Question

Find the value of
$\sin(− 19π /3 )$

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Answer 1

$\large{\textrm{\underline{Given:-}}}$

$\sin(-\dfrac{19\pi}{3})$

$\large{\textrm{\underline{To find:-}}}$

The given value.

$\large{\textrm{\underline{Solution:-}}}$

$f(x)=\sin x$ is a periodic function with the period of $2\pi$. This is because the function is defined on a circle. The radius returns to the same position after revolving $2\pi n$.

$\hookrightarrow\ \ -\dfrac{19\pi}{3}=(-4)\times 2\pi +\dfrac{5\pi }{3}$

Putting this into the function, we get the following.

$\hookrightarrow \ \ \sin\left(-\dfrac{19}{3}\pi \right)$

$\hookrightarrow \ \ =\sin \left(\dfrac{5}{3} \pi \right)\ \text{[Periodicity]}$

$\hookrightarrow \ \ =\sin\left(2\pi-\dfrac{\pi}{3} \right)$

$\hookrightarrow \ \ =\sin\left(-\dfrac{\pi}{3} \right)\ \text{[Periodicity]}$

$\hookrightarrow \ \ =-\sin\left(\dfrac{\pi }{3} \right)\ \text{[Odd Function]}$

$\hookrightarrow \ \ =-\dfrac{\sqrt{3} }{2}\ \text{(\underline{ANSWER.})}$

$\large{\textrm{\underline{Note:-}}}$

The value of the function can be found in the following method. For comparison, I put the second answer below.

Assume we have an angle in the form of $\dfrac{n}{2} \pi \pm\theta$.

➊ First, if $n$ is odd change the function as follows. If even, keep the function.

• $\sin \rightarrow \cos,\ \sin\rightarrow \cos,\ \tan \rightarrow \dfrac{1}{\tan}$

➋ Second, consider $\theta$ as an acute angle.

➌ Lastly, check if the value of the previous function is positive or negative at $\dfrac{n}{2} \pi \pm\theta$. If negative, put the negative sign. If positive, put the positive sign.

$\large{\textrm{\underline{Verification:-}}}$

For example, we have $\sin \left(\dfrac{5\pi}{3} \right)$.

We have $\dfrac{5}{3} \pi =\dfrac{4}{2} \pi-\dfrac{1}{3} \pi$ which is in the form of $\dfrac{n}{2} \pi \pm\theta$.

➊ Here, $n=4$ and $\theta=\dfrac{1}{3} \pi$. Keep the function.

➋ Second, $\theta$ is an acute angle.

➌ $\dfrac{5}{3} \pi$ lies on the 4th quadrant. The previous function, $f(x)=\sin x$ has a negative value at the quadrant. Put negative sign.

$\sin\left(\dfrac{5\pi}{3} \right)=-\sin\left(\dfrac{1}{3} \pi \right)=-\dfrac{\sqrt{3} }{2} \ \text{(\underline{ANSWER.})}$

Answer 2

${ \boxed{\boxed{\begin{array}{cc} \leadsto \bf \: Given \: : \: \: \\ \\ \bf \: \: sin( - \frac{19\pi}{3}) \\ \\ \sf \: we \: have \: to \: evaluate \: that \: \end{array}}}}$

Now,

${ \boxed{\boxed{\begin{array}{c | c} \underline{ \sf \: Process \: \: \: }& \underline{ \sf \: Explanation} \\ \\ \bf \: sin( - \frac{19\pi}{3} )& \\ \\ \bf = - sin( \frac{19\pi}{3} )& \{\bf \because \: sin( - \theta) = - sin \theta \} \\ \\ \bf = - sin(6\pi + \frac{\pi}{3}) & \\ \\ = \bf - sin( \frac{\pi}{3} )& \{\bf \because \: sin(6\pi + \theta) = sin \theta \\ & \bf \: {1}^{st} \: quadrant \: \} \\ \\ \bf = - \frac{ \sqrt{3} }{2} & \{\bf \because \: sin \frac{ {\pi} }{3} = \frac{ \sqrt{3} }{2} \} \\ \\ \end{array}}}}$

$\pink{ \boxed{\boxed{\begin{array}{cc} \leadsto \: \bf \: Final \: answer \: : \\ \\ \to \bf \: sin( - \frac{19\pi}{3}) = - \frac{ \sqrt{3} }{2} \end{array}}}}$