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kalli

  • one year ago

Use the half-angle formula to evaluate tan(17pi/12)

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  1. freckles
    • one year ago
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    well \[\tan(\frac{17\pi}{12}) \\ y=\tan(x) \text{ has period } \pi \\ \tan(\frac{17\pi}{12}-\pi)=\tan(\frac{17\pi}{12}-\frac{12\pi}{12})=\tan(\frac{5\pi}{12}) \\ \text{ now } 2 \cdot \frac{5 \pi}{12}=\frac{5\pi}{6} \\ \text{ so we have that we want to use the half-angle formula on } \\ \tan( \frac{1}{2} \cdot \frac{5 \pi}{6})\]

  2. freckles
    • one year ago
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    do you know the half angle identity for tan?

  3. kalli
    • one year ago
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    no, only for sin

  4. freckles
    • one year ago
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    so you can't use the half angle identity for tan?

  5. freckles
    • one year ago
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    you can only use the one for sin?

  6. freckles
    • one year ago
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    the weird thing is you have tan here not sin

  7. freckles
    • one year ago
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    or do you mean you can use it but you just don't know it?

  8. freckles
    • one year ago
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    http://www.purplemath.com/modules/idents.htm do you see half-angle identities on this page?

  9. freckles
    • one year ago
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    notice the three different ones they wrote for tan(x/2)

  10. freckles
    • one year ago
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    use one of them

  11. kalli
    • one year ago
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    im confused because I don't really know how to do this lol

  12. freckles
    • one year ago
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    well first let's just copy of of the half-angle identities over from that page \[\tan(\frac{1}{2}x)=\frac{\sin(x)}{1+\cos(x)} \\ \text{ and we want to evaluate } \\ \tan(\frac{1}{2} \frac{5\pi}{6})=?\] do you see what to replace x with ?

  13. freckles
    • one year ago
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    compare tan(1/2*5pi/6) to tan(1/2*x)

  14. freckles
    • one year ago
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    x has to be 5pi/6 right?

  15. freckles
    • one year ago
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    so replace the x's in the identity with 5pi/6

  16. kalli
    • one year ago
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    ok one sec

  17. kalli
    • one year ago
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    tan(1/2*5pi/6)= sin(5pi/6) / 1+cos(5pi/6) ??

  18. freckles
    • one year ago
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    ok and ( ) around the bottom

  19. freckles
    • one year ago
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    so sin(5pi/6)=? and cos(5pi/6)=?

  20. kalli
    • one year ago
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    1/2 and -sqrt 3/2 ??

  21. freckles
    • one year ago
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    \[\tan(\frac{1}{2} \cdot \frac{5\pi}{6})=\frac{\frac{1}{2}}{1+\frac{-\sqrt{3}}{2}}\] multiply top and bottom by 2

  22. freckles
    • one year ago
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    \[\frac{\frac{2}{2}}{2+\frac{2(-\sqrt{3})}{2}}\]

  23. freckles
    • one year ago
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    can you simplify from there?

  24. freckles
    • one year ago
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    you know 2/2=1

  25. kalli
    • one year ago
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    Yes one second

  26. kalli
    • one year ago
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    2+ sqrt3

  27. kalli
    • one year ago
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    @freckles is that my answer?

  28. freckles
    • one year ago
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    let me check one sec

  29. freckles
    • one year ago
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    \[\frac{\frac{2}{2}}{2+\frac{2(-\sqrt{3})}{2}} \\ \frac{1}{2-\sqrt{3} } \\ \frac{1}{2 -\sqrt{3}} \cdot \frac{2 +\sqrt{3}}{2+\sqrt{3}} \\ \frac{2+\sqrt{3}}{4-3} \\ \frac{2 +\sqrt{3}}{1} \\ 2+\sqrt{3}\]

  30. freckles
    • one year ago
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    yep looks fabulous

  31. kalli
    • one year ago
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    thanks so much!!!

  32. freckles
    • one year ago
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    np

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