1. ## finding and angle in radians from cosec and tan

i need to find the angle in radians from cosec x = -2 and tan x = √3÷3 −π < x< π

i know that sin x = -1/2 cos x = sin x/tan x

but where do i go from here?

any help would be great!

2. ## Re: finding and angle in radians from cosec and tan

Originally Posted by gurnster76
i need to find the angle in radians from cosec x = -2 and tan x = √3÷3 −π < x< π

i know that sin x = -1/2 cos x = sin x/tan x

but where do i go from here?

any help would be great!
\displaystyle \begin{align*} \csc{x} &= -2 \\ \frac{1}{\sin{x}} &= 2 \\ \sin{x} &= -\frac{1}{2} \end{align*}

So the angle is in either the third or fourth quadrant.

\displaystyle \begin{align*} \tan{x} &= \frac{\sqrt{3}}{3} \\ \tan{x} &= \frac{1}{\sqrt{3}} \end{align*}

and since the tangent function is positive in the first and third quadrants, together, the two pieces of information tell us that the angle is in the third quadrant.

You should also know enough from the special triangles to be able to answer this question.

3. ## Re: finding and angle in radians from cosec and tan

Originally Posted by Prove It
\displaystyle \begin{align*} \csc{x} &= -2 \\ \frac{1}{\sin{x}} &= 2 \\ \sin{x} &= -\frac{1}{2} \end{align*}

So the angle is in either the third or fourth quadrant.

\displaystyle \begin{align*} \tan{x} &= \frac{\sqrt{3}}{3} \\ \tan{x} &= \frac{1}{\sqrt{3}} \end{align*}

and since the tangent function is positive in the first and third quadrants, together, the two pieces of information tell us that the angle is in the third quadrant.

You should also know enough from the special triangles to be able to answer this question.
so if angle sin x = -1/2 then x = -n/6 or -5n/6

tan x = 1/√3 so x = n/6 or -5n/6

so is the angle required -5n/6 radians?

thanks so much for your help!

4. ## Re: finding and angle in radians from cosec and tan

so if angle sin x = -1/2 then x = -n/6 or -5n/6

tan x = 1/√3 so x = n/6 or -5n/6

so is the angle required -5n/6 radians?

thanks so much for your help!