Question

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

Answer 1

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

\[\] $\theta $	$0$	$\dfrac{\pi }{6}$	\[\dfrac{\pi }{4}\]	$\dfrac{\pi }{3}$	$\dfrac{\pi }{2}$
$\sin \theta $	$0$	\[\dfrac{1}{2}\]	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	$1$
\[\cos \theta \]	$1$	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	\[\dfrac{1}{2}\]	$0$

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.