Answer:
[tex]sin 240^\circ = - \dfrac{\sqrt{3}}{2}[/tex]
Step-by-step explanation:
We are given that:
[tex]sin 60^\circ = \dfrac{\sqrt{3}}{2}[/tex]
We need to find
[tex]sin 240^\circ = ?[/tex]
240 is greater than 180 and lesser than 270
OR
[tex]180^\circ<240^\circ<270^\circ[/tex]
So, Angle [tex]240 ^\circ[/tex] lies in the 3rd quadrant and it is well known that value of sine in 3rd quadrant is negative.
Using the property :
[tex]sin(\pi + \theta) = -sin\theta[/tex] or
[tex]sin(180^\circ + \theta) = -sin\theta[/tex]
Here, [tex]\theta\ is\ 60^\circ[/tex].
[tex]sin 240^\circ =sin(180^\circ + 60^\circ) = -sin60^\circ\\\Rightarrow -\dfrac{\sqrt{3}}{2}[/tex]
Hence, the value
[tex]sin 240^\circ = - \dfrac{\sqrt{3}}{2}[/tex]
