Question

How to find the value of $\sin \left({300}^{°}\right)$?

Answer 1

Find the value of given trigonometric function

Given, $\sin \left({300}^{°}\right)$

We know that

$\sin \left(\alpha -\beta \right)=\sin \left(\alpha \right)\cos \left(\beta \right)-\cos \left(\alpha \right)\sin \left(\beta \right)$

Now by substituting the value of $\alpha ={360}^{°}$ and $\beta ={60}^{°}$in the trigonometric identity we get

$$\begin{gathered} \sin \left({360}^{°}-{60}^{°}\right)=\sin \left({360}^{°}\right)\cos \left({60}^{°}\right)-\cos \left({360}^{°}\right)\sin \left({60}^{°}\right) \\ \Rightarrow \sin \left({360}^{°}-{60}^{°}\right)=0-1\times \frac{\sqrt{3}}{2}\left[\because \cos \left({360}^{°}\right)=1,\sin \left({360}^{°}\right)=0,\cos \left({60}^{°}\right)=\frac{1}{2}\mathrm{and}\sin \left({60}^{°}\right)=\frac{\sqrt{3}}{2}\right] \\ \Rightarrow \sin \left({300}^{°}\right)=-\frac{\sqrt{3}}{2} \end{gathered}$$

Hence the value of $\sin \left({300}^{°}\right)$ is $-\frac{\sqrt{3}}{2}$.