Question

Find the value of$$\sin {120}^{\circ }$$.
A) $${\displaystyle \frac{\sqrt{3}}{2}}$$
B) $$-{\displaystyle \frac{\sqrt{3}}{2}}$$
C) $${\displaystyle \frac{1}{2}}$$
D) $$-{\displaystyle \frac{1}{2}}$$

Answer 1

Hint: Draw a unit circle and chart out the trigonometric values on each quadrant of the circle. Find the value of $$\sin {120}^{\circ }$$ or the value of $$\sin {120}^{\circ }$$can be taken from the trigonometric table. Find either sine function or cosine function.

Complete step-by-step answer:
By using a unit circle we can find the value of $$\sin {120}^{\circ }$$. Now let us draw a cartesian plane with $$x=\cos \theta$$ and $$y=\sin \theta$$.
 Let us draw the trigonometric table as well:

	sin	cos	tan	cot	sec	Cosec
0	0	1	0	N.A	1	N.A
30	$${\displaystyle \frac{1}{2}}$$	$${\displaystyle \frac{\sqrt{3}}{2}}$$	$${\displaystyle \frac{1}{\sqrt{3}}}$$	$$\sqrt{3}$$	$${\displaystyle \frac{2\sqrt{3}}{3}}$$	2
45	$${\displaystyle \frac{1}{\sqrt{2}}}$$	$${\displaystyle \frac{1}{\sqrt{2}}}$$	1	1	$$\sqrt{2}$$	$$\sqrt{2}$$
60	$${\displaystyle \frac{\sqrt{3}}{2}}$$	$${\displaystyle \frac{1}{2}}$$	$$\sqrt{3}$$	$${\displaystyle \frac{\sqrt{3}}{3}}$$	2	$${\displaystyle \frac{2\sqrt{3}}{3}}$$
90	1	0	N.A	0	N.A	1

Now let us mark their values in the unit circle.
 

Here, $$x=\cos \theta ,y=\sin \theta$$.
Eg: -$$(\cos {30}^{\circ },\sin {30}^{\circ })=(x,y)$$
$$(x,y)=({\displaystyle \frac{\sqrt{3}}{2}},{\displaystyle \frac{1}{2}})$$.
In the first quadrant, the values of $$\cos \theta$$ and $$\sin \theta$$ are positive.
In the second quadrant, the value of $$\cos \theta$$ is negative and $$\sin \theta$$ is positive.
In the third quadrant, both are negative.
In the fourth quadrant, $$\cos \theta$$ is positive and $$\sin \theta$$ is negative.
By looking into the figure, you can find that $$\sin 60=\sin 120$$.
i.e. $$\sin 60=\sin 120={\displaystyle \frac{\sqrt{3}}{2}}$$
Or if we are directly taking value from the trigonometric table, we need to find the value of $$\sin {120}^{\circ }$$ by using other angles of sin functions such as $${60}^{\circ }$$ and $$\sin {180}^{\circ }$$.
We know that $${180}^{\circ }-{60}^{\circ }={120}^{\circ }$$.
We also know that the trigonometric identity:
$$\sin (180-\theta )=\sin \theta$$.
Put, $$\theta ={120}^{\circ }$$.
$$\begin{array}{rl}& \Rightarrow \sin (180-120)=\sin {120}^{\circ }\\ & \Rightarrow \sin {60}^{\circ }=\sin {120}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}\end{array}$$
From the trigonometric table, find the value of $$\sin {60}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$$.
$$\therefore$$Value of $$\sin {120}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$$
$$\therefore$$Option (a) is the correct answer.
Note:
We can also find the value of $$\sin {120}^{\circ }$$ by using cosine function.
Using the trigonometry formula,
$$\sin (90+\theta )=\cos \theta$$
Thus to find the values of $$\sin {120}^{\circ }$$, put $$\theta ={30}^{\circ }$$
as, $${90}^{\circ }+{30}^{\circ }={120}^{\circ }$$
$$\begin{array}{rl}& \Rightarrow \sin (90+30)=\cos 30\\ & \sin {120}^{\circ }=\cos {30}^{\circ }\end{array}$$
From trigonometric table, value of $$\cos 30={\displaystyle \frac{\sqrt{3}}{2}}$$
$$\therefore \sin {120}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$$.