Question

Given that sin 60 = root3/2 Find the exact value of sin 240

Answer 1

Answer:

$sin 240^\circ = - \dfrac{\sqrt{3}}{2}$

Step-by-step explanation:

We are given that:

$sin 60^\circ = \dfrac{\sqrt{3}}{2}$

We need to find

$sin 240^\circ = ?$

240 is greater than 180 and lesser than 270

OR

$180^\circ$

So, Angle $240 ^\circ$ lies in the 3rd quadrant and it is well known that value of sine in 3rd quadrant is negative.

Using the property :

$sin(\pi + \theta) = -sin\theta$ or

$sin(180^\circ + \theta) = -sin\theta$

Here, $\theta\ is\ 60^\circ$.

$sin 240^\circ =sin(180^\circ + 60^\circ) = -sin60^\circ\\\Rightarrow -\dfrac{\sqrt{3}}{2}$

Hence, the value

$sin 240^\circ = - \dfrac{\sqrt{3}}{2}$