# Given that cos theta = -0.489 where pi< theta < 3 pi/2, find sin theta

Given that $\mathrm{cos}\theta =-0.489$ where $\pi <\theta <\frac{3\pi }{2}$, find $\mathrm{sin}\theta$​​​​​​​

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Obiajulu

Use the Pythagorean identity:

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$

${\mathrm{sin}}^{2}\theta =1-{\mathrm{cos}}^{2}\theta$

Since $\theta$ is Quadrant III, then $\mathrm{sin}\theta <0$ so take the negative root:

$\mathrm{sin}\theta =-\sqrt{1-\mathrm{cos}2\theta }$

$\mathrm{sin}\theta =-{\sqrt{1-0.489}}^{2}$

$\mathrm{sin}\theta \approx -0.872$