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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.