Find cscθ Given sec θ = -2, sin θ >0

[IntegralDerivative]

Homework Statement

Find cscθ given sec θ = -2 sin θ >0

Homework Equations

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

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

 

[SteamKing]

Homework Statement

Find cscθ given sec θ = -2 sin θ >0

Homework Equations

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

The Attempt at a Solution

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

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

 

[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

 

[SteamKing]

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

Looks good.

 

[Ray Vickson]

Homework Statement

Find cscθ given sec θ = -2 sin θ >0

Homework Equations

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

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

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 $2 \sqrt{3}/3 = 2/\sqrt{3}$.

 

[SammyS]

Homework Statement

Find cscθ given sec θ = -2 sin θ >0

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.

 

[Dr Transport]

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