Question

Find the value of $\sin {{120}^{\circ }}$?

Answer 1

Hint: To find the value of $\sin {{120}^{\circ }}$, we are going to write the angle ${{120}^{\circ }}$ as ${{180}^{\circ }}-{{60}^{\circ }}$ in $\sin {{120}^{\circ }}$. Then we are going to use the trigonometry property which states that: $\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta $. Now, the $\theta $ in $\sin \left( {{180}^{\circ }}-{{60}^{\circ }} \right)$ is equal to ${{60}^{\circ }}$. And then we will use the value of sine of ${{60}^{\circ }}$ and solve it.

Complete step by step solution:
In the above problem, we are asked to find the value of $\sin {{120}^{\circ }}$ which we are going to calculate by substituting the angle ${{120}^{\circ }}$ to ${{180}^{\circ }}-{{60}^{\circ }}$. Now, substituting the angle ${{120}^{\circ }}$ to ${{180}^{\circ }}-{{60}^{\circ }}$we get,
$\sin {{120}^{\circ }}=\sin \left( {{180}^{\circ }}-{{60}^{\circ }} \right)$
Now, we know the trigonometry identity which states that $\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta $ so applying this identity in the above sine expression. So, the angle $\theta ={{60}^{\circ }}$ and substituting this value of $\theta $ in $\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta $ we get,
$\sin \left( {{180}^{\circ }}-{{60}^{\circ }} \right)=\sin {{60}^{\circ }}$
Also, we know that the value of $\sin {{60}^{\circ }}$ is equal to $\dfrac{\sqrt{3}}{2}$. Hence, we have calculated the value of $\sin {{120}^{\circ }}$ as $\dfrac{\sqrt{3}}{2}$.

Note: The other way of solving the above problem is that in place of the angle ${{120}^{\circ }}$, you can write ${{90}^{\circ }}+{{30}^{\circ }}$ and substituting this value of angle in $\sin {{120}^{\circ }}$we get,
$\sin {{120}^{\circ }}=\sin \left( {{90}^{\circ }}+{{30}^{\circ }} \right)$
From the trigonometric identities we know that $\sin \left( {{90}^{\circ }}+\theta \right)=\cos \theta $ so using this trigonometric relation in the above expression $\sin \left( {{90}^{\circ }}+{{30}^{\circ }} \right)$ we get,
$\sin \left( {{90}^{\circ }}+{{30}^{\circ }} \right)=\cos {{30}^{\circ }}$
We know the value of trigonometric ratio of $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$. Hence, we are getting the same value which we got in the above solution.
Generally, we get confused about the signs like in which quadrant the value of trigonometric ratio is positive or negative. In the above problem, the angle is ${{120}^{\circ }}$ and we know that this angle lies in the second quadrant and we know that in the second quadrant, sine is positive. That’s why our answer is coming positive.