How to find $\sin\left(-\frac{11\pi}{12}\right)$ from the unit circle?
I Want Above To Be In Unit Circle.... unit circle. like convert $\frac{\pi}{3}$ to $\frac{4\pi}{12}$ to make it same denom... and then do stuff.... i tried to do with $\frac{\pi}{2}$ (convert to $\frac{6\pi}{12}$) and $\frac{\pi}{3}$ (convert to $\frac{4\pi}{12}$) ... but it only gives you $\frac{10\pi}{12}$.. Please Help. I DO NOT want the anwser only.. i want how you did it.. i already have anwser as it is a study guide but i want to know how you did it.. Thanks a lot!
$\endgroup$ 42 Answers
$\begingroup$Since the sine function is odd and fulfills $\sin(t)=\sin(\pi-t)$ we have: $$\sin\left(-\frac{11\pi}{12}\right)=-\sin\left(\frac{11\pi}{12}\right)=-\sin\frac{\pi}{12}$$ and since $\frac{\pi}{12}$ is an acute angle, by the half-angle formula $\sin\frac{t}{2}=\sqrt{\frac{1-\cos t}{2}}$ we get: $$\sin\left(-\frac{11\pi}{12}\right)=-\sqrt{\frac{1-\cos\frac{\pi}{6}}{2}}=\color{red}{-\frac{1}{2}\sqrt{2-\sqrt{3}}}.$$
$\endgroup$ 18 $\begingroup$Since $\sin(-x) = -\sin x$,
$$\sin\left(-\frac{11\pi}{12}\right) = -\sin\left(\frac{11\pi}{12}\right)$$
We can compute $\sin\left(\frac{11\pi}{12}\right)$ using the sum of angles formula for sine.
\begin{align*}
\sin\left(\frac{11\pi}{12}\right) & = \sin\left(\frac{8\pi}{12} + \frac{3\pi}{12}\right)\\
& = \sin\left(\frac{2\pi}{3} + \frac{\pi}{4}\right)\\
& = \sin\left(\frac{2\pi}{3}\right)\cos\left(\frac{\pi}{4}\right) + \cos\left(\frac{2\pi}{3}\right)\sin\left(\frac{\pi}{4}\right)\\
& = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(-\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\\
& = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}\\
& = \frac{\sqrt{6} - \sqrt{2}}{4}
\end{align*}
Hence,
$$\sin\left(-\frac{11\pi}{12}\right) = -\sin\left(\frac{11\pi}{12}\right) = -\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$$
