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Trigonometry Trigonometric functions


Active member
May 13, 2013
find the values of other five trig functions

$\csc\theta=-2\,and\, \cot\theta>0$

my solution



$\displaystyle \sin\theta=-\frac{1}{\sqrt{5}}$





are they correct?


New member
Aug 7, 2013
it's not correct! use Pythagorean Identity involving sin and cos to find x.
Last edited:


Active member
Aug 18, 2013
I may have misunderstood your question, but your restrictions stated that $$\csc\theta=-2$$ whereas the value of your $\sin\theta$ was $-\frac{1}{\sqrt{5}}$, which most certainly does not satisfy $$\sin(x)=\frac{1}{\csc(x)}$$

*Edit* LATEBLOOMER beat me to it.


Staff member
Feb 24, 2012
I just want to add the following:

We are told \(\displaystyle \csc(\theta)<0\) and \(\displaystyle \cot(\theta)>0\), and so we ask ourselves in which quadrant are both of these true. The cosecant function is negative in quadrants 3 and 4, while the cotangent function is positive in quadrants 1 and 3, so we now know $\theta$ must be in quadrant 3, i.e:

\(\displaystyle \pi<\theta<\frac{3\pi}{2}\)

Now, one way we could proceed is to simply solve for $\theta$ directly, and then evaluate the other 5 functions at that angle, or as suggested above, we may employ identities and definitions to get the other functions. It is usually the second of these methods that is preferred, because we won't always be able to easily solve for the angle $\theta$, and so this second method is a more general way to proceed.

If I were to solve this problem using this method, I would first look at:

\(\displaystyle \sin(\theta)=\frac{1}{\csc(\theta)}=?\)

Next, I would use:

\(\displaystyle \cos(\theta)=-\sqrt{1-\sin^2(\theta)}=?\)

Why do we take the negative root here?

Next, we may finish up with definitions:

\(\displaystyle \sec(\theta)=\frac{1}{\cos(\theta)}=?\)

\(\displaystyle \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=?\)

\(\displaystyle \cot(\theta)=\frac{1}{\tan(\theta)}=?\)

There are of course other Pythagorean identities that could be used, so my outline above is just one such way to go.