Trig problem

1. Jun 6, 2016

IntegralDerivative

1. The problem statement, all variables and given/known data
Find cscθ given sec θ = -2 sin θ >0

2. Relevant equations
I do not know where to begin or what equations to use.

3. The attempt at a solution
I am assuming there is a typo in the question and that there should be a comma sec θ = -2, sin θ >0.

If so I got csc θ = 2√3 / 3

2. Jun 6, 2016

SteamKing

Staff Emeritus
Why not? How is the secant defined? The cosecant? Those would be good places to start.

3. Jun 6, 2016

IntegralDerivative

I am assuming there is a typo in the question and that there should be a comma sec θ = -2, sin θ >0.

If so I got csc θ = 2√3 / 3

4. Jun 6, 2016

SteamKing

Staff Emeritus
Looks good.

5. Jun 6, 2016

Ray Vickson

The question as written (no comma) also makes perfectly good sense; it says that $0 < \sec \theta = - 2 \sin \theta$, and you need to use the definition of $\sec$ to get a solvable equation. You can get $\sin \theta$ and $\cos \theta$, then compute $\csc \theta$, which does NOT equal $2 \sqrt{3}/3 = 2/\sqrt{3}$.

Last edited: Jun 7, 2016
6. Jun 7, 2016

SammyS

Staff Emeritus
As Ray points out, the problem as stated makes sense and can be solved. If you are currently studying trig equations in your course, then that's probably not a typo. On the other hand, if you are studying the basics of trigonometric functions early in your course, then it very well may be a typo, similar to what you have concluded.

The statement, $\ \sec θ = -2 \sin θ >0 \,,\$ is really two (or three) statements rolled into one.
• $\ \sec θ >0 \$
• $\ -2 \sin θ >0 \,,\$ so that $\ \sin θ <0 \$
• $\ \sec θ = -2 \sin θ$
The first two tell you what quadrant θ is in.

7. Jun 7, 2016

Dr Transport

recast $sec(\theta)$ as $1/cos(\theta)$ and the solution falls out in about 2 lines...