Question

Find the value of trigonometric expression $\sin {{3060}^{\circ }}$.

Answer 1

Hint:Convert the given trigonometric expression to acute angle form i.e. use trigonometric relation to convert the angle ${{360}^{\circ }}$ to acute angle (${{0}^{\circ }}$ to ${{90}^{\circ }}$). Observe ${{3060}^{\circ }}$ as multiple of angle ${{180}^{\circ }}$. Value of $\sin \left( n\pi +\theta \right)=\sin \theta$, if n is an even integer and $\sin \left( n\pi +\theta \right)=-\sin \theta$, if n is an odd integer. Use this relation to get the value of $\sin {{3060}^{\circ }}$ and value of $\sin {{0}^{\circ }}=0$.

Complete step-by-step answer:
As we know there are certain identities with the help of which we can convert one trigonometric function to another by changing its angle.
There are mainly two ways for conversion: -
(i) Angles written in sum with multiple of $\dfrac{\pi }{2}$. (not multiple of $\pi$).
(ii) Angles written in sum with multiple of $\pi$.
So, we do not change the trigonometric function while changing the angle with the second angle with the second case but trigonometric function will change while using the ($\sin \rightleftarrows \cos$ and $\tan \rightleftarrows \cot$ and $\sec \rightleftarrows \csc$). A sign while converting the term to another expression will be changed only on the basis of the angle it is lying.
We know the expression $\sin \left( n\pi +\theta \right)$ can be evaluated with the help of above fundamental properties for conversion, as
$\sin \left( n\pi +\theta \right)=\sin \theta$, if n is even.
$\sin \left( n\pi +\theta \right)=-\sin \theta$, if n is odd.
Where, we can observe that $\left( n\pi +\theta \right)$ will lie in first quadrant, if n is an even number and $n\pi +\theta$ will lie in ${{3}^{rd}}$ quadrant for n as odd number and sign in front of $\sin \theta$ has been determined only on the basis of quadrant. (in which angle is lying)
So, let us use the above concept to get the value of $\sin {{3060}^{\circ }}$.
As the angle ${{3060}^{\circ }}$ is higher than ${{180}^{\circ }}$. So, we need to get the multiple of ${{180}^{\circ }}$, which is just less than ${{3060}^{\circ }}$ or equal to ${{3060}^{\circ }}$.
So, let us divide ${{3060}^{\circ }}$ by ${{180}^{\circ }}$ as,

& 3060 \\\ & \underline{180} \\\ & 1260 \\\ & \underline{1260} \\\ & \underline{0} \\\ \end{aligned}}\right.}}$$ It means, we can write $$\sin {{3060}^{\circ }}$$ as, $$\begin{aligned} & \sin {{3060}^{\circ }}=\sin \left( 180\times 17+{{0}^{\circ }} \right) \\\ & \sin {{3060}^{\circ }}=\sin \left( 17\pi+{{0}^{\circ }} \right) \\\ \end{aligned}$$ Now, as $$17\pi $$ is an odd multiple of $$\pi $$, it means angle $$\left( 17\pi +\theta \right)$$ will lie in $${{3}^{rd}}$$ quadrant. So, the trigonometric function will not change as the angle is written in the sum of multiple of $$\pi $$. But the sign of it will change because the sine function is negative in the third quadrant. Hence, $$\sin \left( 17\pi +{{0}^{\circ }} \right)$$ can be written as, $$\sin \left( 17\pi +{{0}^{\circ }} \right)=-\sin {{0}^{\circ }}$$ We know the value of $$\sin {{0}^{\circ }}$$ is 0. So, the value of $$\sin \left( 17\pi +{{0}^{\circ }} \right)$$ is 0. Hence, we get, $$\sin {{3060}^{\circ }}=0$$. Note: One may go wrong, if he / she does not know the conversion of higher angles to acute angles. Here, as the angle written with $$17\pi $$ is $${{0}^{\circ }}$$. So, one need not to take care of signs but with any other angle, signs will play an important role.We should remember signs of trigonometric functions in all quadrants.In first quadrant all trigonometric functions are positive ,in 2nd quadrant sine and cosec functions are positive ,in 3rd quadrant Tan and cot functions are positive and in 4th quadrant cos and secant functions are positive.Students have to remember this important concept for solving these types of questions,