If $\cos x = -15/17$ with $x$ in quadrant III, find the exact value of $\sin 2x $ and the quadrant of $2x$

Question

If $\cos x = -15/17$ with $x$ in quadrant III, find the exact value of $\sin 2x $ and the quadrant of $2x$.

I'm stuck on finding the quadrant of $2x$, so far I've done: $$\sin 2x = 2 \sin x\cos x$$

$$(-\frac8{15})(-\frac{15}{17})= $$

$$= \frac{120}{255} \Rightarrow \frac{24}{51}$$

Answer 1

HINT

Note that $-\frac{15}{17}\approx -1$ thus $x\approx \pi$ and $2x \approx 0$ in the first quadrant.

To find $\sin 2x$ let use

• $\sin 2x=2 \sin x \cos x >0$
• $\sin x =-\sqrt{1-\cos^2 x}$

Answer 2

Without sign one has $$\cos x=\dfrac{15}{17}\iff\sin x=\frac {8}{17}$$ then $$\sin 2x=2\cdot\dfrac{15}{17}\cdot\frac {8}{17}=0.830449827$$ Now with signs one has, with $x$ in quadrant III, $$\arccos\frac {15}{17}=28.07248694^{\circ}\Rightarrow x=180^{\circ}+28.07248694^{\circ}=208.07248694^{\circ}$$ Therefore $$2(208.07248694^{\circ})=416.1449738^{\circ}$$ Since $$416.1449738^{\circ}-360^{\circ}=56.1449738^{\circ}$$ the quadrant of $2x$ is the first one.