How do you solve #cot [Arcsin (-12/13)]#?

Question

Anna Andrews · Accepted Answer

#-5/12# against the principal value of #arc sin (-123/13)# and, for the general value, the answer is #+-5/12#.

Let #a = arc sin (-12/13)#. Then, #sin a = -12/13<0#. The principal value of a is in the 4th quadrant. The general value is either in the 4th or in the 3rd.. So. cos a is #sqrt(1-12^2/13^2)=5/13# or #+-5/13#. The given expression is cot a = cos a/sin a and, accordingly, the answer is given..

Kai Anderson · Answer

undefined To solve $ \cot(\arcsin(-\frac{12}{13})) $, first, find the value of $ \arcsin(-\frac{12}{13}) $, then take the cotangent of that value.

Given $ \arcsin(-\frac{12}{13}) $, we need to find the angle whose sine is $ -\frac{12}{13} $. Since sine is negative in the third and fourth quadrants, we're looking for an angle in either the third or fourth quadrant.

Let's call the angle $ \theta $. We know that $ \sin(\theta) = -\frac{12}{13} $. Using the Pythagorean identity $ \sin^2(\theta) + \cos^2(\theta) = 1 $, we can find $ \cos(\theta) $.

$$ \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} $$
$$ \cos(\theta) = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13} $$

Since cosine is also negative in the third quadrant, $ \cos(\theta) = -\frac{5}{13} $.

Now, to find $ \cot(\arcsin(-\frac{12}{13})) $, we use the fact that $ \cot(\theta) = \frac{1}{\tan(\theta)} $, and $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $.

$$ \tan(\theta) = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{5} $$

Thus, $ \cot(\arcsin(-\frac{12}{13})) = \frac{1}{\tan(\arcsin(-\frac{12}{13}))} = \frac{1}{\frac{12}{5}} = \frac{5}{12} $.