Jawaban:
[tex] \csc {1350}^{ \circ} - \tan{30}^{ \circ} \sin{60}^{ \circ} = \frac{1}{2} \\ [/tex]
Penjelasan dengan langkah-langkah:
nilai trigonometri untuk sudut istimewa
[tex] \sf \small \boxed{\boxed{ \begin{array}{ c |c| c |c| c |c } \underline{ \: \: \: \rm x \: \: \: } &\underline{ \: \:{0}^{\circ}\:} & \underline{ \: \:{30}^{\circ}\:} & \underline{\: \: {45}^{\circ}\:} & \underline{ \: \:{60}^{\circ}\:} & \underline{ \: \:{90}^{\circ}\:} \\ \\ \sin( \rm x ) &0& \frac{1}{2} & \frac{1}{2} \sqrt{2} & \frac{1}{2} \sqrt{3} &1\\ \\ \cos(\rm x ) &1& \frac{1}{2} \sqrt{3} & \frac{1}{2} \sqrt{2} & \frac{1}{2} & 0\\ \\ \tan(\rm x ) &0& \frac{1}{3} \sqrt{3} &1& \sqrt{3} & - \\ \\ \csc(\rm x ) & - &2& \sqrt{2} & \frac{2}{3} \sqrt{3} &1 \\ \\ \sec(\rm x ) &1& \frac{2}{3} \sqrt{3} & \sqrt{2} &2& - \\ \\ \cot(\rm x ) & - & \sqrt{3} &1& \frac{1}{3} \sqrt{3} &0 \end{array}}}[/tex]
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[tex] \csc {1350}^{ \circ} - \tan{30}^{ \circ} \sin{60}^{ \circ}[/tex]
[tex] ={ \csc }^{2} {( \cancel{(3 \times 360)} + 270)}^{ \circ} - \tan{30}^{ \circ} \sin{60}^{ \circ}[/tex]
[tex] ={ \csc }^{2} ( {360}^{ \circ} - {90}^{ \circ} )- \tan{30}^{ \circ} \sin{60}^{ \circ}[/tex]
[tex] = - { \csc }^{2} {90}^{ \circ} - \tan{30}^{ \circ} \sin{60}^{ \circ}[/tex]
[tex] = { ( - 1) }^{2} - \frac{1}{3} \sqrt{3} . \frac{1}{2} \sqrt{3}\\ [/tex]
[tex] = 1 - \frac{1}{6} .3\\[/tex]
[tex] = 1 - \frac{1}{2} \\[/tex]
[tex] = \frac{1}{2}\\ [/tex]
[tex]\large\text{$\begin{aligned}&{\rm cosec}^2\,{1350^{\circ}}-\tan30^{\circ}\sin60^{\circ}\\&=\bf\frac{1}{2}\end{aligned}$}[/tex]
Pembahasan
Trigonometri
[tex]\begin{aligned}&{\rm cosec}^2\,{1350^{\circ}}-\tan30^{\circ}\sin60^{\circ}\\&{=\ }{\rm cosec}^2\,{\left(1440^{\circ}-90^{\circ}\right)}-\frac{\sin30^{\circ}}{\cos30^{\circ}}\cdot\sin60^{\circ}\\&\quad\left[\ \begin{aligned}\cos30^{\circ}=\sin60^{\circ}\end{aligned}\right.\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{1}{\sin^2{\left(1440^{\circ}-90^{\circ}\right)}}-\frac{\sin30^{\circ}}{\cancel{\sin60^{\circ}}}\cdot\cancel{\sin60^{\circ}}\\&{=\ }\frac{1}{\sin^2{\left(4\cdot360^{\circ}-90^{\circ}\right)}}-\sin30^{\circ}\\&\quad\left[\ \begin{aligned}&\sin\left(k\cdot360^{\circ}-\alpha\right)=-\sin\alpha\\&\quad k=0,1,2,3,{\dots}\\&\Rightarrow \sin^2\left(k\cdot360^{\circ}-\alpha\right)=\sin^2\alpha\end{aligned}\right.\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{1}{\sin^2{90^{\circ}}}-\sin30^{\circ}\\&{=\ }\frac{1}{1^2}-\frac{1}{2}\\&{=\ }\boxed{\ \bf\frac{1}{2}\ }\end{aligned}[/tex]
[tex]\blacksquare[/tex]
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