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Q...Q Eutopya 100 poin​

Berikut ini adalah pertanyaan dari unknown pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Q...Q Eutopya 100 poin
​
Q...Q Eutopya 100 poin​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawaban: (pada gambar)

PEMBAHASAN: Trigonometri

Nomor 1.

\begin{aligned}&\sin330^{\circ}+2\cdot\cos240^{\circ}-\sin210^{\circ}\\&{=\ }\sin(360^{\circ}-30^{\circ})\\&{\quad}+2\cdot\cos(180^{\circ}+60^{\circ})\\&{\quad}-\sin(180^{\circ}+30^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \sin(360^{\circ}-{x})=-\sin{x}\\&\bullet\ \cos(180^{\circ}+{x})=-\cos{x}\\&\bullet\ \sin(180^{\circ}+{x})=-\sin{x}\end{aligned}\right.\end{aligned}
\begin{aligned}&{=\ }\cancel{-\sin30^{\circ}}+2(-\cos60^{\circ})-\cancel{(-\sin30^{\circ})}\\&{=\ }-2\cos60^{\circ}\\&{=\ }-2\cdot\frac{1}{2}\\&{=\ }\boxed{\ \bf{-}1\ }\end{aligned}

Nomor 2.

\begin{aligned}&\sin(-120^{\circ})+\cos225^{\circ}-\cos(-30^{\circ})\\&{=\ }\sin(-120^{\circ})+\cos(180^{\circ}+45^{\circ})-\cos(-30^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \sin(-{x})=-\sin{x}\\&\bullet\ \cos(-{x})=\cos{x}\\&\bullet\ \cos(180^{\circ}+{x})=-\cos{x}\\\end{aligned}\right.\end{aligned}
\begin{aligned}&{=\ }{-}\sin120^{\circ}+(-\cos45^{\circ})-\cos30^{\circ}\\&{=\ }{-}\sin(90^{\circ}+30^{\circ})-\cos45^{\circ}-\cos30^{\circ}\\&\ \left[\ \begin{aligned}&\bullet\ \sin(90^{\circ}+{x})=\cos{x}\\\end{aligned}\right.\\&{=\ }{-}\cos30^{\circ}-\cos45^{\circ}-\cos30^{\circ}\\&{=\ }{-}2\cos30^{\circ}-\cos45^{\circ}\\&{=\ }{-}2\cdot\frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}\\&{=\ }\boxed{\ \bf-\frac{2\sqrt{3}+\sqrt{2}}{2}\ }\end{aligned}

Nomor 3.

\begin{aligned}&\sin330^{\circ}+\cos660^{\circ}-\tan(-210^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \tan(-{x})=-\tan{x}\\\end{aligned}\right.\\&{=\ }\sin(360^{\circ}-30^{\circ})\\&{\quad}+\cos(2\cdot360^{\circ}-60^{\circ})\\&{\quad}-(-\tan210^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \sin(k\cdot360^{\circ}-{x})=-\sin{x}\\&\bullet\ \cos(k\cdot360^{\circ}-{x})=\cos{x}\\\end{aligned}\right.\end{aligned}
\begin{aligned}&{=\ }-\sin30^{\circ}+\cos60^{\circ}\\&{\quad}+\tan(180^{\circ}+30^{\circ})\\&{=\ }-\sin30^{\circ}+\cos(90^{\circ}-30^{\circ})\\&{\quad}+\tan(180^{\circ}+30^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \cos(90^{\circ}-{x})=\sin{x}\\&\bullet\ \tan(180^{\circ}+{x})=\tan{x}\\\end{aligned}\right.\end{aligned}
\begin{aligned}&{=\ }\cancel{-\sin30^{\circ}}+\cancel{\sin30^{\circ}}+\tan30^{\circ}\\&{=\ }\tan30^{\circ}\ =\ \frac{\sin30^{\circ}}{\cos30^{\circ}}\\&{=\ }\frac{\cancel{\frac{1}{2}}}{\cancel{\frac{1}{2}}\sqrt{3}}\ =\ \frac{1}{\sqrt{3}}\\&{=\ }\boxed{\ \bf\frac{1}{3}\sqrt{3}\ }\end{aligned}

Nomor 4.

\begin{aligned}&\frac{\sin270^{\circ}\cdot\cos135^{\circ}\cdot\tan135^{\circ}}{\sin150^{\circ}\cdot\cos225^{\circ}}\\&\ \left[\ \begin{aligned}&\bullet\ \cos{x}\cdot\tan{x}=\sin{x}\end{aligned}\right.\\&{=\ }\frac{\sin270^{\circ}\cdot\sin135^{\circ}}{\sin150^{\circ}\cdot\cos(90^{\circ}+135^{\circ})}\\&\ \left[\ \begin{aligned}&\bullet\ \cos(90^{\circ}+{x})=-\sin{x}\\\end{aligned}\right.\end{aligned}
\begin{aligned}&{=\ }\frac{\sin270^{\circ}\cdot\sin135^{\circ}}{\sin150^{\circ}\cdot(-\sin135^{\circ})}\\&{=\ }-\frac{\sin270^{\circ}\cdot\cancel{\sin135^{\circ}}}{\sin150^{\circ}\cdot\cancel{\sin135^{\circ}}}\\&{=\ }-\frac{\sin(360^{\circ}-90^{\circ})}{\sin(180^{\circ}-30^{\circ})}\\&{=\ }-\frac{-\sin90^{\circ}}{\sin30^{\circ}}\\&{=\ }-\frac{-1}{\frac{1}{2}}\\&{=\ }\boxed{\ \bf2\ }\end{aligned}

Nomor 5.

\begin{aligned}&\frac{{\rm cosec\,}135^{\circ}\cdot\sec225^{\circ}\cdot\tan405^{\circ}}{\sin{60}^{\circ}\cdot\cos{150}^{\circ}\cdot\tan(-{60}^{\circ})}\\&{=\ }\frac{{\rm cosec\,}(90^{\circ}+45^{\circ})\cdot\sec(180^{\circ}+45^{\circ})\cdot\tan(360^{\circ}+45^{\circ})}{\sin{60}^{\circ}\cdot\cos({90}^{\circ}+{60}^{\circ})\cdot\tan(-{60}^{\circ})}\end{aligned}
\begin{aligned}&\ \left[\ \begin{aligned}&\bullet\ {\rm cosec\,}(90^{\circ}+{x})=\sec{x}\\&\bullet\ \sec(180^{\circ}+{x})=-\sec{x}\\&\bullet\ \tan(k\cdot360^{\circ}+{x})=\tan{x}\\&\bullet\ \cos(90^{\circ}+{x})=-\sin{x}\\&\bullet\ \tan(-{x})=-\tan{x}\end{aligned}\right.\\&{=\ }\frac{\sec45^{\circ}\cdot(-\sec45^{\circ})\cdot\tan45^{\circ}}{\sin{60}^{\circ}\cdot(-\sin{60}^{\circ})\cdot(-\tan{60}^{\circ})}\end{aligned}
\begin{aligned}&{=\ }-\frac{\sec^2{45}^{\circ}\cdot\tan45^{\circ}}{\sin^2{60}^{\circ}\cdot\tan{60}^{\circ}}\\&{=\ }-\sec^2{45}^{\circ}\cdot\tan45^{\circ}\cdot{\rm cosec}^2\,{60}^{\circ}\cdot\cot{60}^{\circ}\\&\ \left[\ \begin{aligned}&\bullet\ \sec^2{x}=1+\tan^2{x}\\&\bullet\ {\rm cosec}^2\,{x}=1+\cot^2{x}\end{aligned}\right.\\&{=\ }-\left(1+\tan^2{45}^{\circ}\right)\tan45^{\circ}\cdot\left(1+\cot^2{60}^{\circ}\right)\cot{60}^{\circ}\end{aligned}
\begin{aligned}&\ \left[\ \begin{aligned}&\bullet\ \tan45^{\circ}=1\\&\bullet\ \cot{60}^{\circ}=\frac{\sqrt{3}}{3}\end{aligned}\right.\\&{=\ }-\left(1+1^2\right)\cdot1\cdot\left(1+\left(\frac{\sqrt{3}}{3}\right)^2\right)\frac{\sqrt{3}}{3}\\&{=\ }-2\left(1+\frac{1}{3}\right)\frac{\sqrt{3}}{3}\\&{=\ }-2\cdot\frac{4}{3}\cdot\frac{\sqrt{3}}{3}\\&{=\ }\boxed{\ \bf-\frac{8}{9}\sqrt{3}\ }\end{aligned}

 

Jawaban: (pada gambar)PEMBAHASAN: TrigonometriNomor 1.[tex]\begin{aligned}&\sin330^{\circ}+2\cdot\cos240^{\circ}-\sin210^{\circ}\\&{=\ }\sin(360^{\circ}-30^{\circ})\\&{\quad}+2\cdot\cos(180^{\circ}+60^{\circ})\\&{\quad}-\sin(180^{\circ}+30^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \sin(360^{\circ}-{x})=-\sin{x}\\&\bullet\ \cos(180^{\circ}+{x})=-\cos{x}\\&\bullet\ \sin(180^{\circ}+{x})=-\sin{x}\end{aligned}\right.\end{aligned}[/tex][tex]\begin{aligned}&{=\ }\cancel{-\sin30^{\circ}}+2(-\cos60^{\circ})-\cancel{(-\sin30^{\circ})}\\&{=\ }-2\cos60^{\circ}\\&{=\ }-2\cdot\frac{1}{2}\\&{=\ }\boxed{\ \bf{-}1\ }\end{aligned}[/tex]Nomor 2.[tex]\begin{aligned}&\sin(-120^{\circ})+\cos225^{\circ}-\cos(-30^{\circ})\\&{=\ }\sin(-120^{\circ})+\cos(180^{\circ}+45^{\circ})-\cos(-30^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \sin(-{x})=-\sin{x}\\&\bullet\ \cos(-{x})=\cos{x}\\&\bullet\ \cos(180^{\circ}+{x})=-\cos{x}\\\end{aligned}\right.\end{aligned}[/tex][tex]\begin{aligned}&{=\ }{-}\sin120^{\circ}+(-\cos45^{\circ})-\cos30^{\circ}\\&{=\ }{-}\sin(90^{\circ}+30^{\circ})-\cos45^{\circ}-\cos30^{\circ}\\&\ \left[\ \begin{aligned}&\bullet\ \sin(90^{\circ}+{x})=\cos{x}\\\end{aligned}\right.\\&{=\ }{-}\cos30^{\circ}-\cos45^{\circ}-\cos30^{\circ}\\&{=\ }{-}2\cos30^{\circ}-\cos45^{\circ}\\&{=\ }{-}2\cdot\frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}\\&{=\ }\boxed{\ \bf-\frac{2\sqrt{3}+\sqrt{2}}{2}\ }\end{aligned}[/tex]Nomor 3.[tex]\begin{aligned}&\sin330^{\circ}+\cos660^{\circ}-\tan(-210^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \tan(-{x})=-\tan{x}\\\end{aligned}\right.\\&{=\ }\sin(360^{\circ}-30^{\circ})\\&{\quad}+\cos(2\cdot360^{\circ}-60^{\circ})\\&{\quad}-(-\tan210^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \sin(k\cdot360^{\circ}-{x})=-\sin{x}\\&\bullet\ \cos(k\cdot360^{\circ}-{x})=\cos{x}\\\end{aligned}\right.\end{aligned}[/tex][tex]\begin{aligned}&{=\ }-\sin30^{\circ}+\cos60^{\circ}\\&{\quad}+\tan(180^{\circ}+30^{\circ})\\&{=\ }-\sin30^{\circ}+\cos(90^{\circ}-30^{\circ})\\&{\quad}+\tan(180^{\circ}+30^{\circ})\\&\ \left[\ \begin{aligned}&\bullet\ \cos(90^{\circ}-{x})=\sin{x}\\&\bullet\ \tan(180^{\circ}+{x})=\tan{x}\\\end{aligned}\right.\end{aligned}[/tex][tex]\begin{aligned}&{=\ }\cancel{-\sin30^{\circ}}+\cancel{\sin30^{\circ}}+\tan30^{\circ}\\&{=\ }\tan30^{\circ}\ =\ \frac{\sin30^{\circ}}{\cos30^{\circ}}\\&{=\ }\frac{\cancel{\frac{1}{2}}}{\cancel{\frac{1}{2}}\sqrt{3}}\ =\ \frac{1}{\sqrt{3}}\\&{=\ }\boxed{\ \bf\frac{1}{3}\sqrt{3}\ }\end{aligned}[/tex]Nomor 4.[tex]\begin{aligned}&\frac{\sin270^{\circ}\cdot\cos135^{\circ}\cdot\tan135^{\circ}}{\sin150^{\circ}\cdot\cos225^{\circ}}\\&\ \left[\ \begin{aligned}&\bullet\ \cos{x}\cdot\tan{x}=\sin{x}\end{aligned}\right.\\&{=\ }\frac{\sin270^{\circ}\cdot\sin135^{\circ}}{\sin150^{\circ}\cdot\cos(90^{\circ}+135^{\circ})}\\&\ \left[\ \begin{aligned}&\bullet\ \cos(90^{\circ}+{x})=-\sin{x}\\\end{aligned}\right.\end{aligned}[/tex][tex]\begin{aligned}&{=\ }\frac{\sin270^{\circ}\cdot\sin135^{\circ}}{\sin150^{\circ}\cdot(-\sin135^{\circ})}\\&{=\ }-\frac{\sin270^{\circ}\cdot\cancel{\sin135^{\circ}}}{\sin150^{\circ}\cdot\cancel{\sin135^{\circ}}}\\&{=\ }-\frac{\sin(360^{\circ}-90^{\circ})}{\sin(180^{\circ}-30^{\circ})}\\&{=\ }-\frac{-\sin90^{\circ}}{\sin30^{\circ}}\\&{=\ }-\frac{-1}{\frac{1}{2}}\\&{=\ }\boxed{\ \bf2\ }\end{aligned}[/tex]Nomor 5.[tex]\begin{aligned}&\frac{{\rm cosec\,}135^{\circ}\cdot\sec225^{\circ}\cdot\tan405^{\circ}}{\sin{60}^{\circ}\cdot\cos{150}^{\circ}\cdot\tan(-{60}^{\circ})}\\&{=\ }\frac{{\rm cosec\,}(90^{\circ}+45^{\circ})\cdot\sec(180^{\circ}+45^{\circ})\cdot\tan(360^{\circ}+45^{\circ})}{\sin{60}^{\circ}\cdot\cos({90}^{\circ}+{60}^{\circ})\cdot\tan(-{60}^{\circ})}\end{aligned}[/tex][tex]\begin{aligned}&\ \left[\ \begin{aligned}&\bullet\ {\rm cosec\,}(90^{\circ}+{x})=\sec{x}\\&\bullet\ \sec(180^{\circ}+{x})=-\sec{x}\\&\bullet\ \tan(k\cdot360^{\circ}+{x})=\tan{x}\\&\bullet\ \cos(90^{\circ}+{x})=-\sin{x}\\&\bullet\ \tan(-{x})=-\tan{x}\end{aligned}\right.\\&{=\ }\frac{\sec45^{\circ}\cdot(-\sec45^{\circ})\cdot\tan45^{\circ}}{\sin{60}^{\circ}\cdot(-\sin{60}^{\circ})\cdot(-\tan{60}^{\circ})}\end{aligned}[/tex][tex]\begin{aligned}&{=\ }-\frac{\sec^2{45}^{\circ}\cdot\tan45^{\circ}}{\sin^2{60}^{\circ}\cdot\tan{60}^{\circ}}\\&{=\ }-\sec^2{45}^{\circ}\cdot\tan45^{\circ}\cdot{\rm cosec}^2\,{60}^{\circ}\cdot\cot{60}^{\circ}\\&\ \left[\ \begin{aligned}&\bullet\ \sec^2{x}=1+\tan^2{x}\\&\bullet\ {\rm cosec}^2\,{x}=1+\cot^2{x}\end{aligned}\right.\\&{=\ }-\left(1+\tan^2{45}^{\circ}\right)\tan45^{\circ}\cdot\left(1+\cot^2{60}^{\circ}\right)\cot{60}^{\circ}\end{aligned}[/tex][tex]\begin{aligned}&\ \left[\ \begin{aligned}&\bullet\ \tan45^{\circ}=1\\&\bullet\ \cot{60}^{\circ}=\frac{\sqrt{3}}{3}\end{aligned}\right.\\&{=\ }-\left(1+1^2\right)\cdot1\cdot\left(1+\left(\frac{\sqrt{3}}{3}\right)^2\right)\frac{\sqrt{3}}{3}\\&{=\ }-2\left(1+\frac{1}{3}\right)\frac{\sqrt{3}}{3}\\&{=\ }-2\cdot\frac{4}{3}\cdot\frac{\sqrt{3}}{3}\\&{=\ }\boxed{\ \bf-\frac{8}{9}\sqrt{3}\ }\end{aligned}[/tex] 

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Last Update: Sat, 13 Aug 22
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