Question

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

Answer 1

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.