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anonymous

  • one year ago

1 last trig problem

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  1. anonymous
    • one year ago
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  2. jim_thompson5910
    • one year ago
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    Try rewriting everything in terms of either sine or cosine

  3. jim_thompson5910
    • one year ago
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    eg: sec(x) = 1/cos(x)

  4. anonymous
    • one year ago
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    ahhhhh ok

  5. anonymous
    • one year ago
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    give me one sec

  6. jim_thompson5910
    • one year ago
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    also, when it comes to proving identities, you should only work on one side

  7. jim_thompson5910
    • one year ago
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    it might be easier to work on the right side

  8. anonymous
    • one year ago
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    \[q \cot q/\cos^{-1} -\sin(\theta)/expsec/\csc ^2\frac{ f1/\sec}{ ? } \]

  9. anonymous
    • one year ago
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    CRAP, i hate uploading equations on this website

  10. anonymous
    • one year ago
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    one sec, let me try again

  11. anonymous
    • one year ago
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    ok I got my answer, but when I try to put it in the equation on this website it messes up

  12. anonymous
    • one year ago
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    so i started working on the right side

  13. jim_thompson5910
    • one year ago
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    After trying out the right side, it's somewhat more complicated than I thought. So I'm going to work on the left side instead \[\Large \frac{\sec(\beta)}{\sin(\beta)}-\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large {\color{red}{\frac{\sec(\beta)}{\sec(\beta)}*}}\frac{\sec(\beta)}{\sin(\beta)}-\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}-\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}-{\color{red}{\frac{\sin(\beta)}{\sin(\beta)}}}*\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}-\frac{\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\sec^2(\beta)-\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] Take a few minutes and look all that over. Then let me know when you're ready for the next step.

  14. anonymous
    • one year ago
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    wait I think I need to prove the right side of the equation not the left

  15. jim_thompson5910
    • one year ago
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    you can start on either side

  16. anonymous
    • one year ago
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    ok i get it

  17. jim_thompson5910
    • one year ago
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    just as long as you only work with one side

  18. anonymous
    • one year ago
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    alright is that it

  19. anonymous
    • one year ago
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    @jim_thompson5910

  20. jim_thompson5910
    • one year ago
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    look at the "Pythagorean Identities" on page 2

  21. jim_thompson5910
    • one year ago
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    what is sec^2 equal to?

  22. anonymous
    • one year ago
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    tan^2(theta)+1

  23. jim_thompson5910
    • one year ago
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    and what is sin^2 equal to?

  24. anonymous
    • one year ago
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    1-cos^2(theta)

  25. jim_thompson5910
    • one year ago
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    So, \[\Large \frac{\sec^2(\beta)-\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\] \[\Large \frac{\tan^2(\beta)+1-(1-\cos^2(\beta))}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]

  26. jim_thompson5910
    • one year ago
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    simplify the numerator for the denominator, use the fact that sin*sec = sin*(1/cos) = sin/cos = ???

  27. anonymous
    • one year ago
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    Ok I did it, can you show me what you got

  28. jim_thompson5910
    • one year ago
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    when you simplified the numerator, what did you get?

  29. anonymous
    • one year ago
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    i wish I could show you, trust me, but I cant

  30. jim_thompson5910
    • one year ago
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    draw it out

  31. jim_thompson5910
    • one year ago
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    or type what you can

  32. anonymous
    • one year ago
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    I got back to where we started,

  33. jim_thompson5910
    • one year ago
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    \[\Large \tan^2(\beta)+1-(1-\cos^2(\beta))\] \[\Large \tan^2(\beta)+1-1+\cos^2(\beta)\] which becomes what?

  34. anonymous
    • one year ago
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    what the other side of the equation is

  35. jim_thompson5910
    • one year ago
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    yeah ultimately the goal is to get the left side equal to the right side

  36. anonymous
    • one year ago
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    what about the denominator

  37. jim_thompson5910
    • one year ago
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    what is sin(x)*sec(x) equal to?

  38. jim_thompson5910
    • one year ago
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    hint: look at "Tangent and Cotangent Identities" and "Reciprocal Identities" on that pdf

  39. anonymous
    • one year ago
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    ok I got it

  40. anonymous
    • one year ago
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    it is sinxsec

  41. jim_thompson5910
    • one year ago
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    sin(x)*sec(x) = sin(x)*(1/cos(x)) = sin(x)/cos(x) = ????

  42. anonymous
    • one year ago
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    tan

  43. jim_thompson5910
    • one year ago
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    yep

  44. anonymous
    • one year ago
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    alright thanks

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