Solve for the trigonometric value of $ an(x) $ when $ cos(x) = frac{1}{2} $ on the unit circle
Answer 1
To solve for $ \tan(x) $ given $ \cos(x) = \frac{1}{2} $ on the unit circle, we must first determine the corresponding $ \sin(x) $. On the unit circle:
$ \cos(x) = \frac{1}{2} $
We know that at $ x = \frac{\pi}{3} $ and $ x = -\frac{\pi}{3} $, $ \cos(x) = \frac{1}{2} $. Correspondingly, $ \sin(x) $ at these points are:
$ \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} $
$ \sin( – \frac{\pi}{3}) = – \frac{\sqrt{3}}{2} $
Using the identity for tangent:
$ \tan(x) = \frac{\sin(x)}{\cos(x)} $
For $ x = \frac{\pi}{3} $:
$ \tan(\frac{\pi}{3}) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} $
For $ x = -\frac{\pi}{3} $:
$ \tan(-\frac{\pi}{3}) = \frac{ – \frac{\sqrt{3}}{2}}{\frac{1}{2}} = – \sqrt{3} $
Answer 2
To determine $ sin(x) $ given $ cos(x) = -1 $ on the unit circle, use the Pythagorean identity:
$ sin^2(x) + cos^2(x) = 1 $
Given $ cos(x) = -1 $:
$ sin^2(x) + (-1)^2 = 1 $
$ sin^2(x) + 1 = 1 $
$ sin^2(x) = 0 $
Thus:
$ sin(x) = 0 $
Answer 3
To find $ cos(x) $ given $ an(x) = -1 $ on the unit circle, recall that:
$ an(x) = frac{sin(x)}{cos(x)} $
If $ an(x) = -1 $, then:
$ frac{sin(x)}{cos(x)} = -1 $
This implies:
$ sin(x) = -cos(x) $
Using the Pythagorean identity:
$ sin^2(x) + cos^2(x) = 1 $
Substitute $ sin(x) = -cos(x) $:
$ (-cos(x))^2
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