Question

State the signs of : (i) $\sin {675}^o$ (ii)$\cos (-\frac{8\pi }{3})$

Answer 1

$\left(i\right)\sin {675}^{\circ }$

$=\sin ({360}^{\circ }+{315}^{\circ })$

$=\sin ({360}^{\circ }+{360}^{\circ }-{45}^{\circ })$

$=\sin ({360}^{\circ }\times 2-{45}^{\circ })$

$\therefore \sin {675}^{\circ }$ is in fourth quadrant.

Hence $\sin {675}^{\circ }$ is negative

$\therefore \sin ({360}^{\circ }\times 2-{45}^{\circ })=-\sin {45}^{\circ }=\frac{-1}{\sqrt{2}}$

$\left(ii\right)\cos \left(\frac{-8\pi }{3}\right)$

$=\cos \left(\frac{8\pi }{3}\right)$ since $-\theta$ is in fourth quadrant and cosine is positive in fourth quadrant.

$=\cos (3\pi -\frac{\pi }{3})$

We know that $(3\pi -\frac{\pi }{3})$ is in second quadrant.

Hence $\cos \left(\frac{-8\pi }{3}\right)=-\cos \frac{\pi }{3}=-\frac{1}{2}$

Hence its sign is negative.