If sin(θ) = − 5/6 with θ in Quadrant III, what is cos(θ)?

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If sin(θ) = − 5/6 with θ in Quadrant III, what is cos(θ)?

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Answer 1 · 2018-01-28T10:25:18

$cos (θ) = - \frac{\sqrt{11}}{6}$ $cos(theta)=-sqrt(11)/6$

Explanation:

$sin (θ) = - \frac{5}{6}$ $sin(theta)=-5/6$

We can build a right triangle in the unit circle,
with the hypotenuse 1 and a side $sin (θ) = - \frac{5}{6}$ $sin(theta)=-5/6$
(I recommend to make a sketch)

Then use Pythagoras (we deal with the sign afterwards)

$cos (θ) = \sqrt{1 - {(- \frac{5}{6})}^{2}}$ $cos(theta)=sqrt(1-(-5/6)^2)$

$= \sqrt{\frac{36}{36} - \frac{25}{36}}$ $=sqrt(36/36-25/36)$

$= \sqrt{\frac{11}{36}}$ $=sqrt(11/36)$

$= \frac{\sqrt{11}}{6}$ $=sqrt(11)/6$

When we are in the third quadrant $cos (θ) < 0$ $cos(theta)<0$

Therefore

$cos (θ) = - \frac{\sqrt{11}}{6}$ $cos(theta)=-sqrt(11)/6$

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If sin(θ) = − 5/6 with θ in Quadrant III, what is cos(θ)?

1 Answer

Explanation:

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