Stuck on Finding the Remaining Trig Functions if sec theta = 2x.

[philthethrill]

I'm taking a Calculus 1 class that starts tomorrow and I have a review assignment that covers various topics from algebra and trig. I have a trig problem that I'm stuck on. It says given sec theta = 2x, find the remaining five trigonometric ratios. I have already found cos theta, 1/2x, but I'm having trouble finding the rest. So far I have tried finding sin theta by using the pythagorean identity sin²theta +cos²theta = 1. Here's what I have so far.

sin²theta + (1/2x)² = 1
sqroot sin²theta = sqroot 1 - 1/4x²

I'm not sure how to simplify that.

I have tried sin theta = sqroot 4x²/4x² -1/4x²
= sqroot (2x-1) (2x+1)/4x²

I started rationalizing and I got sqroot (2x-1) (2x+1)/sqroot 4x² * sqroot 4x²/sqroot 4x² .

I just don't know how to proceed. Am I doing something wrong?

 

[Deleted member 4993]

I'm taking a Calculus 1 class that starts tomorrow and I have a review assignment that covers various topics from algebra and trig. I have a trig problem that I'm stuck on. It says given sec theta = 2x, find the remaining five trigonometric ratios. I have already found cos theta, 1/2x, but I'm having trouble finding the rest. So far I have tried finding sin theta by using the pythagorean identity sin²theta +cos²theta = 1. Here's what I have so far.

sin²theta + (1/2x)² = 1
sqroot sin²theta = sqroot 1 - 1/4x²

sin(Θ) = ±√[1 - 1/(4x²)]

In my opinion, no further simplification is warranted for the given problem.

I'm not sure how to simplify that.

I have tried sin theta = sqroot 4x²/4x² -1/4x²
= sqroot (2x-1) (2x+1)/4x²

I started rationalizing and I got sqroot (2x-1) (2x+1)/sqroot 4x² * sqroot 4x²/sqroot 4x² .

I just don't know how to proceed. Am I doing something wrong?

.