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)

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

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

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

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

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

Lanjutan Materi (10) Turunan Kedua Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\text{H. Turunan Kedua Fungsi Trigonometri}$

Definisi dari bahasan ini adalah jika turunan pertama dari suatu fungsi  $f$ dan dinyatakan dengan  $f^{\prime }$ ada dan terdefinisi untuk setiap nilai  $x$  dalam daerah terdefinisi  $f$, maka turunan kedua dari fungsi  $f$  dinyatakan dengan $f^{″}$ adalah:

$f^{″}(x)=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{f^{\prime }(x+h)-f^{\prime }(x)}{h}={\displaystyle \frac{d}{dx}(f^{\prime }(x))}}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Tentukan turunan kedua dari}\\ & y=\sin x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}y& =\sin x\\ y^{\prime }& =\cos x\\ y^{″}& =-\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukan turunan kedua dari}\\ & y=\sin 2x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}y& =\sin 2x\\ y^{\prime }& =2\cos 2x\\ y^{″}& =-4\sin 2x\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Tentukan turunan kedua dari}\\ & y={\sin }^{2}x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}y& ={\sin }^{2}x\\ y^{\prime }& =2\sin x(-\cos x)=-\sin 2x\\ y^{″}& =-2\cos 2x\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Tentukan turunan kedua dari}\\ & y=\cos x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}y& =\cos x\\ y^{\prime }& =-\sin x\\ y^{″}& =-\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Tentukan turunan kedua dari}\\ & y=\tan x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}y& =\tan x\\ y^{\prime }& ={\sec }^{2}x\\ y^{″}& =2\sec x(\sec x\tan x)=2{\sec }^{2}x\tan x\end{array}\end{array}$

$\begin{array}{l}\\ 6.& \text{Tentukan turunan kedua dari}\\ & y=\cot x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}y& =\cot x\\ y^{\prime }& =-{\csc }^{2}x\\ y^{″}& =-2\csc x(-\csc x\cot x)=2{\csc }^{2}x\cot x\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Tentukan turunan kedua dari}\\ & y=\sec x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}y& =\sec x\\ y^{\prime }& =\sec x\tan x\\ y^{″}& =\sec x\tan x(\tan x)+\sec x({\sec }^{2}x)\\ & =\sec x{\tan }^{2}x+{\sec }^{3}x\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Tentukan turunan kedua dari}\\ & y=\csc x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}y& =\csc x\\ y^{\prime }& =-\csc x\cot x\\ y^{″}& =-(-\csc x\cot x)\cot x+(-\csc x)(-{\csc }^{2}x)\\ & =\csc x{\cot }^{2}x+{\csc }^{3}x\end{array}\end{array}$

$\text{I. Fungsi Naik dan Fungsi Turun}$

Sebelumnya telah diketahui bahwa pada selang terbuka

$\begin{array}{l}\\ & \bullet \text{untuk}{f}^{\prime }(x)>0\text{maka fungsi naik}\\ & \bullet \text{untuk}{f}^{\prime }(x)<0\text{maka fungsi turun}\end{array}$

$\begin{array}{l}\\ & \text{Misalkan}{f}^{\prime }\text{dan}{f}^{″}\text{ada untuk setiap}\\ & \text{titik pada suatu interval yang memuat}\\ & c\text{dengan}{f}^{\prime }(c)=0\\ & \bullet \text{jika}{f}^{″}(c)>0\text{maka}f(c)\text{adalah}\\ & \text{nilai minimum lokal (titik minimum)}\\ & \bullet \text{jika}{f}^{\prime }(c)<0\text{maka}f(c)\text{adalah}\\ & \text{nilai maksimum lokal (titik maksimum)}\\ & \bullet \text{jika}{f}^{″}(c)=0\text{maka nilai stasioner}\\ & \text{belum dapat ditentukan}\end{array}$

$\begin{array}{l}\\ & \text{Titik Belok}\\ \\ & \text{Jika}(c,f(c))\text{adalah titik belok grafik}\\ & f,\text{maka}{f}^{″}(x)=0\text{atau}{f}^{″}\text{tidak ada}\\ & \text{pada}x=c\end{array}$

$\boxed{\text{CONTOH SOAL}}$

Perhatikan lagi contoh pada bagian ini LANJUTAN MATERI 8 berikut

$\begin{array}{l}\\ 1.& \text{Tentukanlah semua titik stasioner}\\ & \text{berikut jenisnya dari fungsi}\\ & f(x)=\sin x+\cos x\text{dengan}\\ & 0\le x\le 2\pi \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Dengan Turunan Pertama}\\ & \text{Diketahui}\\ & f(x)=\sin x+\cos x\\ & f^{\prime }(x)=\cos x-\sin x\\ & \text{Saat}{f}^{\prime }(x)=0,\\ & f^{\prime }(x)=\cos x-\sin x=0\cos x=\sin x\\ & \cos x=\cos \left({\displaystyle \frac{\pi }{2}-x}\right)\\ & x=\pm \left({\displaystyle \frac{\pi }{2}-x}\right)+k.2\pi \\ & \{\begin{array}{ll}x+x& ={\displaystyle \frac{\pi }{2}+k.2\pi ,\text{atau}}\\ x-x& =-{\displaystyle \frac{\pi }{2}+k.2\pi }\end{array}\\ & \text{maka}\\ & \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{4}+k.\pi ,\text{atau}}\\ 0& =-{\displaystyle \frac{\pi }{2}+k.2\pi (\text{tidak memenuhi})}\end{array}\\ & \text{Sehingga ada dua absis yang memenuhi}\\ & \text{sebagai titik STASIONER},\text{yaitu}\\ & x={\displaystyle \frac{\pi }{4}\text{dan}x=\frac{5\pi }{4}}\\ & \text{untuk}x={\displaystyle \frac{\pi }{4}}\\ & f\left({\displaystyle \frac{\pi }{4}}\right)=\sin \left({\displaystyle \frac{\pi }{4}}\right)-\cos \left({\displaystyle \frac{\pi }{4}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}=\sqrt{2}}\\ & \text{untuk}x={\displaystyle \frac{5\pi }{4}}\\ & f\left({\displaystyle \frac{5\pi }{4}}\right)=\sin \left({\displaystyle \frac{5\pi }{4}}\right)+\cos \left({\displaystyle \frac{5\pi }{4}}\right)\\ & =-{\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}}\\ & \text{Jadi titik stasionernya}:\left({\displaystyle \frac{\pi }{4},2}\right)\mathrm{\&}\left({\displaystyle \frac{5\pi }{4},-\sqrt{2}}\right)\\ & \text{Langkah berikutnya gunakanlah titik}\\ & \text{uji di sekitar nilai stasioner yaitu}:\\ & \begin{array}{ccccccccc}& & & & & & & & \\ 0& & {\displaystyle \frac{\pi }{4}}& & \pi & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Selanjutnya}\\ & \text{Untuk}{f}^{\prime }(x)=\cos x-\sin x\\ & x=0\Rightarrow f^{\prime }(0)=\cos 0-\sin 0\\ & =1+0=1>0(\text{positif})\\ & x=\pi \Rightarrow f^{\prime }(\pi )=\cos \pi -\sin \pi \\ & =-1+0=-1<0(\text{negatif})\\ & x=0\Rightarrow f^{\prime }(2\pi )=\cos 2\pi -\sin 2\pi \\ & =1+0=1>0(\text{positif})\\ & \begin{array}{|cccccl|}\hline x& 0& {\displaystyle \frac{\pi }{4}}& \pi & {\displaystyle \frac{5\pi }{4}}& 2\pi \\ f^{\prime }(x)& +& 0& -& 0& +\\ & & --& & & \\ \text{Garfik}& /& & \mathrm{\setminus }& & /\\ & & & & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}& \\ \hline\end{array}\\ & \text{Dari tabel di atas didapatkan}\\ & \{\begin{array}{ll}\left({\displaystyle \frac{\pi }{4},\sqrt{2}}\right)& \text{titik balik maksimum}\\ \left({\displaystyle \frac{5\pi }{4},-\sqrt{2}}\right)& \text{titik balik minimum}\end{array}\end{array}\end{array}$

$.\begin{array}{|c|}\hline \begin{array}{rl}& \text{Dengan Turunan Kedua}\\ & f(x)=\sin x+\cos x\\ & f^{\prime }(x)=\cos x-\sin x\\ & f^{″}(x)=-\sin x-\cos x=-(\sin x+\cos x)\\ & f^{″}\left({\displaystyle \frac{\pi }{4}}\right)=-\left(\sin {\displaystyle \frac{\pi }{4}+\cos {\displaystyle \frac{\pi }{4}}}\right)=-\sqrt{2}<0\\ & \Rightarrow (\text{maksimum atau cekung ke bawah})\\ & f^{″}\left({\displaystyle \frac{5\pi }{4}}\right)=-\left(\sin {\displaystyle \frac{5\pi }{4}+\cos {\displaystyle \frac{5\pi }{4}}}\right)=\sqrt{2}>0\\ & \Rightarrow (\text{minimum atau cekung ke atas})\\ & \text{Dengan}f(x)=\sin x+\cos x,\text{maka}\\ & \bullet \text{nilai maksimumnya}:\sin {\displaystyle \frac{\pi }{4}+\cos \frac{\pi }{4}=\sqrt{2}}\\ & \bullet \text{nilai minimumnya}:\sin {\displaystyle \frac{5\pi }{4}+\cos \frac{5\pi }{4}=-\sqrt{2}}\\ & \text{Jadi, titik maksimumnya}\left({\displaystyle \frac{\pi }{4},\sqrt{2}}\right)\\ & \text{dan nilai minimumnya}\left({\displaystyle \frac{5\pi }{4},-\sqrt{2}}\right)\end{array}\\ \hline\end{array}$

$.\begin{array}{rl}& \text{Untuk TITIK BELOK}\\ & \text{Syarat titik belok adalah}{f}^{″}(x)=0\\ & \text{Diketahui}f(x)=\sin x+\cos x\\ & f^{″}(x)=-(\sin x+\cos x)=0\\ & \iff \sin x+\cos x=0\iff \sin x=-\cos x\\ & \iff {\displaystyle \frac{\sin x}{\cos x}=-1\iff \tan x=-1}\\ & \iff \tan x=\tan {135}^{\circ }\\ & \iff x={135}^{\circ }+k{.180}^{\circ }\\ & \iff k=0\Rightarrow x={135}^{\circ }={\displaystyle \frac{3\pi }{4}}\\ & \iff k=1\Rightarrow x={135}^{\circ }+{180}^{\circ }={315}^{\circ }={\displaystyle \frac{7\pi }{4}}\\ & \text{Adapun titik beloknya pada fungsi}f(x)\\ & \text{adalah}:\\ & \bullet x={\displaystyle \frac{3\pi }{4}}\\ & f\left({\displaystyle \frac{3\pi }{4}}\right)=\sin \left({\displaystyle \frac{3\pi }{4}}\right)+\cos \left({\displaystyle \frac{3\pi }{4}}\right)=0\\ & \text{maka titiknya}\left({\displaystyle \frac{3\pi }{4},0}\right)\\ & \bullet x={\displaystyle \frac{7\pi }{4}}\\ & f\left({\displaystyle \frac{7\pi }{4}}\right)=\sin \left({\displaystyle \frac{7\pi }{4}}\right)+\cos \left({\displaystyle \frac{7\pi }{4}}\right)=0\\ & \text{maka titiknya}\left({\displaystyle \frac{7\pi }{4},0}\right)\end{array}$

$.\text{Berikut Sketsa grafiknya}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah semua titik stasioner}\\ & \text{berikut jenisnya dari fungsi}\\ & f(x)=2\sin x\text{dengan}\\ & 0\le x\le 2\pi \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=2\sin x\\ & f^{\prime }(x)=2\cos x\\ & \text{Syarat titik stasioner}{f}^{\prime }(x)=0\\ & 2\cos x=0\iff \cos x=0\\ & \iff \cos x=\cos {90}^{\circ }\iff x={90}^{\circ }\pm k{.360}^{\circ }\\ & \iff k=0\Rightarrow x={90}^{\circ }\text{yang memenuhi}\\ & \iff k=1\Rightarrow x={270}^{\circ }\text{yang memenuhi}\\ & \text{Turunan kedua fungsi di atas adalah}:\\ & f^{″}(x)=-2\sin x\\ & \text{maka},\\ & \begin{array}{|clll|}\hline \text{Nilai}& \text{Hasil}& \text{Keterangan}& \text{Titik}\\ x={90}^{\circ }& f^{″}({90}^{\circ })=-2\sin {90}^{\circ }=-2<0& \text{Maksimum}& \\ & f({90}^{\circ })=2\sin {90}^{\circ }=2& & ({90}^{\circ },2)\\ x={270}^{\circ }& f^{″}({270}^{\circ })=-2\sin {270}^{\circ }=2>0& \text{Minimum}& \\ & f({270}^{\circ })=2\sin {270}^{\circ }=-2& & ({270}^{\circ },-2)\\ \text{Syarat}& f^{″}(x)=0& \text{Belok}& \\ & \begin{array}{rl}& -2\sin x=0\iff \sin x=0\\ & \iff \sin x=\sin 0^{\circ }\\ & \iff x=\{\begin{array}{ll}{0}^{\circ }& +k{.360}^{\circ }\\ {180}^{\circ }& +k{.360}^{\circ }\end{array}\\ & \text{Yang memenuhi}\\ & x=0^{\circ },{180}^{\circ },\text{dan}{360}^{\circ }\\ & \text{Lalu hasilnya disubstitusikan}\\ & \text{ke persamaan}f(x)=2\sin x\end{array}& \begin{array}{rl}& \\ & \text{Hasilnya}:\end{array}& \begin{array}{rl}& (0^{\circ },0),\\ & ({180}^{\circ },0),\\ & ({360}^{\circ },0)\end{array}\\ \hline\end{array}\end{array}\end{array}$

$\text{J. Selang Kecekungan}$

Lihat keterkaitan materi dan contoh di atas berkaitan dengan selang kecekungan kurva fungsi trigonometri

$\begin{array}{rl}& \text{Misalkan pada suatu selang}(a,b)\\ & \text{terdapat sembarang bilangan real}c\\ & \text{serta turunan kedua fungsi}f\text{ada}\\ & \text{pada selang tersebut}\\ & \bullet \text{saat}{f}^{″}(c)<0,\text{maka kurva}\\ & f\text{cekung ke bawah}\\ & \bullet \text{saat}{f}^{″}(c)>0,\text{maka kurva}\\ & f\text{cekung ke atas}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ & (\text{Perhatikan lagi contoh soal no.1 di atas})\\ & \text{Tentukanlah interval di mana kurva}\\ & \text{cekung ke bawah dan atas dari fungsi}\\ & f(x)=\sin x+\cos x\text{dengan}\\ & 0<x<2\pi \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& f(x)=\sin x+\cos x\\ & f^{\prime }(x)=\cos x-\sin x\\ & f^{″}(x)=-(\sin x+\cos x)\\ & \text{Sebelum menentukan batas kecekungan}\\ & \text{dengan menentukan titik beloknya dulu}\\ & \text{yaitu}:f^{″}(x)=0\\ & \text{Sebelumnya telah dibahas titik beloknya}\\ & \text{fungsi}f\text{di atas mempunyai 2 buah}\\ & \text{titik belok pada selang}0<x<2\pi \\ & x={\displaystyle \frac{3\pi }{4}\text{dan}x={\displaystyle \frac{7\pi }{4}}}\\ & \text{Melihat banyaknya titik belok, maka}\\ & \text{akan terdapat 3 selang kecekungan, yaitu}:\\ & \{\begin{array}{ll}1\bullet & 0<x<{\displaystyle \frac{3\pi }{4}}\\ 2\bullet & {\displaystyle \frac{3\pi }{4}<x<\frac{7\pi }{4}}\\ 3\bullet & {\displaystyle \frac{7\pi }{4}<x<2\pi }\end{array}\\ & \text{Kita ambil titik uji tiap selang di atas}\\ & \text{dan substitusikan ke turunan kedua fungsi}f\\ & f^{″}\left({\displaystyle \frac{\pi }{2}}\right)=-\left(\sin {\displaystyle \frac{\pi }{2}+\cos {\displaystyle \frac{\pi }{2}}}\right)=-1<0\\ & \text{Sehingga pada selang ini, kurva cekung ke bawah}\\ & f^{″}\left(\pi \right)=-(\sin \pi +\cos \pi )=1>0\\ & \text{Sehingga pada selang ini, kurva cekung ke atas}\\ & f^{″}\left({\displaystyle \frac{11\pi }{6}}\right)=-\left(\sin {\displaystyle \frac{11\pi }{6}+\cos {\displaystyle \frac{11\pi }{6}}}\right)={\displaystyle \frac{1}{2}-\frac{1}{2}\sqrt{3}<0}\\ & \text{Sehingga pada selang ini, kurva cekung ke bawah}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Kurnia, N., dkk. 2018. Jelajah Matematika 3 SMA Kelas XII Peminatan MIPA. Bogor: YUDHISTIRA
2. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

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)

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

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

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

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

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

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

Lanjutan Materi (9) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

MENYELESAIKAN MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI TRIGONOMETRI

(APLIKASI TITIK STASIONER)

$\begin{array}{l}\\ 1.& \text{Masalah Maksimum minimum}\\ 2.& \text{Kecepatan dan percepatan}\end{array}$

Aplikasi dari titik stasioner ini yang sering muncul dalam kasus maksimum-manimum khususnya berkaitan dengan fungsi trigonometri di samping juga masalah kecepatan dan percepatan. Berikut ilustrasi contoh-contohnya

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Perhatikanlah gambar berikut}\\ & \text{Gambar di bawah menunjukkan}\\ & \text{trapesium PQRS dengan}\\ & PS=RS=QR=4cm\text{dan}\\ & \mathrm{\angle }SPQ=\mathrm{\angle }RQB=2\theta \text{radian}\\ & \text{dengan}\theta \text{sudut lancip}\end{array}$

$\begin{array}{l}\\ .& \text{a}.\text{Nyatakanlah luas trapesium}\\ & \text{dalam fungsi}\theta \\ & \text{b}.\text{Tentukanlah besar}\theta \text{agar}\\ & \text{luas trapesium maksimum}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}& \text{Luas Trapesium}\\ & ={\displaystyle \frac{1}{2}\times \text{jmlh sisi sjjr}\times \text{tinggi}}\\ & {\text{L}}_{Trapesium}=\left({\displaystyle \frac{PQ+SR}{2}}\right)\times SA\\ & {\text{L}}_T=\left({\displaystyle \frac{PA+4+BQ+4}{2}}\right)\times 4\sin 2\theta \\ & =(4\cos 2\theta +4+4\cos 2\theta +4)\times 2\sin 2\theta \\ & =(8+8\cos 2\theta )\times 4\sin \theta \cos \theta \\ & =\left(8(1+\cos 2\theta )\right)\times 4\sin \theta \cos \theta \\ & =\left(8(2{\cos }^2\theta )\right)\times 4\sin \theta \cos \theta \\ & =64\sin \theta {\cos }^3\theta \\ \text{b}& \text{Supaya luas maksimum adalah}\\ & \text{nilai stasioner fungsi luas}=0\text{yaitu}:\\ & {\text{L}}_T^{{}^{\prime }}=0\\ & \begin{array}{rl}& {\text{L}}_T=U.V\{\begin{array}{ll}U& =64\sin \theta \\ V& ={\cos }^3\theta \end{array}\\ & {\text{L}}_T^{{}^{\prime }}=U^{\prime }V+U{V}^{\prime }\\ & =64\cos \theta .{\cos }^3\theta +64\sin \theta (-3{\cos }^2\theta \sin \theta )\\ & =64{\cos }^2\theta ({\cos }^2\theta -3{\sin }^2\theta )\end{array}\\ & \text{Karena syarat luas maksimum}\\ & {\text{L}}_T^{{}^{\prime }}=0,\text{maka}\\ & 64{\cos }^2\theta ({\cos }^2\theta -3{\sin }^2\theta )=0\\ & \begin{array}{r}\\ 64{\cos }^2\theta =0& \text{V}& ({\cos }^2\theta -3{\sin }^2\theta )=0\\ \cos \theta =0& & {\cos }^2\theta =3{\sin }^2\theta \\ \theta ={\displaystyle \frac{\pi }{2}}& & {\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }={\displaystyle \frac{1}{3}}}\\ & & {\tan }^2\theta ={\displaystyle \frac{1}{3}}\\ & & \tan \theta =\sqrt{{\displaystyle \frac{1}{3}}}\\ & & \tan \theta ={\displaystyle \frac{1}{3}\sqrt{3}}\\ & & \theta ={\displaystyle \frac{\pi }{6}={30}^{\circ }}\end{array}\\ & \text{Jadi},\theta ={\displaystyle \frac{\pi }{6}}\end{array}\end{array}$

$\begin{array}{l}\\ 2& \text{Sebuah partikel bergerak mengikuti}\\ & \text{sebuah lintasan yang dinyatakan dalam}\\ & s=6\cos 3t+{\sin }^{2}t+t^2+5\text{dalam meter}\\ & \text{Jika waktu yang ditempuh dalam}t\text{detik}\\ & \text{tentukanlah kecepatan saat}t={\displaystyle \frac{\pi }{2}\text{detik}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}v={\displaystyle \frac{ds}{dt}}\\ & \text{maka},\\ & v=-18\sin 3t+2\sin t\cos t+2t\\ & \text{Kecepatan saat}t={\displaystyle \frac{\pi }{2}\text{detik}}\\ & v=-18\sin 3t+\sin 2t+2t\\ & v=-18\sin 3\left(\frac{\pi }{2}\right)+\sin 2\left(\frac{\pi }{2}\right)+2\left(\frac{\pi }{2}\right)\\ & =-18(-1)+0+\pi \\ & =18+\pi \end{array}\end{array}$

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.

 

Lanjutan Materi (8) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

 MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI TRIGONOMETRI

$\text{G. Nilai Stasioner}$

Jika fungsi  $y=f(x)$  kontinu dan diferensiabel di  $x=f^{\prime }(a)=0$ , maka fungsi tersebut mempunyai nilai stasioner di  $x=a$.

$\begin{array}{l}\\ \text{a}.& \text{Suatu fungsi memiliki nilai stasioner}\\ & \text{adalah}{f}^{\prime }(x)=0\text{untuk suatu nilai}x\\ \text{b}& \text{Jika fungsi}f(x)\text{mempunyai nilai}f(a)\\ & \text{di}x=a,\text{maka titik}(a,f(a))\text{adalah}\\ & \mathbf{\text{titik stasioner}}\end{array}$

Selanjutnya titik stasioner disebut juga dengan titik kritis atau titik ekstrim dan titik stasioner ini terbagi dalam 3 macam

• titik maksimum
• titik minimum, dan 
• titik belok

Sebagai ilustrasi pada fungsi trigonometri, perhatikanlah ilustrasi fungsi sinus berikut

$\begin{array}{l}\\ \underset{\begin{array}{c}\Downarrow \\ \overbrace{\begin{array}{ccc}\begin{array}{rl}& \text{Maksimum}\\ & \Downarrow \\ & \text{Nilai}\\ & \text{maksimum}\\ & =f(a)\\ & \text{titiknya}\\ & =(a,f(a))\\ & \text{atau}\\ & f^{″}(x)<0\\ & \text{Pada contoh di atas}\\ & \text{Titik A,C,E}\\ & \end{array}& \begin{array}{rl}& \text{Minimum}\\ & \Downarrow \\ & \text{Nilai}\\ & \text{minimum}\\ & =f(a)\\ & \text{titiknya}\\ & =(a,f(a))\\ & \text{atau}\\ & f^{″}(x)>0\\ & \text{Pada contoh di atas}\\ & \text{Titik B,D}\\ & \end{array}& \begin{array}{rl}& \text{Belok}\\ & \Downarrow \\ & \text{Nilai}\\ & \text{belok}\\ & =f(a)\\ & \text{titiknya}\\ & =(a,f(a))\\ & \text{atau}\\ & f^{″}(x)=0\\ & \text{Pada contoh di atas}\\ & \text{Titik}(-\pi ,0)\\ & (0^{\circ },0),(\pi ,0),(2\pi ,0)\end{array}\end{array}}\end{array}}{\begin{array}{c}\text{Stasioner}\\ f^{\prime }(x)=0\text{saat}x=a\end{array}}\end{array}$

Sebagai catatan bahwa, nilai maksimum dan minimum yang telah di dapatkan sampai dengan memasukkan titik ujinya adalah sebenarnya titik maksimum atau minimum LOKAL dalam selang yang diberikan. Supaya menjadi nilai maksimum atau minimum mutlak, maka nilai-nilai dari nilai stasioner ini harus dibandingkan dengan nilai-nilai FUNGSI pada titik-titik ujung intervalnya yang diberikan tersebut.

$\boxed{\text{CONTOH SOAL}}$

Pada contoh soal LANJUTAN MATERI (7) lihat di sini tentang fungsi naik fungsi turun

$\begin{array}{l}\\ 1.& \text{Tentukanlah semua titik stasioner}\\ & \text{berikut jenisnya dari fungsi}\\ & f(x)=\sin x+\cos x\text{dengan}\\ & 0\le x\le 2\pi \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin x+\cos x\\ & f^{\prime }(x)=\cos x-\sin x\\ & \text{Saat}{f}^{\prime }(x)=0,\\ & f^{\prime }(x)=\cos x-\sin x=0\cos x=\sin x\\ & \cos x=\cos \left({\displaystyle \frac{\pi }{2}-x}\right)\\ & x=\pm \left({\displaystyle \frac{\pi }{2}-x}\right)+k.2\pi \\ & \{\begin{array}{ll}x+x& ={\displaystyle \frac{\pi }{2}+k.2\pi ,\text{atau}}\\ x-x& =-{\displaystyle \frac{\pi }{2}+k.2\pi }\end{array}\\ & \text{maka}\\ & \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{4}+k.\pi ,\text{atau}}\\ 0& =-{\displaystyle \frac{\pi }{2}+k.2\pi (\text{tidak memenuhi})}\end{array}\\ & \text{Sehingga ada dua absis yang memenuhi}\\ & \text{sebagai titik STASIONER},\text{yaitu}\\ & x={\displaystyle \frac{\pi }{4}\text{dan}x=\frac{5\pi }{4}}\\ & \text{untuk}x={\displaystyle \frac{\pi }{4}}\\ & f\left({\displaystyle \frac{\pi }{4}}\right)=\sin \left({\displaystyle \frac{\pi }{4}}\right)-\cos \left({\displaystyle \frac{\pi }{4}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}=\sqrt{2}}\\ & \text{untuk}x={\displaystyle \frac{5\pi }{4}}\\ & f\left({\displaystyle \frac{5\pi }{4}}\right)=\sin \left({\displaystyle \frac{5\pi }{4}}\right)+\cos \left({\displaystyle \frac{5\pi }{4}}\right)\\ & =-{\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}}\\ & \text{Jadi titik stasionernya}:\left({\displaystyle \frac{\pi }{4},2}\right)\mathrm{\&}\left({\displaystyle \frac{5\pi }{4},-\sqrt{2}}\right)\\ & \text{Langkah berikutnya gunakanlah titik}\\ & \text{uji di sekitar nilai stasioner yaitu}:\\ & \begin{array}{ccccccccc}& & & & & & & & \\ 0& & {\displaystyle \frac{\pi }{4}}& & \pi & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Selanjutnya}\\ & \text{Untuk}{f}^{\prime }(x)=\cos x-\sin x\\ & x=0\Rightarrow f^{\prime }(0)=\cos 0-\sin 0\\ & =1+0=1>0(\text{positif})\\ & x=\pi \Rightarrow f^{\prime }(\pi )=\cos \pi -\sin \pi \\ & =-1+0=-1<0(\text{negatif})\\ & x=0\Rightarrow f^{\prime }(2\pi )=\cos 2\pi -\sin 2\pi \\ & =1+0=1>0(\text{positif})\\ & \begin{array}{|cccccl|}\hline x& 0& {\displaystyle \frac{\pi }{4}}& \pi & {\displaystyle \frac{5\pi }{4}}& 2\pi \\ f^{\prime }(x)& +& 0& -& 0& +\\ & & --& & & \\ \text{Garfik}& /& & \mathrm{\setminus }& & /\\ & & & & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}& \\ \hline\end{array}\\ & \text{Dari tabel di atas didapatkan}\\ & \{\begin{array}{ll}\left({\displaystyle \frac{\pi }{4},\sqrt{2}}\right)& \text{titik balik maksimum}\\ \left({\displaystyle \frac{5\pi }{4},-\sqrt{2}}\right)& \text{titik balik minimum}\end{array}\end{array}\end{array}$

$\begin{array}{rl}& \text{Sebagai CATATAN bahwa}:\\ & \text{Nilai ujung intervalnya adalah}:\\ & \{\begin{array}{ll}x=0& \Rightarrow f(0)=\sin 0+\cos 0=0+1=1\\ & \text{titiknya}(0,1)\\ x=2\pi & \Rightarrow f(2\pi )=\sin 2\pi +\cos 2\pi =0+1=1\\ & \text{dan titiknya}(2\pi ,1)\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah semua titik stasioner}\\ & \text{berikut jenisnya dari fungsi}\\ & f(x)=\sin 2x\text{dengan}\\ & 0\le x\le 2\pi \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin 2x\\ & f^{\prime }(x)=2\cos 2x\\ & \text{Stasioner saat}{f}^{\prime }(x)=0\\ & 2\cos 2x=0\\ & \cos 2x=0\\ & \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi }\\ & x=\pm {\displaystyle \frac{\pi }{4}+k.\pi }\\ & \text{saat}k=0,\Rightarrow x={\displaystyle \frac{\pi }{4}}\\ & \text{saat}k=1,\Rightarrow x={\displaystyle \frac{5\pi }{4}\text{dan}x={\displaystyle \frac{3\pi }{4}}}\\ & \text{saat}k=2,\Rightarrow x={\displaystyle \frac{7\pi }{4}}\\ & \text{Nilai stasionernya dari absis di atas}:\\ & \bullet f\left({\displaystyle \frac{\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{\pi }{4}}\right)=1\\ & \bullet f\left({\displaystyle \frac{3\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{3\pi }{4}}\right)=-1\\ & \bullet f\left({\displaystyle \frac{5\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{5\pi }{4}}\right)=1\\ & \bullet f\left({\displaystyle \frac{7\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{7\pi }{4}}\right)=-1\\ & \\ & \text{SILAHKAN LANJUTKAN SENDIRI}\end{array}\end{array}$

Lanjutan Materi (7) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI TRIGONOMETRI

$\text{F. Fungsi Naik dan Fungsi Turun}$

Dalam menentukan interval-interval di mana fungsi naik atau turun perhatikan dulu ilustrasi berikut ini

Fungsi di atas adalah fungsi $y=f(x)=\sin x$  untuk  $0<x<\pi$  yang memepunya sumbu simetri di  $x={\displaystyle \frac{\pi }{2}=0,5\pi }$. Semua garis singgung yang berada di sebelah kiri sumbu simetri akan mempunyai nilai positif dan semunya garis singgung yang berada di sebelah kanan sumbu simetri bernilai negatif tetapi garis singgung yang tepat pada sumbu simetri memiliki nilai nol.

Pada bahasan sebelumnya-lihat di sini-telah dijelaskan bahwa gradien suatu garis singgung seperti disinggung di atas merupakan nilai dari turunan fungsi pada titik singgung tersebut.

Perhatikanlah gambar ilustrasi berikut

Untuk  

$\bullet m=f^{\prime }(x)>0(\text{tanda positif})$

$\bullet m=f^{\prime }(x)=0$

$\bullet m=f^{\prime }(x)<0(\text{tanda negatif})$

Selanjutnya perhatikan tabel berikut

$\begin{array}{|cll|}\hline \text{Interval}& \text{Nilai}& \text{Keterangan}\\ x<{\displaystyle \frac{\pi }{2}}& f^{\prime }(x)>0& \text{fungsi}f\text{naik}\\ x={\displaystyle \frac{\pi }{2}}& f^{\prime }(x)=0& \text{tidak naik/turun}\\ x>{\displaystyle \frac{\pi }{2}}& f^{\prime }(x)<0& \text{fungsi}f\text{turun}\\ \hline\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah interval ketika fungsi}\\ & f(x)=\sin x+\cos x\text{dengan}\\ & 0<x<2\pi \\ & \text{a}.\text{naik}\\ & \text{b}.\text{turun}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin x+\cos x\\ & f^{\prime }(x)=\cos x-\sin x\\ & \text{Saat}{f}^{\prime }(x)=0,\\ & f^{\prime }(x)=\cos x-\sin x=0\cos x=\sin x\\ & \cos x=\cos \left({\displaystyle \frac{\pi }{2}-x}\right)\\ & x=\pm \left({\displaystyle \frac{\pi }{2}-x}\right)+k.2\pi \\ & \{\begin{array}{ll}x+x& ={\displaystyle \frac{\pi }{2}+k.2\pi ,\text{atau}}\\ x-x& =-{\displaystyle \frac{\pi }{2}+k.2\pi }\end{array}\\ & \text{maka}\\ & \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{4}+k.\pi ,\text{atau}}\\ 0& =-{\displaystyle \frac{\pi }{2}+k.2\pi (\text{tidak memenuhi})}\end{array}\\ & \text{Sehingga nilai}x\text{yang memenuhi}:\\ & x={\displaystyle \frac{\pi }{4}\text{dan}x={\displaystyle \frac{5}{4}\pi }}\\ & \begin{array}{c}\\ & & & & & & & \\ 0& & {\displaystyle \frac{\pi }{4}}& & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Pilih titik uji bebas, misalkan}\\ & x={\displaystyle \frac{\pi }{6},x=\frac{\pi }{3},\text{dan}x={\displaystyle \frac{3\pi }{2}}}\\ & \text{untuk}x={\displaystyle \frac{\pi }{6}}\\ & f^{\prime }(x)=\cos \left({\displaystyle \frac{\pi }{6}}\right)-\sin \left({\displaystyle \frac{\pi }{6}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}(\text{positif})}\\ & \text{untuk}x={\displaystyle \frac{\pi }{3}}\\ & f^{\prime }(x)=\cos \left({\displaystyle \frac{\pi }{3}}\right)-\sin \left({\displaystyle \frac{\pi }{3}}\right)\\ & ={\displaystyle \frac{1}{2}-\frac{1}{2}\sqrt{3}(\text{negatif})}\\ & \text{dan untuk}x={\displaystyle \frac{3\pi }{2}}\\ & f^{\prime }(x)=\cos \left({\displaystyle \frac{3\pi }{2}}\right)-\sin \left({\displaystyle \frac{3\pi }{2}}\right)\\ & =0-(-1)=1(\text{positif})\\ & \begin{array}{c}\\ & & & & & & & \\ & ++& & -& -& & ++& \\ 0& & {\displaystyle \frac{\pi }{4}}& & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Berdasarkan garis bilangan di atas}\\ & \bullet f\text{naik pada}:0<x<{\displaystyle \frac{\pi }{4}}\\ & \text{atau}{\displaystyle \frac{5\pi }{4}<x<2\pi }\\ & \bullet f\text{turun pada}:{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{5\pi }{4}}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah interval ketika fungsi}\\ & f(x)={\cos }^{2}x\text{dengan}\\ & 0^{\circ }<x<{360}^{\circ }\\ & \text{a}.\text{naik}\\ & \text{b}.\text{turun}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& f(x)={\cos }^{2}x\\ & f^{\prime }(x)=2\cos x(-\sin x)=-2\sin x\cos x\\ & =-\sin 2x\\ & \text{Saat}{f}^{\prime }(x)=0\\ & -\sin 2x=0\\ & \sin 2x=0\\ & \sin 2x=\sin 0^{\circ }\\ & \begin{array}{ll}\begin{array}{rl}2x& =0^{\circ }+k{.360}^{\circ }\\ x& =0^{\circ }+k{.180}^{\circ }\\ k& =0\Rightarrow x_1={180}^{\circ }\\ k& =1\Rightarrow x_2={360}^{\circ }\end{array}& \begin{array}{rl}2x& ={180}^{\circ }+k{.360}^{\circ }\\ x& ={90}^{\circ }+k{.180}^{\circ }\\ k& =0\Rightarrow x_3={90}^{\circ }\\ k& =1\Rightarrow x_4={270}^{\circ }\end{array}\end{array}\\ & \text{Lalau kita buat diagram nilai}{f}^{\prime }(x)\text{nya}\\ & \begin{array}{c}\\ & & & & & & & & & & & \\ & --& & ++& & & --& & & ++& \\ 0^{\circ }& & {90}^{\circ }& & & {180}^{\circ }& & & {270}^{\circ }& & {360}^{\circ }\end{array}\\ & \text{Berdasar garis bilangan di atas}\\ & (\text{untuk mengecek gunakan titik uji})\\ & \text{maka fungsi}f(x)\\ & \bullet \text{naik}{90}^{\circ }<x<{180}^{\circ }\text{dan}{270}^{\circ }<x<{360}^{\circ }\\ & \bullet \text{turun}{0}^{\circ }<x<{90}^{\circ }\text{dan}{180}^{\circ }<x<{270}^{\circ }\end{array}\end{array}$

DAFTAR PUSTAKA

1. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA