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anonymous

  • one year ago

Find the exact value of ....

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  1. anonymous
    • one year ago
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    \[\tan (\frac{ 17\pi }{ 12 })\]

  2. freckles
    • one year ago
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    how many 12's are in 17?

  3. anonymous
    • one year ago
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    I know I have to use the equation \[\tan(a+b)=(\frac{ \tan(a)-\tan(b) }{ 1+\tan(a)\tan(b) })\]

  4. anonymous
    • one year ago
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    1

  5. freckles
    • one year ago
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    right so 17=12(1)+5 \[\tan(\frac{17}{12} \pi) \\ \tan([1+\frac{5}{12}]\pi) \\ \tan(\pi+\frac{5\pi}{12}) =\tan(\frac{5\pi}{12} ) \text{ since tangent function has period } \pi \] so now we are looking at tan(5pi/12) instead you said you hated the radian stuff so you can concert 5pi/12 to degrees if you want

  6. freckles
    • one year ago
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    \[\frac{5\pi}{12} \cdot \frac{180^o}{\pi} = \frac{5 }{12} \cdot 180^o =\frac{5 \cdot 180}{12}^o =5(15)^o=75^o \\ \text{ and } 75^o=30^o+45^o\]

  7. anonymous
    • one year ago
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    could I have done that with the 17pi/12 too or would that be too hard ?

  8. freckles
    • one year ago
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    \[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1-\tan(30^o)\tan(45^o)}\]

  9. freckles
    • one year ago
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    you could have

  10. freckles
    • one year ago
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    \[\frac{17 \pi}{12} \cdot \frac{180^o}{\pi}=17(15)^o=255^o \\ \text{ and then instead use } \\ 255^o=225^o+30^o\]

  11. freckles
    • one year ago
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    or you could have chosen 120+135 for 255

  12. freckles
    • one year ago
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    there or other choices you just want to choose the ones that are on the unit circle

  13. anonymous
    • one year ago
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    do I change the degrees into the points ? Like from the unit circle

  14. freckles
    • one year ago
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    you mean change tan(30) and tan(45) using unit circle? if so yes

  15. anonymous
    • one year ago
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    yeah , that's what I mean . And sin tan is sin/cos it'd be tan(30) = \[\frac{ \frac{ 1 }{ 2 } }{ \frac{ \sqrt{3} }{ 2 } }\]

  16. freckles
    • one year ago
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    thats right

  17. freckles
    • one year ago
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    which can be written as 1/sqrt(3) or sqrt(3)/3

  18. anonymous
    • one year ago
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    SO I'd multiply the denominator and numerator by 2/rad(3) ?

  19. anonymous
    • one year ago
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    to get that

  20. freckles
    • one year ago
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    yep

  21. freckles
    • one year ago
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    and you can rationalize the denominator if you prefer which is how I got also or sqrt(3)/3 but you can leave as 1/sqrt(3) for now if you prefer

  22. freckles
    • one year ago
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    and tan(45)=?

  23. anonymous
    • one year ago
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    it'd be 1 right ?

  24. freckles
    • one year ago
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    yes tan(45) is 1 \[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1-\tan(30^o)\tan(45^o)} \\ \text{ inserting values we just found: } \\ \tan(30^o+45^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}(1)} \\ \tan(75^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}}\] or we could say: \[\tan(\frac{17\pi}{12})=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}}\]

  25. freckles
    • one year ago
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    so your teacher probably doesn't want you to leave your answer in compound fraction from

  26. freckles
    • one year ago
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    that is tiny fractions inside a big fraction

  27. freckles
    • one year ago
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    you can chose to clear the "tiny fractions" by multiply the denominator and numerator of the big fraction by the denominators of the "tiny fractions" so you only need to get rid of the compound action by multiply top and bottom of main fraction by sqrt(3)

  28. freckles
    • one year ago
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    \[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)-\sqrt{3}(\frac{1}{\sqrt{3}})} \]

  29. freckles
    • one year ago
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    \[\sqrt{3}(1)=\sqrt{3} \\ \text{ and } \\ \sqrt{3} \cdot \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3}}=1\]

  30. anonymous
    • one year ago
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    couldn't I multiply by 1+sqrt 3 ?

  31. anonymous
    • one year ago
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    and then use foil

  32. freckles
    • one year ago
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    wait you mean 1+1/sqrt(3)?

  33. anonymous
    • one year ago
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    I rationalized the denominator to just have the sqrt 3

  34. freckles
    • one year ago
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    we can rationalize the denominator after what I have... or if you prefer to rationalize first then get rid of the compound fraction action you may do that but you will want to multiply top and bottom by the conjugate of what we have on the bottom which is going to be 1+1/sqrt(3)

  35. anonymous
    • one year ago
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    What I have right now is \[\frac{ 1+\sqrt{3} }{ 1-\sqrt{3} }\]

  36. freckles
    • one year ago
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    that is close to what you should have your bottom should be sqrt(3)-1

  37. anonymous
    • one year ago
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    why is -1 ?

  38. anonymous
    • one year ago
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    why is it *

  39. freckles
    • one year ago
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    tan(30)=1/sqrt(3) and tan(45)=1 \[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1-\tan(30^o)\tan(45^o)} \\ \text{ inserting values we just found: } \\ \tan(30^o+45^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}(1)} \\ \tan(75^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \] \[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)-\sqrt{3}(\frac{1}{\sqrt{3}})} \\ \tan(\frac{17\pi}{12})=\frac{1+\sqrt{3}}{\sqrt{3}-1}\]

  40. freckles
    • one year ago
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    into rationalize the denominator we multiply top and bottom by sqrt(3)+1

  41. anonymous
    • one year ago
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    oooh okay , I rationalized the 1/sqrt 3 before plugging it all in

  42. freckles
    • one year ago
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    ok and you winded up with sqrt(3)/3 when you did that right?

  43. freckles
    • one year ago
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    \[\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{1 \cdot \sqrt{3}}{ \sqrt{3} \cdot \sqrt{3}}=\frac{\sqrt{3}}{3}\]

  44. freckles
    • one year ago
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    this does not equal just sqrt(3)

  45. anonymous
    • one year ago
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    lol oops , I just canceled them out

  46. freckles
    • one year ago
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    \[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)-\sqrt{3}(\frac{1}{\sqrt{3}})} \\ \tan(\frac{17\pi}{12})=\frac{1+\sqrt{3}}{\sqrt{3}-1} \] so basically one last step you multiply top and bottom by sqrt(3)+1

  47. anonymous
    • one year ago
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    \[\sqrt{3}+\sqrt{3} = 2\sqrt{3}\] ??

  48. anonymous
    • one year ago
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    so it would be \[2+\sqrt{3}\]

  49. anonymous
    • one year ago
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    for the final answer

  50. freckles
    • one year ago
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    \[\frac{1+\sqrt{3}}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{(1+\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3})^2-1}\] \[\frac{1+\sqrt{3}+\sqrt{3}+3}{3-1}=\frac{4+2 \sqrt{3}}{2}= \frac{4}{2}+\frac{2}{2} \sqrt{3} \\ =2+1 \sqrt{3}=2+\sqrt{3}\] yes you are right

  51. anonymous
    • one year ago
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    thank you ! I get this a lot more than the last problems lmao

  52. freckles
    • one year ago
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    cool stuff!:)

  53. anonymous
    • one year ago
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    what about sec(pi/8) ?

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