Aturan Sinus dan Aturan Cosinus

A. Pendahuluan

Aturan sinus maupun aturan cosinus keduanya sangat bermanfaat berkaitan dengan unsur segitiga baik siku-siku maupun segitiga bebas dalam penentuan besar sudut dalam segitiga tersebut maupun panjang sisi yang diingin. Dalam hal penentuan besar sudut atau menentukan permasalahan panjang salah satu sisi segitiga jika nantinya sudut diketahui, terkadang besar sudutnya tidak cuma lancip, dibanyak soal dimunculkan sudut tumpul. Oleh karenanya ada baiknya pembaca mengetahui nilai perbandingan trigonometri diberbagai kuadran dan nilai sudut-sudut istimewa dalam trigonometri serta tak lupa juga beberapa identitas trigonometri.

$\begin{array}{c}\sin \alpha ={\displaystyle \frac{BC}{AB}\iff \csc \alpha ={\displaystyle \frac{AB}{BC}={\displaystyle \frac{1}{\sin \alpha }}}}\\ \\ \cos \alpha ={\displaystyle \frac{AC}{AB}\iff \sec \alpha ={\displaystyle \frac{AB}{AC}={\displaystyle \frac{1}{\cos \alpha }}}}\\ \\ \tan \alpha ={\displaystyle \frac{BC}{AC}\iff \cot \alpha ={\displaystyle \frac{AC}{BC}={\displaystyle \frac{1}{\tan \alpha }}}}\end{array}$.

$\begin{array}{|ccccccccc|}\hline {\alpha }^0& 0^0& {30}^0& {45}^0& {60}^0& {90}^0& {180}^0& {270}^0& {360}^0\\ \sin {\alpha }^0& 0& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}\sqrt{3}}& 1& 0& -1& 0\\ \cos {\alpha }^0& 1& {\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}}& 0& -1& 0& 1\\ \tan {\alpha }^0& 0& {\displaystyle \frac{1}{3}\sqrt{3}}& 1& \sqrt{3}& TD& 0& TD& 0\\ \hline\end{array}$.

$\begin{array}{rl}& \text{Macam-Macam Identitas Trigonometri Dasar}\\ & 1.\csc \alpha ={\displaystyle \frac{1}{\sin \alpha }5.\tan \alpha ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}}\\ & 2.\sec \alpha ={\displaystyle \frac{1}{\cos \alpha }6.{\tan }^2\alpha +1={\sec }^2\alpha }\\ & 3.\cot \alpha ={\displaystyle \frac{1}{\tan \alpha }7.{\cot }^2\alpha +1={\csc }^2\alpha }\\ & 4.\cot \alpha ={\displaystyle \frac{\cos \alpha }{\sin \alpha }8.{\sin }^2\alpha +{\cos }^2=1}\end{array}$.

B. Aturan Sinus

$\begin{array}{|c|}\hline {\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R}\\ \hline\end{array}$.

C. Aturan Cosinus

$\begin{array}{|c|}\hline \begin{array}{rl}\bullet & \cos \mathrm{\angle }A={\displaystyle \frac{b^2+c^2-a^2}{2bc}}\\ \bullet & \cos \mathrm{\angle }B={\displaystyle \frac{a^2+c^2-a^2}{2ac}}\\ \bullet & \cos \mathrm{\angle }C={\displaystyle \frac{a^2+b^2-a^2}{2ab}}\end{array}\\ \hline\end{array}$.

D. Luas Segitiga

$\begin{array}{rl}\mathbf{\text{Luas}}\mathrm{△}ABC& =\frac{1}{2}bc.\sin \mathrm{\angle }A\\ & =\frac{1}{2}ac.\sin \mathrm{\angle }B\\ & =\frac{1}{2}ab.\sin \mathrm{\angle }C\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui}\mathrm{△}ABC\text{dengan panjang sisi}\\ & AC=10cm\text{dan}BC=16cm\text{serta luas}\\ & \mathrm{△}ABC=40c{m}^2,\text{maka besar}\mathrm{\angle }ACB\\ & \text{jika sudutnya lancip adalah}\cdots \\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}\{\begin{array}{c}AC=10cm\\ BC=16cm\\ L_{\mathrm{△}}=40c{m}^2\end{array},\text{maka}\\ & \begin{array}{rl}{L}_{\mathrm{△}ABC}& =\frac{1}{2}.AC.BC.\sin \mathrm{\angle }ACB\\ 40& =\frac{1}{2}.10.16.\sin \mathrm{\angle }ACB\\ 40& =80.\sin \mathrm{\angle }ACB\\ \frac{40}{80}& =\sin \mathrm{\angle }ACB\\ \sin \mathrm{\angle }ACB& =\frac{1}{2}\\ \sin \mathrm{\angle }ACB& =\sin {30}^0\\ \mathrm{\angle }ACB& ={30}^0\end{array}.\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Jika }\\ & AB+3=BC+2=CD+1=AD=4cm,\\ & \text{maka}\cos \mathrm{\angle }BAD\text{adalah}\cdots \\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan kembali ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \text{Langkah awal kita gunakan garis bantu BD}\\ & \text{untuk nantinya kita mendapatkan nilai}cos\\ & \text{dari sudut A, yaitu}:\\ & \begin{array}{rl}B{D}^2& =B{A}^2+D{A}^2-2.BA.DA.\cos \mathrm{\angle }A\\ & =1^2+4^2-2.1.4.\cos \mathrm{\angle }A\\ & =17-8\cos \mathrm{\angle }A\\ B{D}^2& =B{C}^2+D{C}^2-2.BC.DC.\cos \mathrm{\angle }C\\ & =2^2+3^2-2.2.3.\cos \mathrm{\angle }C\\ & =13-12\cos \mathrm{\angle }C\end{array}\\ & \text{Perlu diketahui bahwa}\\ & \mathrm{\angle }A+\mathrm{\angle }C=\mathrm{\angle }B+\mathrm{\angle }C={180}^0\\ & \text{karena ABCD segiempat talibusur, sehingga}\\ & \mathrm{\angle }C={180}^0-\mathrm{\angle }A\\ & \begin{array}{rl}B{D}^2& =B{D}^2\\ 17-8\cos \mathrm{\angle }A& =13-12\cos \mathrm{\angle }C\\ 12\cos \mathrm{\angle }C-8\cos \mathrm{\angle }A& =13-17\\ 12(\cos ({180}^0-\mathrm{\angle }A))-8\cos \mathrm{\angle }A& =-4\\ 12(-\cos \mathrm{\angle }A)-8\cos \mathrm{\angle }A& =-4\\ -12\cos \mathrm{\angle }A-8\cos \mathrm{\angle }A& =-4\\ -20\cos \mathrm{\angle }A& =-4\\ \cos \mathrm{\angle }A& =\frac{-4}{-20}\\ \cos \mathrm{\angle }A& =\frac{1}{5}\end{array}\end{array}$.

Ketidaksamaan

 Pada sebuah segitiga ABC dengan panjang sisi AB = c , AC = b,  dan BC = a, dari ketiga sisi ini maka akan berlaku pertidaksamaan umum yang melibatkan ketiga sisinya sebagai berikut:

$\begin{array}{rl}& a+b>c\\ & a+c>b\\ & b+c>a\end{array}$

Identitas-Identitas Aljabar yang Menakjubkan

 Banyak sekali keunikan-keunikan saat kita mencoba melihat identitas-identitas aljabar yang sudah ditemukan sampai saat ini. Tentu semuanya sangat membantu ketika kita menyelesaikan suatu problem yang mengarah ke sana. Kadang sebagian ada yang menyebutkan dengan manipulasi aljabar.

Berikut bentuk dasar dari identitas-identitas aljabar tersebut

$\begin{array}{rl}& a^2-b^2=(a-b)(a+b)\\ & a^3+b^3=(a+b)(a^2-ab+b^2)\\ & a^3-b^3=(a-b)(a^2+ab+b^2)\\ & (a+b{)}^2=a^2+2ab+b^2\\ & (a-b{)}^2=a^2-2ab+b^2\\ & (a+b{)}^3=a^3+b^3+3ab(a+b)\\ & (a-b{)}^3=a^3-b^3-3ab(a-b)\\ & (a+b+c{)}^2=a^2+b^2+c^2+2(ab+ac+bc)\\ & (a+b+c{)}^3=a^3+b^3+c^3+3(a+b)(a+c)(b+c)\\ & a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)\\ & a^3+b^3+c^3-3abc={\displaystyle \frac{1}{2}(a+b+c)((a-b{)}^2+(a-c{)}^2+(b-c{)}^2)}\\ & abc=(a+b+c)(ab+ac+bc)-(a+b)(a+c)(b+c)\\ & \mathit{\text{Sophie Germain}}:a^4+4b^4=(a^2-2ab+2b^2)(a^2+2ab+2b^2)\end{array}$

Contoh Soal 13 Turunan Fungsi Trigonometri (Bagian 3)

 $\begin{array}{l}\\ 61.& \text{(UM UNBRAW)}\\ & \text{Nilai maksimum dari fungsi}\\ & f(x)=4{\cos }^{2}x+14{\sin }^{2}x+24\sin x\cos x+10\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 6\\ \text{b}.& 24\\ \text{c}.& 26\\ \text{d}.& 32\\ \text{e}.& 92\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)=4{\cos }^{2}x+14{\sin }^{2}x+24\sin x\cos x+10\\ & f(x)=4{\cos }^{2}x+4{\sin }^{2}x+10{\sin }^{2}x+12\sin 2x+10\\ & f(x)=4+5(1-\cos 2x)+12\sin 2x+10\\ & f(x)=19-5\cos 2x+12\sin 2x\\ & f(x)=19+12\sin 2x-5\cos 2x\\ & f(x)=19+\sqrt{{12}^2+(-5{)}^2}\cos (2x-\theta )\\ & f(x)=19+13\cos (2x-\theta )\\ & \text{Karena nilai}\cos (2x-\theta )=\pm 1,\text{maka}\\ & f(x{)}_{maks}=19+13=32\end{array}\end{array}$

Contoh Soal 12 Turunan Fungsi Trigonometri (Bagian 3)

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

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

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

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

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

DAFTAR PUSTAKA

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

Contoh Soal 11 Turunan Fungsi Trigonometri (Bagian 3)

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

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

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

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

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

Contoh Soal 10 Turunan Fungsi Trigonometri (Bagian 3)

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

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

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

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

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

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}$