Question

The value of the trigonometric ratio $\sin 120{}^\circ $ is:
a) $\dfrac{1}{2}$
b) $\dfrac{\sqrt{3}}{2}$
c) $\dfrac{-1}{2}$
d) $\dfrac{-\sqrt{3}}{2}$

Answer 1

Hint: Try to simplify the expression that is given in the question using the property that $\sin \left( 180{}^\circ -\alpha \right)=\sin \alpha $ , $\sin \left( 90{}^\circ +\alpha \right)=\cos \alpha $ , and other similar formulas.

Complete step-by-step solution -
Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.

Next, let us see the graph of cosx.

Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . So, we can say that the fundamental period of the cosine function and the sine function is $2{{\pi }^{c}}=360{}^\circ $
We will now solve the expression given in the question.
$\sin 120{}^\circ $
$=\sin \left( 180{}^\circ -60{}^\circ \right)$
Now we know $\sin \left( \pi +x \right)=-\sin x\text{ and sin}\left( \pi -x \right)=\sin x$ . On putting these values in our expression, we get
$\sin 60{}^\circ $
Now, the value of $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ . Putting this in our expression, we get
$\dfrac{\sqrt{3}}{2}$
Therefore, the value of $\sin 120{}^\circ $ is equal to $\dfrac{\sqrt{3}}{2}$ . Hence, the answer to the above question is option (b).

Note: It is useful to remember the graph of the trigonometric ratios along with the signs of their values in different quadrants. For example: sine is always positive in the first and the second quadrant while negative in the other two. Also, you need to remember the properties related to complementary angles and trigonometric ratios.