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anonymous

  • one year ago

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 ??

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  1. UsukiDoll
    • one year ago
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    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.

  2. UsukiDoll
    • one year ago
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    that's sum and difference identities.

  3. anonymous
    • one year ago
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    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

  4. anonymous
    • one year ago
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    or \(cos(2\theta) = cos^2-sin^2\)

  5. zepdrix
    • one year ago
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    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 :)

  6. zepdrix
    • one year ago
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    What's the entire question? +_+ It's kind of hard to suggest which route you should take without seeing it

  7. anonymous
    • one year ago
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    This is the problem \( \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2} \)

  8. anonymous
    • one year ago
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    This is the problem \( \Huge \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2} \)

  9. anonymous
    • one year ago
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    I was thinking

  10. anonymous
    • one year ago
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    \[ \Huge \frac{1}{\cos^2\theta-\sin^2\theta} = \frac{\csc^2\theta}{\csc^2\theta-2} \] Going with it this way

  11. zepdrix
    • one year ago
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    \[\Large\rm \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2}\] \[\Large\rm \frac{1}{\cos(2\theta)} = \frac{\csc^2\theta}{\csc^2\theta-2}\] \[\Large\rm \frac{1}{1-2\sin^2 \theta} = \frac{\csc^2\theta}{\csc^2\theta-2}\]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.

  12. zepdrix
    • one year ago
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    See how the 2's match up? This one should get us on track.

  13. zepdrix
    • one year ago
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    The next step is a little tricky ;) Ok with that step though? :O

  14. anonymous
    • one year ago
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    Yes! I forgot about that one. Do we \( \Large\rm \frac{1}{1-2\sin^2 \theta}\times\frac{1+2\sin^2 \theta}{1+2\sin^2 \theta} \)

  15. zepdrix
    • one year ago
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    Conjugate? Hmm no that's going to give us a 4 in the bottom, i don't think we want that

  16. zepdrix
    • one year ago
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    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

  17. anonymous
    • one year ago
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    Multiply by its reciprocal

  18. zepdrix
    • one year ago
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    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!

  19. zepdrix
    • one year ago
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    \[\Large\rm \frac{1}{1-2\sin^2 \theta} \color{orangered}{\left(\frac{\frac{1}{\sin^2\theta}}{\frac{1}{\sin^2\theta}}\right)}= \frac{\csc^2\theta}{\csc^2\theta-2}\]

  20. anonymous
    • one year ago
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    Yes that is what I got and you are correct. That is weird :-D

  21. zepdrix
    • one year ago
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    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

  22. anonymous
    • one year ago
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    Cool, I got it. Thank you!!!

  23. anonymous
    • one year ago
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    Working it out the rest of the way

  24. zepdrix
    • one year ago
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    cool :)

  25. anonymous
    • one year ago
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    I think the ones with the sec and csc gives me the most trouble

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