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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: $$\sin 2x = 2 \sin x\cos x$$

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

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

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  • $\begingroup$ Where does $-18/15$ come from? $\endgroup$ – Lord Shark the Unknown Feb 25 '18 at 19:50
  • $\begingroup$ note that $$\sin^2(x)+\cos^2(x)=1$$ for all real $x$ $\endgroup$ – Dr. Sonnhard Graubner Feb 25 '18 at 19:53
  • $\begingroup$ Sorry, not -18/15, -8/15. $\endgroup$ – Bill Feb 25 '18 at 19:56
  • $\begingroup$ Where does $-8/15$ come from? $\endgroup$ – Lord Shark the Unknown Feb 25 '18 at 19:56
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    $\begingroup$ ... $\sin x = -8/17,$ not $-8/15$ $\endgroup$ – Will Jagy Feb 25 '18 at 20:01
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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

  • $\sin 2x=2 \sin x \cos x >0$
  • $\sin x =-\sqrt{1-\cos^2 x}$
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  • $\begingroup$ Thank you again gimusi. $\endgroup$ – Bill Feb 25 '18 at 20:21
  • $\begingroup$ $x$ is not $\approx \pi$. Regards. $\endgroup$ – Piquito Feb 25 '18 at 20:36
  • $\begingroup$ @Piquito I've used to express that it is near to $pi$, which symbol do you suggest to use? $\endgroup$ – gimusi Feb 25 '18 at 20:38
  • $\begingroup$ I understand now what you wanted to say (for me a difference of $28^{\circ}$ is not for take the symbol $\approx$). Greetings. $\endgroup$ – Piquito Feb 25 '18 at 21:58
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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.

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