Question

Find the value of the trigonometric function: $\sin \left( -\dfrac{11\pi }{3} \right)$.

Answer 1

Hint: In this question, we can use the concept that all trigonometric ratios of an angle θ and of the sum of any multiple of 2π and θ $\left( n\times 2\pi +\theta \right)$ are equal. So, here we may convert $\sin \left( -\dfrac{11\pi }{3} \right)$ into the form of $\sin \left( n\times 2\pi +\theta \right)$ where n is any integer, and then equalize it to $\sin \theta $ and get our required answer.

Complete step-by-step answer:
In this given question, we are asked to find the value of the trigonometric function: $\sin \left( -\dfrac{11\pi }{3} \right)$.
As we know, all trigonometric ratios of an angle θ and of the sum of multiples of 2π and θ $\left( n\times 2\pi +\theta \right)$ are equal. Statement……. (1.1)
So, here we can convert the given trigonometric ratio of $\sin \left( -\dfrac{11\pi }{3} \right)$ into the form of $\sin \left( n\times 2\pi +\theta \right)$ where n is any integer, that is $\sin \left( -2\times 2\pi +\dfrac{\pi }{3} \right)$, where n is -2.
It gives us $\sin \left( -\dfrac{11\pi }{3} \right)=\sin \left( -2\times 2\pi +\dfrac{\pi }{3} \right)..........(1.1)$
Now, as per statement 1.1 we can write equation 1.1 as:
$\sin \left( -\dfrac{11\pi }{3} \right)=\sin \left( -2\times 2\pi +\dfrac{\pi }{3} \right)=\sin \left( \dfrac{\pi }{3} \right)..........(1.2)$.
Now, we know that the value of $\left( \dfrac{\pi }{3} \right)$ corresponds to ${{60}^{\circ }}$.
So, $\sin \left( \dfrac{\pi }{3} \right)=\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}..........(1.3)$
Hence, from equation 1.4, we get the value of $\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}$.
So, from equation 1.2, 1.3 and 1.4, we get $\sin \left( -\dfrac{11\pi }{3} \right)=\left( \dfrac{\sqrt{3}}{2} \right)$.
Therefore, from equation 1.4 we have got our answer to the question as the value of $\sin \left( -\dfrac{11\pi }{3} \right)$ as $\left( \dfrac{\sqrt{3}}{2} \right)$.

Note: In these types of questions, we should try to convert the given angles as the sum of a multiple of $2\pi $ and an angle. This is because all trigonometric ratios are equal if we add or subtract an angle by a multiple of $2\pi $ and therefore we can solve the ratios by using the values of the trigonometric ratios of angles between 0 and $2\pi $.