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The angle $\theta$ is located in Quadrant III and $\cos(\theta) = -\frac{5}{8}$. What is the exact value of $\sin(\theta)$?

I'm at a loss. The fraction to me implies sides of a right triangle but when we get into quadrants that aren't Quadrant I, I don't understand how the angles and sides are related anymore since the angle is "larger" and wraps around the circle more instead of just being an angle in a triangle.

Edit: Is it $-\frac{\sqrt{39}}{8}$? (I did $\sqrt{8^2-5^2} = \sqrt{39}$ to get the length of the other "side" of the triangle and since it's Quadrant III I assume this to be negative, and so sin is just the ratio of that over the hypotenuse instead?)

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Guide:

Information that can help you solve the problem:

$$\sin^2(\theta) +\cos^2(\theta)=1$$

and also in the third quadrant:

$$\sin(\theta) < 0$$

Reading task about trigonometry function, especially the part about unit circle definition.

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  • $\begingroup$ So $\sin(\theta) = \pm \sqrt{1 - \cos^2(-\frac{5}{8})}$ and I take the negative solution for whatever this comes out to be, but I'm not sure how to evaluate this. $\endgroup$ – user9348401 Feb 12 '18 at 7:59
  • $\begingroup$ you seems to do well. $\endgroup$ – Siong Thye Goh Feb 12 '18 at 8:01
  • $\begingroup$ Comparing decimals it does seem to be my original answer of $-\frac{\sqrt{39}}{8}$ but how do I get this exact form from that expression? $\endgroup$ – user9348401 Feb 12 '18 at 8:05
  • $\begingroup$ what do you mean by exact form? $-\frac{\sqrt{39}}{8}$ is an exact form isn't it? $\endgroup$ – Siong Thye Goh Feb 12 '18 at 8:07
  • $\begingroup$ Yes but I don't know how to get there from $\sin(\theta) = \pm \sqrt{1 - \cos^2(-\frac{5}{8})}$ $\endgroup$ – user9348401 Feb 12 '18 at 8:15

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