How do you evaluate $\sin \left( {\dfrac{{2\pi }}{3}} \right)$ ?

Hint: To evaluate this value we use the identity \[\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)\] and substitute the value of x and solve using trigonometric property \[\cos \left( { - x} \right) = \cos x\], and also remember the trigonometric table of values 30 and 60 degrees of cosine and sine trigonometric functions.

Complete step-by-step answer:
The objective of the problem is to evaluate the value of $\sin \left( {\dfrac{{2\pi }}{3}} \right)$
Given $\sin \left( {\dfrac{{2\pi }}{3}} \right)$
To solve this we convert the sine trigonometric function into cosine trigonometric function by using the identity \[\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)\].
Now convert sine function into cosine function using the above mentioned formula , we get
\[\sin \left( {\dfrac{{2\pi }}{3}} \right) = \cos \left( {\dfrac{\pi }{2} - \dfrac{{2\pi }}{3}} \right)\]
On solving the above equation we get
\[ \Rightarrow \cos \left( {\dfrac{{3\pi - 4\pi }}{6}} \right) = \cos \left( { - \dfrac{\pi }{6}} \right)\]
We know that cosine of negative value is positive cosine function. That is \[\cos \left( { - x} \right) = \cos x\]
Now we get
$\cos \left( { - \dfrac{\pi }{6}} \right) = \cos \left( {\dfrac{\pi }{6}} \right)$
The trigonometric values of sine and cosine functions are as follows



From the above table we observed that the values of sine and cosine trigonometric functions are equal at 45 degrees that is at \[\dfrac{\pi }{4}\]. The value of sine function at 0 and the value of cosine function at $\dfrac{\pi }{2}$ are equal. And the value of cosine at 0 and the value of sine function at $\dfrac{\pi }{2}$ are equal. The value of the sine function at $\dfrac{\pi }{6}$ and the value of cosine function at $\dfrac{\pi }{3}$ are equal. And the value of the sine function at $\dfrac{\pi }{3}$ and cosine function of $\dfrac{\pi }{6}$ are equal.
Now by using trigonometric table and their values substitute the values in $\cos \left( {\dfrac{\pi }{6}} \right)$
 On substituting we get
$\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}$.
Therefore, the value of \[\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)\] is $\dfrac{{\sqrt 3 }}{2}$.
Thus , on evaluating $\sin \left( {\dfrac{{2\pi }}{3}} \right)$ the value is $\dfrac{{\sqrt 3 }}{2}$.

Note: \[\left( {\dfrac{\pi }{2} - x} \right)\] is called trigonometric functions of complementary angles. The sine of an angle $\theta $ is the cosine of its complementary angle. The cosine of an angle $\theta $ is the same as the sine of its complementary angle. Similarly we can pair the secant with cosecant and tangent with cotangent.