anonymous
  • anonymous
What is sec(a + B) If cos(a+B) = cos a cos B - sin a sin B then does sec(a+ B) = 1 / cos(a+B) = 1 / cos a cosB - sin a sin B ??
Mathematics
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schrodinger
  • schrodinger
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UsukiDoll
  • UsukiDoll
I'm wondering if that exists... usually secant is 1/cosine... but I haven't seen a sec(a+b) unless I'm living under a rock.
UsukiDoll
  • UsukiDoll
that's sum and difference identities.
anonymous
  • anonymous
Well, I know we have cos(a+B) = cos a cos B - sin a sin B which turns to \(cos(2\theta) \) and I have a problem I am working on that has \(sec(2\theta) \) so I was wondering

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anonymous
  • anonymous
or \(cos(2\theta) = cos^2-sin^2\)
zepdrix
  • zepdrix
sec(a+ B) = 1 / cos(a+B) = 1 / (cos a cosB - sin a sin B) Yah this looks fine ^ But if you wanted to relate sec(A+B) to a bunch of secA and secB stuff, that would be a little bit more work :)
zepdrix
  • zepdrix
What's the entire question? +_+ It's kind of hard to suggest which route you should take without seeing it
anonymous
  • anonymous
This is the problem \( \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2} \)
anonymous
  • anonymous
This is the problem \( \Huge \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2} \)
anonymous
  • anonymous
I was thinking
anonymous
  • anonymous
\[ \Huge \frac{1}{\cos^2\theta-\sin^2\theta} = \frac{\csc^2\theta}{\csc^2\theta-2} \] Going with it this way
zepdrix
  • zepdrix
\[\Large\rm \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2}\] \[\Large\rm \frac{1}{\cos(2\theta)} = \frac{\csc^2\theta}{\csc^2\theta-2}\] \[\Large\rm \frac{1}{1-2\sin^2 \theta} = \frac{\csc^2\theta}{\csc^2\theta-2}\]Recall that you can write the cosine double angle in like 3 different ways. I think THIS is the double angle identity that we want to use.
zepdrix
  • zepdrix
See how the 2's match up? This one should get us on track.
zepdrix
  • zepdrix
The next step is a little tricky ;) Ok with that step though? :O
anonymous
  • anonymous
Yes! I forgot about that one. Do we \( \Large\rm \frac{1}{1-2\sin^2 \theta}\times\frac{1+2\sin^2 \theta}{1+2\sin^2 \theta} \)
zepdrix
  • zepdrix
Conjugate? Hmm no that's going to give us a 4 in the bottom, i don't think we want that
zepdrix
  • zepdrix
What is csc(theta) in relation to sin(theta)? \(\Large\rm \csc(\theta)=\frac{1}{\sin(\theta)}\), ya? Hmm so how can we `end up with` a 1/sin^2x in the top there? :d
anonymous
  • anonymous
Multiply by its reciprocal
zepdrix
  • zepdrix
Well if want to turn a 1 into 1/sin^2 then we should probably divide by sin^2. So let's divide top `and` bottom by sin^2(theta). I know it's a weird step!
zepdrix
  • zepdrix
\[\Large\rm \frac{1}{1-2\sin^2 \theta} \color{orangered}{\left(\frac{\frac{1}{\sin^2\theta}}{\frac{1}{\sin^2\theta}}\right)}= \frac{\csc^2\theta}{\csc^2\theta-2}\]
anonymous
  • anonymous
Yes that is what I got and you are correct. That is weird :-D
zepdrix
  • zepdrix
It probably would have made more sense to start with the right side and turn it into the left. The weird step in the middle would had been more straight forward :) This is fine though
anonymous
  • anonymous
Cool, I got it. Thank you!!!
anonymous
  • anonymous
Working it out the rest of the way
zepdrix
  • zepdrix
cool :)
anonymous
  • anonymous
I think the ones with the sec and csc gives me the most trouble

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