Answer:
[tex]-\dfrac{1}{2}[/tex]
Step-by-step explanation:
Given:
[tex]\sin 30^{\circ}=\dfrac{1}{2}[/tex]
Find: [tex]\sin 210^{\circ}[/tex]
I method: Use formula
[tex]\sin (\pi +\alpha)=-\sin \alpha[/tex]
Then
[tex]210^{\circ}=180^{\circ}+30^{\circ}[/tex]
and
[tex]\sin 210^{\circ}=\sin (180^{\circ}+30^{\circ})=-\sin 30^{\circ}=-\dfrac{1}{2}[/tex]
II method: Use formula
[tex]\sin(\alpha +\beta)=\sin \alpha\cos \beta+\sin \beta\cos \alpha[/tex]
Then
[tex]\sin 210^{\circ}=\sin (180^{\circ}+30^{\circ})\\ \\=\sin 180^{\circ}\cos 30^{\circ}+\sin 30^{\circ}\cos 180^{\circ}\\ \\=0\cdot \cos 30^{\circ}+\sin 30^{\circ}\cdot (-1)\\ \\=-\sin 30^{\circ}\\ \\=-\dfrac{1}{2}[/tex]
