# How do you find the sine and cosine from the tangent?

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.

• You forgot that $\sqrt{x^2}$ is not equal to $x$, it is equal to $|x|$. Mar 3 '12 at 4:27
• @ArturoMagidin, Is my current answer correct, $-\sqrt{\frac{1}{3.25}}$? Mar 3 '12 at 4:37
• 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? Mar 3 '12 at 4:39
• Why it doesn't seem correct? This diagram might help. Mar 3 '12 at 4:51
• @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. Mar 3 '12 at 4:54

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?
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?
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?