Solving Trig Equations: Find in terms of k the value of cosec x.

[Captain Sunshine]

Hi all, this is my first post here, and it's because I'm at a complete dead end. I've been trying to solve the following trig identity for the last three hours, and I've got nothing. I was hoping someone would be kind enough to point me in the right direction, as I can't seem to make any headway here.
Here's the question, verbatim:

The angle x is acute, and sec x = k. Find in terms of k the value of cosec x.

And the answer I've yet to reach without any suspect steps in my working:

k/√(k^2 - 1)

Thanks for any and all help in advance, I'd really appreciate it. Clearly I'm not understanding something fundamental here, and I'd really like to get to grips with this before my frustration levels get any higher.

 

[Deleted member 4993]

Hi all, this is my first post here, and it's because I'm at a complete dead end. I've been trying to solve the following trig identity for the last three hours, and I've got nothing. I was hoping someone would be kind enough to point me in the right direction, as I can't seem to make any headway here.
Here's the question, verbatim:

The angle x is acute, and sec x = k. Find in terms of k the value of cosec x.

And the answer I've yet to reach without any suspect steps in my working:

k/√(k^2 - 1)

Thanks for any and all help in advance, I'd really appreciate it. Clearly I'm not understanding something fundamental here, and I'd really like to get to grips with this before my frustration levels get any higher.

sec(x) = k → cos(x) = ? → sin(x) = ? → cosec(x) = ?

 

[Captain Sunshine]

Thank you for your reply; it helped simplify the steps a lot. I think maybe I was over thinking this. Here's what I have:

Sec(x) = k = 1/cos(x) .˙. Cos(x) = 1/k


Sin^2(x) + cos^2(x) = 1

Sin^2(x)/cos(x) + cos(x) = 1/cos(x) = k (divide by cos(x))

Sin^2(x)/cos(x) + cos(x) = k


Sin^2(x)/cos^2(x) + 1 = k/(1/k) = k^2 (divide by cos(x))

Sin^2(x)/cos^2(x) = k2 – 1 (subtract 1 from both sides)

Sin(x)/cos(x) = √(k2 – 1) (square root)

Sin(x) = √(k2 – 1)/ k (multiply by cos(x))

1/sin(x) = k/√(k2 -1) (invert)


Is this legit? Or is there a much simpler way of proving this? Either way, I owe you one.

 

[Ishuda]

Hi all, this is my first post here, and it's because I'm at a complete dead end. I've been trying to solve the following trig identity for the last three hours, and I've got nothing. I was hoping someone would be kind enough to point me in the right direction, as I can't seem to make any headway here.
Here's the question, verbatim:

The angle x is acute, and sec x = k. Find in terms of k the value of cosec x.

And the answer I've yet to reach without any suspect steps in my working:

k/√(k^2 - 1)

Thanks for any and all help in advance, I'd really appreciate it. Clearly I'm not understanding something fundamental here, and I'd really like to get to grips with this before my frustration levels get any higher.

As Subhotosh Khan pointed out, a good way to solve problems like this is to 'draw a right angle triangle' and then write down the length of the sides. You have to pay attention to quadrants but for an example, lets just treat everything as positive and so first quadrant (acute angles):
tan(x) = a/b
means a right triangle similar to side opposite = a, side adjacent = b. So hypotenuse = c = (a² + b²)^1/2. Thus
sin(x) = a/c
cos(x) = b/c
etc.

Another:
sin(x) = a
means a right triangle similar to side opposite = a, hypotenuse = 1. So side adjacent = b = (1 - a²)^1/2 Thus
sin(x) = a
cos(x) = b
tan(x) = a/b
etc.