{dan}\: \: \color{red}x=\frac{7\pi }{4}\\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0<x<2\pi \: \: \textrm{saja}\\ &\bullet \color{red}\textrm{Selang}\: \: 0<x<\displaystyle \frac{\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=30^{\circ}=\frac{\pi }{6}\\ &\Rightarrow f''(30^{\circ})=-4\cos 2\left ( 30^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{4}<x<\displaystyle \frac{3\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=120^{\circ}=\displaystyle \frac{2\pi }{3}\\ &\Rightarrow f''(120^{\circ})=-4\cos 2\left ( 90^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=210^{\circ}=\frac{7\pi }{6}\\ &\Rightarrow f''(210^{\circ})=-4\cos 2\left ( 210^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=300^{\circ}=\displaystyle \frac{5\pi }{3}\\ &\Rightarrow f''(300^{\circ})=-4\cos 2\left ( 300^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi ,\: \: \color{black}\textrm{misal}\: \: x=330^{\circ}=\frac{11\pi }{6}\\ &\Rightarrow f''=-4\cos 2\left ( 330^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 58.&\textrm{Diketahui fungsi}\: \: f(x)=\sin ^{2}x\: \: \textrm{dengan}\\ &0<x<2\pi .\: \textrm{Kurva fungsi tersebut akan}\\ &\textrm{cekung ke bawah pada interval}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\: \: \textrm{atau}\: \: \frac{5\pi }{4}<x<\frac{7\pi }{4}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\: \: \textrm{atau}\: \: \frac{7\pi }{4}<x<2\pi\\ \textrm{c}.&\displaystyle 0<x<\frac{\pi }{2}\: \: \textrm{atau}\: \: \frac{3\pi }{4}<x<\frac{5\pi }{4}\\ \textrm{d}.&\displaystyle \frac{\pi }{4}<x<\frac{3\pi }{4}\\ \textrm{e}.&\displaystyle 0<x<\frac{\pi }{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{black}\begin{aligned}&f(x)=\color{red}\sin ^{2}x\\ &f'(x)=2\sin x\cos x=\sin 2x\\ & f''(x)=2\cos 2x\\ &\color{purple}\textrm{Syarat belok}\: \: f''(x)=0\\ &2\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi \: \: \Leftrightarrow \: \: x=\displaystyle \frac{\pi }{4}+k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi }{4} \: ,\: x=\frac{5\pi }{4}\: \: \color{black}\textrm{dan}\: \: \color{red}x=\frac{7\pi }{4}\\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0<x<2\pi \: \: \textrm{saja}\\ &\bullet \color{red}\textrm{Selang}\: \: 0<x<\displaystyle \frac{\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=30^{\circ}=\frac{\pi }{6}\\ &\Rightarrow f''(30^{\circ})=2\cos 2\left ( 30^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{4}<x<\displaystyle \frac{3\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=120^{\circ}=\displaystyle \frac{2\pi }{3}\\ &\Rightarrow f''(120^{\circ})=2\cos 2\left ( 90^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=210^{\circ}=\frac{7\pi }{6}\\ &\Rightarrow f''(210^{\circ})=2\cos 2\left ( 210^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{5\pi }{4}<x<\frac{7\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=300^{\circ}=\displaystyle \frac{5\pi }{3}\\ &\Rightarrow f''(300^{\circ})=2\cos 2\left ( 300^{\circ} \right )=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi ,\: \: \color{black}\textrm{misal}\: \: x=330^{\circ}=\frac{11\pi }{6}\\ &\Rightarrow f''=2\cos 2\left ( 330^{\circ} \right )=2>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 59.&\textrm{Diketahui fungsi}\: \: f(x)=2\sin x-2\cos x\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Kurva akan cekung ke atas}\\ &\textrm{pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{3\pi }{4}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x <\displaystyle \frac{5\pi }{4}\\ \textrm{c}.& \displaystyle \frac{3\pi }{4}<x<2\pi \\ \textrm{d}.&0<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{3\pi }{4}<x<\frac{5\pi }{4}\\ \color{red}\textrm{e}.&0<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{5\pi }{4}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&f(x)=\color{red}2\sin x-2\cos x\color{blue}\\ &f'(x)=2\cos x+2\sin x\\ & f''(x)=-2\sin x+2\cos x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-2\sin x+2\cos x=0\Leftrightarrow \sin x=\cos x\\ &\Leftrightarrow \tan x=1\\ &\Leftrightarrow x=\displaystyle \frac{\pi }{4}+k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{5\pi }{4}\\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0<x<2\pi \: \: \textrm{saja}\\ &\color{black}\textrm{Sebagai gambaran saja}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{4}<x<\displaystyle \frac{3\pi }{4},\: \: \color{black}\textrm{misal}\: \: x=90^{\circ}=\displaystyle \frac{\pi }{2}\\ &\Rightarrow f''(90^{\circ})=-2\sin 90^{\circ}+2\cos 90^{\circ}=-2<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 60.&\textrm{Diketahui fungsi}\: \: f(x)=\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ &\textrm{dengan}\: \: 0<x<2\pi .\: \textrm{Kurva fungsi tersebut}\\ &\textrm{akan cekung ke atas pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0<x<\frac{\pi }{6}\: \: \textrm{atau}\: \: \frac{\pi }{2}<x<\frac{5\pi }{6}\\ \color{red}\textrm{b}.&\color{magenta}\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\: \: \textrm{atau}\: \: \frac{5\pi }{6}<x<\pi\\ \textrm{c}.&\displaystyle \frac{\pi }{6}<x<\frac{\pi }{2}\: \: \textrm{atau}\: \: \frac{3\pi }{4}<x<\frac{5\pi }{6}\\ \textrm{d}.&\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \frac{3\pi }{4}<x<\frac{5\pi }{6}\\ \textrm{e}.&\displaystyle \frac{\pi }{6}<x<\frac{\pi }{4}\: \: \textrm{atau}\: \: \frac{5\pi }{6}<x<\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&f(x)=\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ &f'(x)=3\cos \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ & f''(x)=-9\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-9\sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )=0\Leftrightarrow \sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )=0\\ &\Leftrightarrow \sin \left ( 3x+\displaystyle \frac{\pi }{2} \right )=\sin 0\\ &\Leftrightarrow \left ( 3x+\displaystyle \frac{\pi }{2} \right )=0+k.2\pi \: \: \Leftrightarrow \: \: \left ( 3x+\displaystyle \frac{\pi }{2} \right )=\displaystyle \pi+k.2\pi \\ &\Leftrightarrow 3x=-\displaystyle \frac{\pi }{2}+k.2\pi \: \: \Leftrightarrow \: \: 3x=\displaystyle \frac{\pi }{2}+k.2\pi \\ &\Leftrightarrow x=-\displaystyle \frac{\pi }{6}+k.\frac{2\pi}{3} \: \: \Leftrightarrow \: \: x=\displaystyle \frac{\pi }{6}+k.\displaystyle \frac{2\pi}{3} \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{6},\: \: x=\displaystyle \frac{\pi }{2},\: x=\frac{5\pi }{6} \: ,\: x=\frac{7\pi }{6},\\ & \color{black}\textrm{dan}\: \: \color{red}x=\frac{3\pi }{2},\: \color{black}\textrm{serta}\: \: \color{red}x=\frac{11\pi}{6} \\ &\qquad \color{red}\textrm{Ingat bahwa domain}\: \: 0\leq x\leq 2\pi \: \: \textrm{saja}\\ &\color{black}\textrm{Sebagai GAMBARAN saja, diberikan 2 nilai selang}\\ &\bullet \color{red}\textrm{Selang}\: \: 0<x<\displaystyle \frac{\pi }{6},\: \: \color{black}\textrm{misal}\: \: x=15^{\circ}=\frac{\pi }{12}\\ &\Rightarrow f''(15^{\circ})=-9\sin \left ( 3\left ( \displaystyle \frac{\pi }{12} \right )+\displaystyle \frac{\pi }{2} \right )=-\displaystyle \frac{9}{2}\sqrt{2}<0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke bawah}\\ &\bullet \color{red}\textrm{Selang}\: \: \displaystyle \frac{\pi }{6}<x<\displaystyle \frac{\pi }{2},\: \: \color{black}\textrm{misal}\: \: x=60^{\circ}=\displaystyle \frac{\pi }{3}\\ &\Rightarrow f''(60^{\circ})=-9\sin \left ( 3\left ( \displaystyle \frac{\pi }{3} \right )+\displaystyle \frac{\pi }{2} \right )=9>0\\ &\qquad \color{black}\textrm{pada selang ini kurva cekung ke atas} \end{aligned} \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)

$\begin{array}{ll}\\ 51.&\textrm{Diketahui}\: \: f(x)=\cos ^{2}2x\: .\: \textrm{Jika}\\ &f''(x)=a\sin ^{2}bx+c\cos ^{2}dx,\: \textrm{nilai untuk}\\ &\displaystyle \frac{a-b}{c-d}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{5}{3}\\ \textrm{b}.&\displaystyle \frac{2}{3}\\ \color{red}\textrm{c}.&-\displaystyle \frac{3}{5}\\ \textrm{d}.&-\displaystyle \frac{6}{5}\\ \textrm{e}.&-\displaystyle \frac{9}{5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{purple}\begin{aligned}&f(x)=\cos ^{2}2x\\ &f'(x)=2\cos 2x(-\sin 2x)(2)\\ &\: \qquad =-4\sin 2x\cos 2x\\ &\color{blue}f''(x)=-4\cos 2x.(2).\cos 2x-4\sin 2x.(-\sin 2x)(2)\\ &\: \: \quad\quad=8\sin ^{2}2x-8\cos ^{2}2x\\ &\textrm{Bandingkan dengan}\\ &\color{red}f''(x)=a\sin ^{2}bx+c\cos ^{2}dx\\ &\textrm{maka},\: \: a=8,\: b=2,\: c=-8,\: d=2\\ &\textrm{Jadi},\: \displaystyle \frac{a-b}{c-d}=\frac{8-2}{-8-2}=-\frac{3}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 52.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{\cos x}{\sin x+\cos x}\: .\: \textrm{Jika}\\ &f''(x)=\displaystyle \frac{m\cos 2x}{\left ( \sin 2x+n \right )^{2}}\: ,\: \textrm{nilai dari}\\ &m.n=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&5\\ \textrm{d}.&8\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&f(x)=\displaystyle \frac{\cos x}{\sin x+\cos x}\\ &f'(x)=\displaystyle \frac{-\sin x(\sin x+\cos x)-\cos x(\cos x-\sin x)}{(\sin x+\cos x)^{2}}\\ &\, \qquad =\displaystyle \frac{-\sin ^{2}x-\cos ^{2}x+0}{\sin ^{2}+2\sin x\cos x+\cos ^{2}x}\\ &\, \qquad=\displaystyle \frac{-1}{1+\sin 2x}\\ &f''(x)=\displaystyle \frac{0-((-1).2\cos 2x)}{\left ( \sin 2x+1 \right )^{2}}=\frac{2\cos 2x}{\left ( \sin 2x+1 \right )^{2}}\\ &\color{red}\textrm{Bandingkan dengan yang diketahui}\\ &\color{black}f''(x)=\displaystyle \frac{m\cos 2x}{\left ( \sin 2x+n \right )^{2}}\\ &\begin{cases} m &=2 \\ n &=2 \end{cases}\\ &\textrm{Jadi},\: \: m.n=2.1=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 53.&\textrm{Salah satu titik belok dari fungsi}\\ & f(x)=\sin 2x\: \: \textrm{dengan}\: \: 0\leq x\leq 2\pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left ( \displaystyle \frac{\pi }{4},0 \right )\\ \color{red}\textrm{b}.&\left ( \displaystyle \frac{\pi }{2},0 \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{4},1 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{\pi }{2},1 \right )\\ \textrm{e}.&\left ( \pi ,1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&f(x)=\sin 2x\\ &f'(x)=2\cos 2x\Rightarrow f''(x)=-4\sin 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &-4\sin 2x=0\Leftrightarrow \sin 2x=0\\ &\Leftrightarrow \sin 2x=\sin 0\\ &\Leftrightarrow 2x=0+k.2\pi \: \: \textrm{atau}\: \: 2x=\pi +k.2\pi \\ &\Leftrightarrow x=0+k.\pi \: \: \textrm{atau}\: \: x=\frac{\pi}{2} +k.\pi \\ &\Leftrightarrow \color{black}x=0,\: \color{red}x=\displaystyle \frac{\pi }{2},\: x=\pi \: ,\: x=\displaystyle \frac{3\pi }{2}\: \: \textrm{atau}\: \: \color{black}x=2\pi\\ &\bullet f\left ( \displaystyle \frac{\pi }{2} \right )=\sin 2\left ( \displaystyle \frac{\pi }{2} \right )=0\Rightarrow \color{red}\left ( \displaystyle \frac{\pi }{2},0 \right )\\ &\bullet f\left ( \displaystyle \pi \right )=\sin 2\left ( \displaystyle \pi \right )=0\Rightarrow \color{red}\left ( \displaystyle \pi ,0 \right )\\ &\bullet f\left ( \displaystyle \frac{3\pi }{2} \right )=\sin 2\left ( \displaystyle \frac{3\pi }{2} \right )=0\Rightarrow \color{red}\left ( \displaystyle \frac{3\pi }{2},0 \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 54.&\textrm{Diketahui fungsi}\: \: f(x)=-3\cos 2x+1\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Salah satu koordinat titik belok}\\ &\textrm{dari fungsi}\: \: f(x)\: \: \textrm{tersebut}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left ( \displaystyle \frac{\pi }{2},2 \right )\\ \textrm{b}.&\left ( \displaystyle \frac{2\pi }{3},\frac{5}{2} \right )\\ \textrm{c}.&\left ( \displaystyle \frac{3\pi}{2} ,4 \right )\\ \color{red}\textrm{d}.&\left ( \displaystyle \frac{5\pi }{4},1 \right )\\ \textrm{e}.&\left ( \displaystyle \frac{5\pi }{3},\frac{5}{2} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&f(x)=-3\cos 2x+1\: \: \color{black}\textrm{untuk}\: \: \color{red}0<x<2\pi \\ &f'(x)=6\sin 2x\Rightarrow f''(x)=12\cos 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &12\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi\Leftrightarrow x=\pm \frac{\pi}{4} +k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi}{4} \: ,\: x=\displaystyle \frac{5\pi }{4}\: \: \textrm{atau}\: \: x=\frac{7\pi }{4}\\ &\bullet f\left ( \displaystyle \frac{\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{\pi }{4},1 \right )\\ &\bullet f\left ( \displaystyle \frac{3\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{3\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{3\pi }{4},1 \right )\\ &\bullet f\left ( \displaystyle \frac{5\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{5\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{5\pi }{4},1 \right )\\ &\bullet f\left ( \displaystyle \frac{7\pi }{4} \right )=-3\cos 2\left ( \displaystyle \frac{7\pi }{4} \right )+1=1\Rightarrow \color{red}\left ( \displaystyle \frac{7\pi }{4},1 \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 55.&\textrm{Diketahui fungsi}\: \: f(x)=\sin^{2} x+2\: \: \textrm{dengan}\\ &0<x<2\pi \: .\: \textrm{Salah satu koordinat titik belok}\\ &\textrm{dari fungsi}\: \: f(x)\: \: \textrm{tersebut}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left ( \displaystyle \frac{\pi }{4},\frac{5}{2} \right )\\ \textrm{b}.&\left ( \displaystyle \frac{\pi }{3},\frac{11}{4} \right )\\ \textrm{c}.&\left ( \displaystyle \pi ,2 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{4\pi }{3},\frac{11}{4} \right )\\ \textrm{e}.&\left ( \displaystyle \frac{11\pi }{6},\frac{9}{4} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&f(x)=\sin^{2} x+2\: \: \color{black}\textrm{untuk}\: \: \color{red}0<x<2\pi \\ &f'(x)=2\sin x\cos x\Rightarrow f'(x)=\sin 2x\\ &f''(x)=2\cos 2x\\ &\color{black}\textrm{Syarat belok}\: \: f''(x)=0\\ &2\cos 2x=0\Leftrightarrow \cos 2x=0\\ &\Leftrightarrow \cos 2x=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow 2x=\pm \displaystyle \frac{\pi }{2}+k.2\pi\Leftrightarrow x=\pm \frac{\pi}{4} +k.\pi \\ &\Leftrightarrow \color{red}x=\displaystyle \frac{\pi }{4},\: x=\frac{3\pi}{4} \: ,\: x=\displaystyle \frac{5\pi }{4}\: \: \textrm{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 \color{red}\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 \color{red}\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 \color{red}\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 \color{red}\left ( \displaystyle \frac{7\pi }{4},\frac{5}{2} \right ) \end{aligned} \end{array}$

Contoh Soal 10 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{ll}\\ 46.&\textrm{Turunan kedua dari}\: \: f(x)=x^{3}-\sin 3x\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&6x^{2}+9\sin 3x\\ \textrm{b}.&3x^{2}+6\sin 3x\\ \textrm{c}.&3x-9\sin 3x\\ \color{red}\textrm{d}.&6x+9\sin 3x\\ \textrm{e}.&9x-6\sin 3x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}f(x)&=x^{3}-\sin 3x\\ f'(x)&=3x^{2}-3\cos 3x\\ f''(x)&=6x+9\sin 3x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 47.&\textrm{Diketahui fungsi}\: \: g(x)=\displaystyle \frac{1-\cos x}{\sin x}\: . \textrm{Nilai}\\ &\textrm{turunan kedua saat}\: \: x=\displaystyle \frac{\pi}{4}\: \: \textrm{adalah}\: .... \\ &\begin{array}{llll}\\ \textrm{a}.&\sqrt{2}+4\\ \textrm{b}.&2\sqrt{2}-3\\ \textrm{c}.&2\sqrt{2}+3\\ \color{red}\textrm{d}.&3\sqrt{2}-4\\ \textrm{e}.&3\sqrt{2}+4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}g(x)&=\displaystyle \frac{1-\cos x}{\sin x}\\ g'(x)&=\displaystyle \frac{\sin x(\sin x)-\cos x(1-\cos x)}{\sin ^{2}x}\\ &=\displaystyle \frac{\sin ^{2}x-\cos x+\cos ^{2}x}{\sin ^{2}x}\\ &=\displaystyle \frac{1-\cos x}{\sin ^{2}x}\\ g''(x)&=\displaystyle \frac{\sin x(\sin ^{2}x)-2\sin x\cos x(1-\cos x)}{\sin ^{4}x}\\ &=\displaystyle \frac{\sin x(\sin ^{2}x)-\sin 2x(1-\cos x)}{\sin ^{4}x}\\ &=\color{red}\displaystyle \frac{\sin \displaystyle \frac{\pi }{4}(\sin ^{2}\displaystyle \frac{\pi }{4})-\sin 2\displaystyle \frac{\pi }{4}(1-\cos \displaystyle \frac{\pi }{4})}{\sin ^{4}\displaystyle \frac{\pi }{4}}\\ &=\color{black}\displaystyle \frac{\left ( \displaystyle \frac{1}{\sqrt{2}} \right )\left ( \displaystyle \frac{1}{\sqrt{2}} \right )^{2}-1.\left ( 1-\left ( \displaystyle \frac{1}{\sqrt{2}} \right ) \right )}{\left ( \displaystyle \frac{1}{\sqrt{2}} \right )^{4}}\\ &=\color{black}\displaystyle \frac{\displaystyle \frac{1}{2}\displaystyle \frac{1}{\sqrt{2}}-1+\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle \frac{1}{4}}\times \displaystyle \frac{4}{4}\\ &=\displaystyle \frac{\displaystyle \frac{2}{\sqrt{2}}-4+\frac{4}{\sqrt{2}}}{1}\\ &=\displaystyle \frac{6}{\sqrt{2}}-4=3\sqrt{2}-4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Turunan kedua fungsi}\: \: f(x)=\sin ^{2}x-\cos ^{2}x\\ &\textrm{adalah}\: \: f''(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&6\sin 2x\\ \color{red}\textrm{b}.&4\cos 2x\\ \textrm{c}.&2\cos 2x\\ \textrm{d}.&-2\cos 2x\\ \textrm{e}.&-4\cos 2x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{purple}\begin{aligned}f(x)&=\sin ^{2}x-\cos ^{2}x\\ f'(x)&=2\sin x\cos x-2\cos x(-\sin x)\\ &=2\sin x\cos x+2\sin x\cos x\\ &=2(2\sin x\cos x)=\color{black}2\sin 2x\\ f''(x)&=\color{red}2.2\cos 2x=4\cos 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 49.&\textrm{Diketahui}\: \: f(x)=\sqrt{\sin x}\: .\: \textrm{Jika}\: \: f''(x)\\ &\textrm{adalah turunan keduafungsi}\: \: f,\: \textrm{maka}\\ &\textrm{nilai dari}\: \: f''\left ( \displaystyle \frac{\pi }{2} \right )\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\displaystyle \frac{1}{2}\\ \textrm{b}.&-\displaystyle \frac{1}{4}\\ \textrm{c}.&0\\ \textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}f(x)&=\color{black}\sqrt{\sin x}=\sin ^{\frac{1}{2}}x\\ f'(x)&=\displaystyle \frac{1}{2}\sin ^{-\frac{1}{2}}x.\cos x=\displaystyle \frac{\cos x}{2\sin ^{\frac{1}{2}}x}\\ f''(x)&=\color{red}\displaystyle \frac{-\sin x\left ( 2\sin ^{\frac{1}{2}}x \right )-\cos x\left ( 2.\displaystyle \frac{1}{2}\sin ^{-\frac{1}{2}}x.\cos x \right )}{4\sin x}\\ &=\displaystyle \frac{-2\sin x\sqrt{\sin x}-\displaystyle \frac{\cos ^{2}x}{\sqrt{\sin x}}}{4\sin x}\\ f''\left ( \displaystyle \frac{\pi }{2} \right )&=\color{black}\displaystyle \frac{-2\sin \displaystyle \frac{\pi }{2}.\sqrt{\sin \displaystyle \frac{\pi }{2}}-\displaystyle \frac{\cos ^{2}\displaystyle \frac{\pi }{2}}{\sin \displaystyle \frac{\pi }{2}}}{4\sin \displaystyle \frac{\pi }{2}}\\ &=\displaystyle \frac{-2.1.1-0}{4.1}=-\frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Jika}\: \: f(x)=\tan ^{2}(3x-2) \: \: \textrm{maka}\: \: f''(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&36\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &-18\sec ^{4}(3x-2)\\ \textrm{b}.&36\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+18\sec ^{2}(3x-2)\\ \color{red}\textrm{c}.&36\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+18\sec ^{4}(3x-2)\\ \textrm{d}.&18\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+36\sec ^{4}(3x-2)\\ \textrm{e}.&18\tan ^{2}(3x-2)\sec ^{2}(3x-2)\\ &+18\sec ^{4}(3x-2) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}f(x)&=\tan ^{2}(3x-2)\\ f'(x)&=2\tan (3x-2)\sec ^{2}(3x-2)(3)\\ &=6\tan (3x-2)\sec ^{2}(3x-2)\\ f''(x)&=6\sec ^{2}(3x-2).(3)\sec ^{2}(3x-2)\\ &+6\tan (3x-2).2\sec (3x-2).\sec (3x-2)\tan (3x-2)(3)\\ &=18\sec ^{4}(3x-2)\\ &+36\tan ^{2}(3x-2)\sec ^{2}(3x-2) \end{aligned} \end{array}$

Contoh Soal 9 Turunan Fungsi Trigonometri (Bagian 2)

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

$\begin{array}{ll}\\ 42.&\textrm{Sebuah benda duhubungkan dengan}\\ &\textrm{pegas dan bergerak sepanjang sumbu}\\ &\textrm{X dengan formula persamaan}:\\ &\qquad x=\sin 2t+\sqrt{3}\cos 2t\\ &\textrm{Jarak terjauh dari titik}\: \: O\: \: \textrm{yang dapat}\\ &\textrm{dicapai oleh benda tersebut adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \color{red}\textrm{b}.&2\\ \textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui gerak benda yang bergerak}\\ &\textrm{mengikuti formula}:\\ &\qquad \color{red}x=\sin 2t+\sqrt{3}\cos 2t\\ &\color{black}\textrm{Jarak terjauh dicapai saat}\: \: x'=\displaystyle \frac{dx}{dt}=0\\ &\: \: \color{red}x'=2\cos 2t-2\sqrt{3}\sin 2t=0\\ &\: \: \color{black}\Leftrightarrow \: 2\cos 2t=2\sqrt{3}\sin 2t\\ &\: \: \color{black}\Leftrightarrow \: \displaystyle \frac{\sin 2t}{\cos 2t}=\displaystyle \frac{1}{3}\sqrt{3}\\ &\: \: \color{black}\Leftrightarrow \: \tan 2t=\tan 30^{\circ}\\ &\: \: \color{black}\Leftrightarrow \: 2t=30^{\circ}+k.180^{\circ}\\ &\: \: \color{black}\Leftrightarrow \: t=15^{\circ}+k.90^{\circ}\begin{cases} k=0, &t=15^{\circ} \\ k=1, &t=105^{\circ} \\ k=2, &t=195^{\circ} \\ k=3, &t=285^{\circ}\\ k=4, &t=375^{\circ}\\ &\textrm{dst} \end{cases}\\ &\textrm{Ambil}\: \: t=15^{\circ},\: \textrm{maka nilai}\\ &x-\textrm{nya adalah}:\\ &\quad \color{red}x=\sin 2t+\sqrt{3}\cos 2t\\ &\quad \Leftrightarrow \: \color{black}x=\sin 2(15^{\circ})+\sqrt{3}\cos 2(15^{\circ})\\ &\quad \Leftrightarrow \: \color{black}x=\displaystyle \frac{1}{2}+\sqrt{3}\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &\quad \Leftrightarrow \: \color{red}x=\displaystyle \frac{1}{2}+\frac{3}{2}=2 \end{aligned} \end{array}$

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

$.\: \qquad\color{blue}\begin{aligned}&\textrm{Misalkan koordinat titik}\: \: P\left ( \displaystyle \frac{2\pi }{3},k \right )\\ & \color{black}\textrm{maka},\\ &x_{p}=\displaystyle \frac{2\pi }{3},\: y_{p}=k=\sin \displaystyle \frac{2\pi }{3}=\displaystyle \frac{1}{2}\sqrt{3}\\ &\textrm{Persamaan garis singgung di titik P}:\\ &\color{red}y=m_{x_{p}}\left ( x-x_{p} \right )+y_{p}\\ &\color{black}\begin{cases} \left ( \displaystyle \frac{2\pi }{3},k \right ) &=\color{red}\left ( \displaystyle \frac{2\pi }{3},\frac{1}{2}\sqrt{3} \right ) \\ m_{x_{p}}=\displaystyle \frac{dy}{dx} &=y'=\cos x\\ \qquad m_{x_{p}}&=\cos \left ( \displaystyle \frac{2\pi }{3} \right )=\color{red}-\frac{1}{2} \end{cases}\\ &\textrm{Sehingga persamaan garis singgungnya}\\ &\color{red}y=\left ( -\displaystyle \frac{1}{2} \right )\left ( x-\displaystyle \frac{2\pi }{3} \right )+\displaystyle \frac{1}{2}\sqrt{3}\\ &\Leftrightarrow \color{red}2y=-x+\displaystyle \frac{2\pi }{3}+\displaystyle \frac{1}{2}\sqrt{3}\\ &\bullet\quad \textrm{memotong sumbu-X, maka}\: \: y_{A}=0\\ &\qquad \color{red}2y_{A}=-x_{A}+\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\qquad \color{black}0=-x_{A}+\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\qquad \color{black}x_{A}=\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\bullet\quad \textrm{memotong sumbu-Y, maka}\: \: x_{B}=0\\ &\qquad \color{red}2y_{B}=-x_{B}+\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\qquad \color{black}2y_{B}=0+\displaystyle \frac{2\pi }{3}+\sqrt{3}\\ &\qquad \color{black}y_{B}=\displaystyle \frac{\pi }{3}+\displaystyle \frac{1}{2}\sqrt{3}\\ &\bullet\quad \color{blue}\textrm{Luas}\: \: \triangle AOB=\left [ AOB \right ]=\color{red}\displaystyle \frac{x_{A}.y_{B}}{2}\\ &\qquad \color{black}=\displaystyle \frac{\left ( \displaystyle \frac{2\pi }{3}+\sqrt{3} \right ).\left ( \displaystyle \frac{\pi }{3}+\displaystyle \frac{1}{2}\sqrt{3} \right )}{2}\\ &\qquad \color{black}=\displaystyle \frac{1}{6}\left ( 2\pi +3\sqrt{3} \right ).\displaystyle \frac{1}{6}\left ( 2\pi +3\sqrt{3} \right )\\ &\qquad \color{red}=\displaystyle \frac{1}{36}\left ( 2\pi +3\sqrt{3} \right )^{2} \end{aligned}$

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

$\begin{array}{l} .\: \qquad&\\ &\begin{array}{llll} \color{red}\textrm{a}.&\displaystyle \frac{\pi }{3}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}\\ \textrm{c}.&\displaystyle \frac{\pi }{5}\\ \textrm{d}.&\displaystyle \frac{\pi }{6}\\ \textrm{e}.&\displaystyle \frac{\pi }{8} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Supaya memuat dapat maksimum}\\ &\textrm{maka luas penampang haruslah}\\ &\color{black}\textrm{MAKSIMUM, yaitu} \end{aligned} \end{array}$

gambar 1

gambar 2

$.\: \qquad\color{blue}\begin{aligned}&\textrm{Luas penampang}=\textrm{Luas Trapesium}\\ &\color{red}\textrm{dengan}\color{black}\begin{cases} t &=3\sin \theta \\ n &=3\cos \theta \end{cases}\\ &\textrm{Luas Penampang}=\color{red}\displaystyle \frac{1}{2}\left ( \sum \textrm{sisi sejajar} \right )\times t\\ &\Leftrightarrow \: L=\displaystyle \frac{1}{2}\left ( 6+2n \right )\times t\\ &\Leftrightarrow \: L=\left ( 3+n \right )\times t\\ &\Leftrightarrow \: L=\left ( 3+3\cos \theta \right )\times 3\sin \theta \\ &\Leftrightarrow \: L=9\sin \theta +9\sin \theta \cos \theta \\ &\Leftrightarrow \: L=9\sin \theta +\displaystyle \frac{9}{2}\sin 2\theta \\ &\color{black}\textrm{Suapa luas penampang}\: \: \color{red}\textrm{MAKSIMUM}\\ &\color{black}\textrm{maka}\: \: L'=\displaystyle \frac{dL}{d\theta }=0\\ &\Leftrightarrow \: L'=9\cos \theta +9\cos 2\theta =0\\ &\Leftrightarrow \: 9\cos \theta +9\cos 2\theta =0\\ &\Leftrightarrow \: 9\cos \theta +9\left ( 2\cos ^{2}\theta -1 \right ) =0\\ &\Leftrightarrow \: 2\cos^{2} \theta +\cos \theta -1=0\\ &\Leftrightarrow \: \left (\cos \theta +1 \right )\left ( 2\cos \theta -1 \right )=0\\ &\Leftrightarrow \: \cos \theta =-1\: \: \color{black}\textrm{atau}\: \: \color{blue}2\cos \theta =1\\ &\Leftrightarrow \: \cos \theta =-1\: \: \color{black}\textrm{atau}\: \: \color{blue}\cos \theta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \theta =\cos \pi \: \: \color{black}\textrm{atau}\: \: \color{blue}\cos \theta =\cos \displaystyle \frac{\pi }{3}\\ &\Leftrightarrow \: \theta =\pi \: \: \color{black}\textrm{atau}\: \: \color{red}\theta =\displaystyle \frac{\pi }{3} \end{aligned}$

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

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}{ll}\\ 36.&\textrm{Titik stasioner fungsi}\: \: f(x)=\cos 3x\\ & \textrm{pada}\: \: 0\leq x\leq \pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(0,1),\left ( \displaystyle \frac{\pi }{4},1 \right ),\left ( \displaystyle \frac{\pi }{3},1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{2},-1 \right )\\ \textrm{b}.&(0,1),\left ( \displaystyle \frac{\pi }{3},1 \right ),\left ( \displaystyle \frac{\pi }{2},-1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \pi ,-1 \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{6},-1 \right ),\left ( \displaystyle \frac{\pi }{3},1 \right ),\left ( \displaystyle \frac{\pi }{2},-1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{2\pi }{3},1 \right )\\ \textrm{d}.&\left ( \displaystyle \frac{\pi }{6},1 \right ),\left ( \displaystyle \frac{\pi }{3},-1 \right ),\left ( \displaystyle \frac{\pi }{2},1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{2\pi }{3},-1 \right ) \\ \color{red}\textrm{e}.&(0,1),\left ( \displaystyle \frac{\pi }{3},-1 \right ),\left ( \displaystyle \frac{2\pi }{3},1 \right ),\: \: \textrm{dan}\: \: \left ( \displaystyle \pi ,-1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\\ & f(x)=\cos 3x\Rightarrow \color{red}f'(x)=-3\sin 3x\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &-\sin 3x=0\Leftrightarrow \sin 3x=0\Leftrightarrow \sin 3x=\sin 0\\ &\Leftrightarrow 3x=0+k.2\pi \: \: \textrm{atau}\: \: 3x=\pi +k.2\pi \\ &\Leftrightarrow x=k.\displaystyle \frac{2\pi }{3}\: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{3} +k.\frac{2\pi}{3} \\ &\Leftrightarrow k=0\Rightarrow x=0\: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{3}\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{2\pi }{3}\: \: \textrm{atau}\: \: x=\displaystyle \pi \\ &\color{red}\textrm{Sekarang kita tentukan nilai dan titiknya}\\ &x=0\Rightarrow f(0)=\cos 3(0)=1\rightarrow (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 \\ &\qquad=-1\rightarrow \left ( \displaystyle \frac{\pi }{3},-1 \right )\\ &\textrm{dan seterusnya} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Titik stasioner fungsi}\: \: f(x)=\sin \left ( 2x-\displaystyle \frac{\pi }{6} \right )\\ &\textrm{pada}\: \: 0\leq x\leq \pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(0,1)\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{6},-1 \right )\\ \textrm{b}.&\left ( \displaystyle \frac{\pi }{6},1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{3},-1 \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{4},-1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{2},-1 \right )\\ \color{red}\textrm{d}.&\left ( \displaystyle \frac{\pi }{3},1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{5\pi }{6},-1 \right )\\ \textrm{e}.&\left ( \displaystyle \frac{\pi }{2},-1 \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \pi ,1 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\\ & f(x)=\sin \left ( 2x-\displaystyle \frac{\pi }{6} \right )\Rightarrow \color{red}f'(x)=2\cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &2\cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )=0\Leftrightarrow \cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )=0\\ &\color{black}\cos \left ( 2x-\displaystyle \frac{\pi }{6} \right )=\cos \displaystyle \frac{\pi }{2}\\ &\Leftrightarrow \left ( 2x-\displaystyle \frac{\pi }{6} \right )=\pm \displaystyle \frac{\pi }{2}+k.2\pi \\ &\Leftrightarrow x=\displaystyle \frac{\pi }{12}\pm \frac{\pi }{4}+k.\pi \begin{cases} x & =\displaystyle \frac{\pi }{3}+k.\pi \\ x & =-\displaystyle \frac{\pi }{6}+k.\pi \end{cases}\\ &\Leftrightarrow k=0\Rightarrow \begin{cases} x & =\displaystyle \frac{\pi }{3} \\ x & =-\displaystyle \frac{\pi }{6}\: \: \color{red}(\textrm{tm}) \end{cases}\\ &\Leftrightarrow k=1\Rightarrow \begin{cases} x & =\displaystyle \frac{4\pi }{3}\: \: \color{red}\textrm{tm} \\ x & =\displaystyle \frac{5\pi }{6} \end{cases}\\ &\color{red}\textrm{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 \\ &\qquad=1\rightarrow \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 )\\ &\qquad=\sin \frac{3\pi}{2}=-1\rightarrow \left ( \color{black}\displaystyle \frac{5\pi }{6},-1 \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Nilai}\: \: x\: \: \textrm{pada titik stasioner}\\ &\textrm{fungsi}\: \: f(x)=x+\sin x\: \: \textrm{untuk}\\ &0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&90^{\circ}\\ \textrm{b}.&135^{\circ}\\ \textrm{c}.&150^{\circ}\\ \color{red}\textrm{d}.&180^{\circ}\\ \textrm{e}.&360 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\\ & f(x)=x+\sin x\Rightarrow \color{red}f'(x)=1+\cos x\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &1+\cos =0\Leftrightarrow \cos x=-1\\ &\Leftrightarrow \cos x=\cos 180^{\circ}\\ &\Leftrightarrow x=\pm 180^{\circ}+k.360^{\circ}\\ &\Leftrightarrow k=0\Rightarrow x=\begin{cases} 180^{\circ} & \color{black}\textrm{mungkin} \\ -180^{\circ} & \color{red}\textrm{tidak mungkin} \end{cases}\\ &\Leftrightarrow k=1\Rightarrow x=\begin{cases} 540^{\circ} & \color{red}\textrm{tidak mungkin} \\ 180^{\circ} & \color{black}\textrm{mungkin} \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Nilai}\: \: y\: \: \textrm{pada titik stasioner}\\ &\textrm{fungsi}\: \: f(x)=4\cos x+\cos 2x\\ &\textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-5\: \: \textrm{dan}\: \: 3\\ \textrm{b}.&-4\: \: \textrm{dan}\: \: 2\\ \color{red}\textrm{c}.&-3\: \: \textrm{dan}\: \: 5\\ \textrm{d}.&-2\: \: \textrm{dan}\: \: 4\\ \textrm{e}.&3\: \: \textrm{dan}\: \: 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\\ & f(x)=4\cos x+\cos 2x\\ &\Rightarrow \color{red}f'(x)=-4\sin x-2\sin 2x\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &-4\sin x-2\sin 2x=0\\ &\Leftrightarrow -4\sin x-4\sin x\cos x=0\\ &\Leftrightarrow -4\sin x \left ( 1+\cos x \right )=0\\ &\Leftrightarrow \sin x \left ( 1+\cos x \right )=0\\ &\Leftrightarrow \color{black}\sin x=0\: \: \color{red}\textrm{atau}\: \: \color{black}1+\cos x=0\\ &\Leftrightarrow \color{black}\sin x=0\: \: \color{red}\textrm{atau}\: \: \color{black}\cos x=-1\\ &\Leftrightarrow \color{black}\sin x=\sin 0^{\circ}\: \: \color{red}\textrm{atau}\: \: \color{black}\cos x=\cos 180^{\circ}\\ &\Leftrightarrow x=\begin{cases} 0^{\circ} +k.360^{\circ} \\ 180^{\circ} +k.360^{\circ} \end{cases}\: \textrm{atau}\: \: x=\begin{cases} 180^{\circ} +k.360^{\circ} \\ -180^{\circ} +k.360^{\circ} \end{cases}\\ &\Leftrightarrow k=0\Rightarrow x=0^{\circ}\: \textrm{atau}\: 180^{\circ}\\ &\color{red}\textrm{Nilai}\: \: y-\textrm{nya}\\ &\color{black}x=0^{\circ}\Rightarrow f(0^{\circ})\\ &\qquad=\color{black}4\cos 0^{\circ}+\cos 2(0^{\circ})=\color{red}4+1=5\\ &\color{black}x=180^{\circ}\Rightarrow f(180^{\circ})\\ &\qquad=\color{black}4\cos 180^{\circ}+\cos 2(180^{\circ})=\color{red}-4+1=-3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Nilai stasioner fungsi}\\ &\quad\quad\quad f(x)=\displaystyle \frac{\sin x}{2-\cos x}\\ &\textrm{untuk}\: \: 0\leq x\leq 2\pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left ( \displaystyle \frac{\pi }{2},\frac{1}{2} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{2},-\frac{1}{2} \right )\\ \textrm{b}.&\left ( \displaystyle \frac{\pi }{3},\frac{1}{2}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{\pi }{3},-\frac{1}{2}\sqrt{3} \right )\\ \textrm{c}.&\left ( \displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{2\pi }{3},-\frac{1}{3}\sqrt{3} \right )\\ \color{red}\textrm{d}.&\left ( \displaystyle \frac{\pi }{3},\frac{1}{3}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{5\pi }{3},-\frac{1}{3}\sqrt{3} \right )\\ \textrm{e}.&\left ( \displaystyle \frac{\pi }{4},\frac{1}{4}\sqrt{3} \right )\: \: \textrm{dan}\: \: \left ( \displaystyle \frac{3\pi }{4},-\frac{1}{4}\sqrt{3} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\\ & f(x)=\displaystyle \frac{\sin x}{2-\cos x}\Rightarrow \color{red}f'(x)=\displaystyle \frac{2\cos x-1}{(2-\cos x)^{2}}\\ &\textrm{Stasioner fungsi}\: \: f\: \: \textrm{saat}\: \: \color{black}f'(x)=0\: \: \textrm{maka},\\ &\displaystyle \frac{2\cos x-1}{(2-\cos x)^{2}}=0\Leftrightarrow 2\cos x-1=0\\ &\Leftrightarrow \cos x=\displaystyle \frac{1}{2}\Leftrightarrow \cos x=\cos \displaystyle \frac{\pi }{3}\\ &\Leftrightarrow x=\pm \displaystyle \frac{\pi }{3}+k.2\pi \\ &\Leftrightarrow k=0\Rightarrow x=\pm \displaystyle \frac{\pi }{3}\Leftrightarrow x=\begin{cases} \displaystyle \frac{\pi }{3} & \color{black}\textrm{memenuhi} \\ -\displaystyle \frac{\pi }{3} & \color{red}\textrm{tidak memenuhi} \end{cases}\\ &\Leftrightarrow k=1\Rightarrow x=\pm \displaystyle \frac{\pi }{3}+2\pi \Leftrightarrow x=\begin{cases} \displaystyle \frac{7\pi }{3} & \color{red}\textrm{tidak memenuhi} \\ \displaystyle \frac{5\pi }{3} & \color{black}\textrm{memenuhi} \end{cases}\\ &\color{black}\textrm{Titiknya adalah}\\ &\color{black}x=\displaystyle \frac{\pi }{3}\Rightarrow f\left ( \displaystyle \frac{\pi }{3} \right )\\ &\qquad=\color{black}\displaystyle \frac{\sin \displaystyle \frac{\pi }{3}}{2-\cos \displaystyle \frac{\pi }{3}}=\color{red}\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{3}}{2-\displaystyle \frac{1}{2}}=\displaystyle \frac{1}{3}\sqrt{3}\\ &\qquad \left ( \displaystyle \frac{\pi }{3},\displaystyle \frac{1}{3}\sqrt{3} \right )\\ &\color{black}x=\displaystyle \frac{5\pi }{3}\Rightarrow f\left ( \displaystyle \frac{5\pi }{3} \right )\\ &\qquad=\color{black}\displaystyle \frac{\sin \displaystyle \frac{5\pi }{3}}{2-\cos \displaystyle \frac{5\pi }{3}}=\color{red}\displaystyle \frac{-\displaystyle \frac{1}{2}\sqrt{3}}{2-\displaystyle \frac{1}{2}}=-\displaystyle \frac{1}{3}\sqrt{3}\\ &\qquad \left ( \displaystyle \frac{5\pi }{3},-\displaystyle \frac{1}{3}\sqrt{3} \right ) \end{aligned} \end{array}$

Contoh Soal 7 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{ll}\\ 31.&\textrm{Fungsi}\: \: f(x)=\sin x-\cos x\: \: \textrm{dengan}\\ &0<x<2\pi \: \: \textrm{naik pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0<x<\displaystyle \frac{\pi }{4}\\ \textrm{b}.&\displaystyle \frac{\pi }{4}<x<2\pi \\ \textrm{c}.&\displaystyle \frac{3\pi }{4}<x<\displaystyle \frac{7\pi }{4} \\ \color{red}\textrm{d}.&0<x<\displaystyle \frac{3\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{7\pi }{4}<x<2\pi \\ \textrm{e}.&0<x<\displaystyle \frac{\pi }{4}\: \: \textrm{atau}\: \: \displaystyle \frac{3\pi }{4}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sin x-\cos x\\ &\color{black}\textrm{Fungsi}\: \: f\: \: \textrm{naik, jika}\: \: f'(x)>0\\ &\textrm{Selanjutnya}\\ &f'(x)=\cos x+\cos x=0\\ &\sin x=-\cos x\Leftrightarrow \displaystyle \frac{\sin x}{\cos x}=-1\\ &\Leftrightarrow \tan x=-1\\ &\Leftrightarrow \tan x=\tan \displaystyle \frac{3\pi }{4}\\ &\Leftrightarrow x=\displaystyle \frac{3\pi }{4}\pm k.\pi \\\ &\Leftrightarrow k=0\Rightarrow x=\displaystyle \frac{3\pi }{4}\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{3\pi }{4}\pm \pi =\frac{7\pi }{4}\\ &\Leftrightarrow k=2\Rightarrow x=\displaystyle \frac{3\pi }{4}\pm 2\pi =\color{red}\textrm{tm}\\ &\color{black}\begin{array}{ccccccccc}\\ &&&&&&&\\ &++&&--&&&++&\\\hline 0&&\color{red}\displaystyle \frac{3\pi }{4}&&&\color{red}\displaystyle \frac{7\pi }{4}&&2\pi \end{array}\\ &\textrm{ambil titik uji}\: \: x=\displaystyle \frac{1}{2}\pi \\ &\textrm{untuk}\: \: x=\displaystyle \frac{1}{2}\pi \Rightarrow f'\left ( \displaystyle \frac{1}{2}\pi \right )\\ &\quad =\cos \displaystyle \frac{1}{2}\pi +\sin \displaystyle \frac{1}{2}\pi =0+1=1\: \: \color{red}(\textrm{positif})\\ &\textrm{untuk}\: \: x=\displaystyle \frac{3}{2}\pi \Rightarrow f'\left ( \displaystyle \frac{3}{2}\pi \right )\\ &\quad =\cos \displaystyle \frac{3}{2}\pi +\sin \displaystyle \frac{3}{2}\pi =0-1=-1\: \: \color{red}(\textrm{negatif})\\ &\textrm{untuk}\: \: x=\displaystyle \frac{11}{6}\pi \Rightarrow f'\left ( \displaystyle \frac{11}{6}\pi \right )\\ &\quad =\cos \displaystyle \frac{11}{6}\pi +\sin \displaystyle \frac{11}{6}\pi =\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\: \: \color{red}(\textrm{positif}) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Fungsi}\: \: f(x)=\sin^{2} x\: \: \textrm{dengan}\\ &0<x<2\pi \: \: \textrm{naik pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{\pi }{2}<x<\pi \: \: \textrm{atau}\: \: \displaystyle \frac{3\pi }{2}<x<2\pi \\ \textrm{b}.&\displaystyle \frac{2\pi }{3}<x<\pi \\ \color{red}\textrm{c}.&0<x<\displaystyle \frac{\pi }{2}\: \: \textrm{atau}\: \: \pi <x<\displaystyle \frac{3\pi }{2} \\ \textrm{d}.&\displaystyle \frac{4\pi }{3}<x<2\pi \\ \textrm{e}.&\displaystyle \frac{\pi }{3}<x<\pi \: \: \textrm{atau}\: \: \displaystyle \frac{4\pi }{3}<x<2\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sin^{2} x\\ &\color{black}\textrm{Fungsi}\: \: f\: \: \textrm{naik, jika}\: \: f'(x)>0\\ &\textrm{Selanjutnya}\\ &f'(x)=2\sin x\cos x=\sin 2x=0\\ &\Leftrightarrow \sin 2x=0\\ &\Leftrightarrow \sin 2x=\sin 0\\ &\Leftrightarrow 2x=\pm k.2\pi \: \: \textrm{atau}\: \: 2x=\pi \pm k.2\pi \\ &\Leftrightarrow x=\pm k.\pi \: \: \textrm{atau}\: \: x=\frac{\pi}{2} \pm k.\pi \\ &\Leftrightarrow k=0\Rightarrow x=0\: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{2}\\ &\Leftrightarrow k=1\Rightarrow x=\pi \: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{2}+ \pi =\frac{3\pi }{2}\\ &\Leftrightarrow k=2\Rightarrow x= 2\pi \: \: \textrm{atau}\: \: x=\displaystyle \frac{\pi }{2}+2\pi =\displaystyle \frac{5}{2}\pi \: \color{red}(\textrm{tm})\\ &\color{black}\begin{array}{ccccccccc}\\ &&&&&&&&\\ &++&&--&&++&&--&\\\hline 0&&\color{red}\displaystyle \frac{\pi }{2}&&\pi &&\color{red}\displaystyle \frac{3\pi }{2}&&2\pi \end{array} \\ &\textrm{ambil titik uji}\: \: x=\displaystyle \frac{1}{6}\pi \\ &\textrm{untuk}\: \: x=\displaystyle \frac{1}{6}\pi \Rightarrow f'\left ( \displaystyle \frac{1}{6}\pi \right )\\ &\quad =\sin 2\left (\displaystyle \frac{1}{6}\pi \right )=\sin \displaystyle \frac{1}{3}\pi =\frac{1}{2}\: \: \color{red}(\textrm{positif})\\ &\textrm{untuk}\: \: x=\displaystyle \frac{3}{4}\pi \Rightarrow f\left ( \displaystyle \frac{3}{4}\pi \right )\\ &\quad =\sin 2\left (\displaystyle \frac{3}{4}\pi \right )=-1\: \: \color{red}(\textrm{negatif}) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Fungsi}\: \: f(x)=\cos ^{2}2x\: \: \textrm{untuk}\\ &0^{\circ}<x<360^{\circ}\: \: \textrm{turun pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&45^{\circ}<x<90^{\circ}\\ \textrm{b}.&135^{\circ}<x<180^{\circ}\\ \textrm{c}.&225^{\circ}<x<270^{\circ}\\ \color{red}\textrm{d}.&270^{\circ}<x<300^{\circ}\\ \textrm{e}.&315^{\circ}<x<360^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&f(x)=\cos ^{2}2x\\ &\color{black}\textrm{Fungsi}\: \: f\: \: \textrm{turun, jika}\: \: f'(x)<0\\ &f'(x)=2\cos 2x(-\sin 2x)(2)=-2\sin 4x\\ &\textrm{Selanjutnya}\\ &\Leftrightarrow -2\sin 4x=0\Leftrightarrow \sin 4x=0\Leftrightarrow \sin 4x=\sin 0^{\circ}\\ &\Leftrightarrow \begin{cases} 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{cases}\\ &\Leftrightarrow k=0\Rightarrow x=0^{\circ}\: \: \textrm{atau}\: \: x=45^{\circ}\\ &\Leftrightarrow k=1\Rightarrow x=90^{\circ}\: \: \textrm{atau}\: \: x=135^{\circ}\\ &\Leftrightarrow k=2\Rightarrow x=180^{\circ}\: \: \textrm{atau}\: \: x=225^{\circ}\\ &\Leftrightarrow k=3\Rightarrow x=270^{\circ}\: \: \textrm{atau}\: \: x=315^{\circ}\\ &\Leftrightarrow k=4\Rightarrow x=360^{\circ}\: \: \textrm{atau}\: \: x=405^{\circ}\: \: \color{red}(\textrm{tm})\\ &\textrm{Gunakan titik uji pada}\: \: x=30^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(30^{\circ})=-2\sin 4(30^{\circ})=-\sqrt{3}\: \: \color{red}(\textrm{negatif})\\ &\textrm{Gunakan titik uji pada}\: \: x=60^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(60^{\circ})=-2\sin 4(60^{\circ})=\sqrt{3}\: \: \color{red}(\textrm{positif})\\ &\textrm{Gunakan titik uji pada}\: \: x=120^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(120^{\circ})=-2\sin 4(120^{\circ})=-\sqrt{3}\: \: \color{red}(\textrm{negatif})\\ &\textrm{Gunakan titik uji pada}\: \: x=150^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'(150^{\circ})=-2\sin 4(150^{\circ})=\sqrt{3}\: \: \color{red}(\textrm{positif})\\ &\color{black}\textrm{dan seterusnya}\: ...\\ &\color{black}\begin{array}{ccccccccccc}\\ &&&&&&&&\\ &--&&++&&--&&++\\\hline 0&&\color{red}\displaystyle 45^{\circ}&&90^{\circ}&&\color{red}135^{\circ}&&180^{\circ}\\ &--&&++&&--&&++\\\hline 180^{\circ}&&\color{red}\displaystyle 225^{\circ}&&270^{\circ}&&\color{red}315^{\circ}&&360^{\circ} \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&(\textbf{SBMPTN 2015})\\ &\textrm{Fungsi}\: \: f(x)=2\sqrt{\sin ^{2}x+\displaystyle \frac{x\sqrt{3}}{2}}\\ & \textrm{pada}\: \: 0<x<\pi \: \: \textrm{turun pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{5\pi }{12}<x<\frac{11\pi }{12}\\ \textrm{b}.&\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}\\ \color{red}\textrm{c}.&\displaystyle \frac{2\pi }{3}<x<\frac{5\pi }{6}\\ \textrm{d}.&\displaystyle \frac{3\pi }{4}<x<\pi \\ \textrm{e}.&\displaystyle \frac{3\pi }{4}<x<\frac{3\pi }{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \: f(x)=2\sqrt{\sin ^{2}x+\displaystyle \frac{x\sqrt{3}}{2}}\\ &\color{black}\textrm{Fungsi}\: \: f\: \: \textrm{turun, jika}\: \: f'(x)<0\\ &f'(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\Leftrightarrow \sin 2x=-\displaystyle \frac{1}{2}\sqrt{3}\\ &\Leftrightarrow \sin 2x=\sin \displaystyle \frac{4\pi }{3}\\ &\Leftrightarrow 2x=\displaystyle \frac{4\pi }{3}+ k.2\pi \: \: \textrm{atau}\: \: 2x=\pi -\frac{4\pi }{3}+k.2\pi \\ &\Leftrightarrow x=\displaystyle \frac{2\pi }{3}+ k.\pi \: \: \textrm{atau}\: \: x= -\frac{\pi }{6}+k.\pi \\ &\Leftrightarrow k=0\Rightarrow x=\displaystyle \frac{2\pi }{3}\: \: \textrm{atau}\: \: x=-\displaystyle \frac{\pi }{6}\: \: \color{red}(\textrm{tm})\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{5\pi }{3}\: \: \textrm{atau}\: \: x=\displaystyle \frac{5\pi }{6}\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{\pi }{2}=\color{black}90^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\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}}}=\color{red}+\\ &\qquad \color{red}(\textrm{positif})\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{3\pi }{4}=\color{black}135^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\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}}}=\color{red}-\\ &\qquad \color{red}(\textrm{negatif})\\ &\color{black}\begin{array}{cccccc}\\ &&&&\\ &++&&--\\\hline 0&&\color{red}\displaystyle \displaystyle \frac{2\pi }{3}&&\color{red}\displaystyle \frac{5\pi }{6} \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Fungsi}\: \: f(x)=\sqrt{\sin ^{2}x+\displaystyle \frac{x}{2}}\\ & \textrm{dengan}\: \: x>0 \: \: \textrm{turun pada interval}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{5\pi }{12}<x\leq \frac{13\pi }{12}\\ \color{red}\textrm{b}.&\displaystyle \frac{7\pi }{12}<x<\frac{11\pi }{12}\\ \textrm{c}.&\displaystyle \frac{\pi }{12}<x<\frac{5\pi }{12}\\ \textrm{d}.&\displaystyle \frac{7\pi }{6}<x\leq \displaystyle \frac{13\pi }{6} \\ \textrm{e}.&\displaystyle \frac{7\pi }{6}<x\leq \frac{11\pi }{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \: f(x)=\sqrt{\sin ^{2}x+\displaystyle \frac{x}{2}}\\ &\color{black}\textrm{Fungsi}\: \: f\: \: \textrm{turun, jika}\: \: f'(x)<0\\ &f'(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\Leftrightarrow \sin 2x=-\displaystyle \frac{1}{2}\\ &\Leftrightarrow \sin 2x=\sin \displaystyle \frac{7\pi }{6}\\ &\Leftrightarrow 2x=\displaystyle \frac{7\pi }{6}+ k.2\pi \: \: \textrm{atau}\: \: 2x=\pi -\frac{7\pi }{6}+k.2\pi \\ &\Leftrightarrow x=\displaystyle \frac{7\pi }{12}+ k.\pi \: \: \textrm{atau}\: \: x= -\frac{\pi }{12}+k.\pi \\ &\Leftrightarrow k=0\Rightarrow x=\displaystyle \frac{7\pi }{12}\: \: \textrm{atau}\: \: x=-\displaystyle \frac{\pi }{12}\: \: \color{red}(\textrm{tm})\\ &\Leftrightarrow k=1\Rightarrow x=\displaystyle \frac{19\pi }{12}\: \: \textrm{atau}\: \: x=\displaystyle \frac{11\pi }{12}\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{\pi }{2}=\color{black}90^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\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}}}=\color{red}+\\ &\qquad \color{red}(\textrm{positif})\\ &\textrm{Gunakan titik uji pada}\: \: x=\displaystyle \frac{3\pi }{4}=\color{black}135^{\circ}\\ &\bullet \quad \textrm{untuk}\: \: f'\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}}}=\color{red}-\\ &\qquad \color{red}(\textrm{negatif})\\ &\color{black}\begin{array}{cccccc}\\ &&&&\\ &++&&--\\\hline 0&&\color{red}\displaystyle \displaystyle \frac{7\pi }{12}&&\color{red}\displaystyle \frac{11\pi }{12} \end{array} \end{aligned} \end{array}$

Contoh Soal 6 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{ll}\\ 25.&\textrm{Persamaan garis singgung pada kurva}\\ &y=3\sin x\: \: \textrm{pada titik yang berabsis}\: \: \displaystyle \frac{\pi }{3}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\displaystyle \frac{2}{3}\left ( x-\frac{\pi }{3} \right )-\frac{2\sqrt{2}}{3}\\ \textrm{b}.&y=\displaystyle \frac{2}{3}\left ( x-\frac{\pi }{3} \right )+\frac{2\sqrt{2}}{3}\\ \textrm{c}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )-\frac{3\sqrt{3}}{2}\\ \color{red}\textrm{d}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )+\frac{3\sqrt{3}}{2}\\ \textrm{e}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )-\frac{3\sqrt{2}}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&y=3\sin x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &y_{0}=3\sin \left ( \color{red}\displaystyle \frac{\pi }{3} \right )=3\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )=\frac{3\sqrt{3}}{2}\\ &\color{black}\textrm{kita cari gradien}\: \: \color{blue}m\: \: \color{black}\textrm{saat}\: \: \color{blue}y',\: \: \textrm{yaitu}:\\ &m=y'=3\cos x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &m=3\cos \left ( \color{red}\displaystyle \frac{\pi }{3} \right )=3\left ( \displaystyle \frac{1}{2} \right )=\frac{3}{2}\\  &\color{black}\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &y=\displaystyle \frac{3}{2}\left ( x-\displaystyle \frac{\pi }{3} \right )+\displaystyle \frac{3\sqrt{3}}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 26.&\textrm{Kurva}\: \: y=\sin x+\cos x\: \: \textrm{untuk}\\ &0<x<\pi \: \: \textrm{memotong sumbu X}\\ &\textrm{di titik A. Persamaan garis}\\ &\textrm{singgung di titik A adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )\\ \textrm{b}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{\pi }{2} \right )\\ \color{red}\textrm{c}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right )\\ \textrm{d}.&y=\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )\\ \textrm{e}.&y=\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\color{black}\textrm{Kurva memotong sumbu X}\\ &\textrm{di titik A, berarti}\: \: \color{red}y=0\\ &\sin x+\cos x=\color{red}0\\ &\sin x=-\cos x\\ &\displaystyle \frac{\sin x}{\cos x}=-1\Leftrightarrow \tan x=-1\\ &\tan x=\tan \left ( \displaystyle \frac{3\pi }{4} \right )\Rightarrow x=\displaystyle \frac{3\pi }{4}\\ &\color{red}\textrm{Jadi, titik A-nya}:\: \: \left ( \displaystyle \frac{3\pi }{4},0 \right )\\ &\textrm{dan nilai gradien}\: \: m=y',\: \: \textrm{yaitu}:\\ &m=\cos x-\sin x\\ &m=\cos \left ( \displaystyle \frac{3\pi }{4} \right )-\sin \left ( \displaystyle \frac{3\pi }{4} \right )\\ &m=-\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}\\ &\color{black}\textrm{Persamaan garis singgung di A}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) +0\\ &\Leftrightarrow \: y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sec ^{2}x\: \: \textrm{pada titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{3}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4\\ \color{red}\textrm{b}.&y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4\\ \textrm{c}.&y=-8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4\\ \textrm{d}.&y=-8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4\\ \textrm{e}.&y=4\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&y=\sec^{2} x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &y_{0}=\sec^{2} \left ( \color{red}\displaystyle \frac{\pi }{3} \right )=(2)^{2}=4\\ &\color{black}\textrm{kita cari gradien}\: \: \color{blue}m\: \: \color{black}\textrm{saat}\: \: \color{blue}y',\: \: \textrm{yaitu}:\\ &m=y'=2\sec^{2} x\tan x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &m=2\sec^{2} \left ( \color{red}\displaystyle \frac{\pi }{3} \right )\tan \left ( \color{red}\displaystyle \frac{\pi }{3} \right )\\ &\quad=2(4)\sqrt{3}=\color{red}8\sqrt{3}\\ &\color{black}\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 29.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sec x\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{4}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\sqrt{3}x-\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}\\ \textrm{b}.&y=\sqrt{3}x+\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}\\ \color{red}\textrm{c}.&y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}\\ \textrm{d}.&y=\sqrt{2}x+\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}\\ \textrm{e}.&y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&y=\sec x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{4}\\ &y_{0}=\sec \left ( \color{red}\displaystyle \frac{\pi }{4} \right )=\sqrt{2}\\ &\color{black}\textrm{kita cari gradien}\: \: \color{blue}m\: \: \color{black}\textrm{saat}\: \: \color{blue}y',\: \: \textrm{yaitu}:\\ &m=y'=\sec x\tan x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{4}\\ &m=\sec \left ( \color{red}\displaystyle \frac{\pi }{4} \right )\tan \left ( \displaystyle \frac{\pi }{4} \right )=\sqrt{2}.1=\sqrt{2}\\ &\color{black}\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )+\sqrt{2}\\ &\Leftrightarrow \: \color{red}y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sin x+\cos x\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{2}\: \: \textrm{akan memotong sumbu}\\ &\textrm{Y dengan ordinatnya berupa}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{\pi }{2}+1\\ \textrm{b}.&\displaystyle \frac{\pi }{2}-1\\ \textrm{c}.&1-\displaystyle \frac{\pi }{2}\\ \textrm{d}.&2+\displaystyle \frac{\pi }{2}\\ \textrm{e}.&2-\displaystyle \frac{\pi }{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&y=\sin x+\cos x,\: \: \color{red}\textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{2}\\ &y_{0}=\sin \left ( \color{red}\displaystyle \frac{\pi }{2} \right )+\cos \left ( \displaystyle \frac{\pi }{2} \right )=1+0=1\\ &\color{black}\textrm{kita cari gradien}\: \: \color{blue}m\: \: \color{black}\textrm{saat}\: \: \color{blue}y',\: \: \textrm{yaitu}:\\ &m=\cos x-\sin x\\ &m=\cos \left ( \displaystyle \frac{\pi }{2} \right )-\sin \left ( \displaystyle \frac{\pi }{2} \right )\\ &m=0-1=-1\\ &\color{black}\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=-1\left ( x-\displaystyle \frac{\pi }{2} \right )+1\\ &\Leftrightarrow \: \color{red}y=-x+\displaystyle \frac{\pi }{2}+1\\ &\textrm{Ordinat garis singgungnya saat}\\ &\textrm{memotong sumbu-Y adalah}:\: \: x=0,\\ &\textrm{maka}\\ &\color{red}y=-0+\displaystyle \frac{\pi }{2}+1=\displaystyle \frac{\pi }{2}+1 \end{aligned} \end{array}$