2
$\begingroup$

I'm given the problem:

If $\cot(\theta) = 1.5$ and $\theta$ is in quadrant 3, what is the value of $\sin(\theta)$?

I looked at all the related answers I could find on here, but I haven't been able to piece together the answer I need from them.

I know that $\sin^2\theta + \cos^2\theta = 1$, $ \cot^2\theta + 1 = \csc^2\theta $, and $\csc^2\theta = \frac{1}{\sin^2\theta}$

Substituting 3.25 for $\cot^2\theta + 1$ and $\frac{1}{\sin^2\theta}$ for $\csc^2\theta$ I get:

$3.25 = \frac{1}{\sin^2\theta}$

then

$\sin\theta = -\sqrt{\frac{1}{3.25}}$

This doesn't seem correct though. Can anyone help please?

edit: Sorry, meant to make that answer negative.

$\endgroup$
6
  • $\begingroup$ You forgot that $\sqrt{x^2}$ is not equal to $x$, it is equal to $|x|$. $\endgroup$ Mar 3 '12 at 4:27
  • $\begingroup$ @ArturoMagidin, Is my current answer correct, $-\sqrt{\frac{1}{3.25}}$? $\endgroup$
    – mowwwalker
    Mar 3 '12 at 4:37
  • $\begingroup$ And why do you think that it "does not seem correct"? $\frac{1}{3.25} = \frac{4}{13}$; is that the problem, the representation of this number? $\endgroup$ Mar 3 '12 at 4:39
  • $\begingroup$ Why it doesn't seem correct? This diagram might help. $\endgroup$
    – Eelvex
    Mar 3 '12 at 4:51
  • $\begingroup$ @ArturoMagidin, Yes, it's the format of the answer. In the lessons, when going over the 45-45-90 triangles, the ratio of the sides were given as $\frac{\sqrt{2}}{2}$ so there wasn't an irrational number in the denominator. I was hoping for a cleaner answer, but it seems that that's the best I'm going to get. $\endgroup$
    – mowwwalker
    Mar 3 '12 at 4:54
2
$\begingroup$

Indeed, we know that $$1.5 = \cot\theta = \frac{\cos\theta}{\sin\theta}$$ hence $$1.5\sin\theta = \cos\theta.$$ Squaring both sides we have $$2.25\sin^2\theta = \cos^2\theta$$ and since $\cos^2\theta = 1-\sin^2\theta$ we have $$\begin{align*} 2.25\sin^2\theta &= 1-\sin^2\theta\\ 2.25\sin^2\theta + \sin^2\theta &= 1\\ 3.25\sin^2\theta &= 1. \end{align*}$$ From this, you can figure out the value of $\sin^2\theta$. Taking square roots will tell you something about the absolute value of $\sin\theta$.

Now... why did they tell you $\theta$ was in the third quadrant?

$\endgroup$
1
$\begingroup$

If $\cot\theta=1.5$, then $\tan\theta=\frac23$. This means that if $\theta$ were in the first quadrant, it would be one of the angles of a right triangle whose legs measure $2$ and $3$ and whose hypotenuse measures $\sqrt{2^2+3^2}=\sqrt{13}$. Specifically, it would be the angle opposite the side of length $2$. Sketch the triangle, and you’ll see that in that case we’d have $$\sin\theta=\frac2{\sqrt{13}}\;.$$

But $\theta$ is in the third quadrant, not the first; what effect does this have on its sine?

$\endgroup$
0
$\begingroup$

Consider the point $P=(-3,-2)$, and the (reflex) angle $\theta$ formed by the positive $x$-axis and the line from the origin to $P$. Can you see why this is the $\theta$ in the problem? Can you work out the other functions from this diagram?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.