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anonymous

  • one year ago

find the exact value of sec(pi/8)

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  1. freckles
    • one year ago
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    \[\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}\]

  2. freckles
    • one year ago
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    1/2*pi/4 means half of pi/4 think half angle identity

  3. anonymous
    • one year ago
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    I'm confused where the half angle identity comes in

  4. freckles
    • one year ago
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    well we know pi/4 is on the unit circle and half of pi/4 is pi/8 so since pi/8 is in the first quadrant we know cos to be positive \[\cos(\frac{\pi}{8}) \\ =\cos(\frac{1}{2} \cdot \frac{\pi}{4})=\sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}}\]

  5. anonymous
    • one year ago
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    I haven't seen that equation before

  6. freckles
    • one year ago
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    \[\cos(\frac{\theta}{2})= \pm \sqrt{\frac{1+\cos(\theta)}{2}}\]? you haven't seen this one?

  7. freckles
    • one year ago
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    have you seen this one: \[\cos^2(\theta)=\frac{1}{2}(1+\cos(2 \theta))\]

  8. anonymous
    • one year ago
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    maybe I wrote it down for the wrong thing

  9. anonymous
    • one year ago
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    My notes are all wrong lol

  10. freckles
    • one year ago
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    http://www.purplemath.com/modules/idents.htm here are a lot of trig identities with the corresponding names

  11. freckles
    • one year ago
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    maybe you can use this to fix your notes

  12. anonymous
    • one year ago
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    \[\cos \frac{ x }{ 2 }=\sqrt{\frac{ 1+sinx }{ 2 }}\] that's what I have

  13. anonymous
    • one year ago
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    +- in the front

  14. freckles
    • one year ago
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    err yes that sin(x) thing is definitely suppose to be cos(x)

  15. anonymous
    • one year ago
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    Apex is terrible

  16. freckles
    • one year ago
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    \[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \]

  17. freckles
    • one year ago
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    there is still a couple things for you to do

  18. freckles
    • one year ago
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    find cos(pi/4) using unit circle

  19. freckles
    • one year ago
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    and simplify

  20. anonymous
    • one year ago
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    how'd you get cos pi/4 ?

  21. freckles
    • one year ago
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    x is pi/4

  22. freckles
    • one year ago
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    i replace x with pi/4

  23. freckles
    • one year ago
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    \[\frac{1}{2} \cdot \frac{\pi}{4} =\frac{\pi}{8} \\ \text{ so } x=\frac{\pi}{4}\]

  24. anonymous
    • one year ago
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    oooh okay

  25. anonymous
    • one year ago
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    cos pi/4 is sqrt2/2

  26. anonymous
    • one year ago
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    and 45 degrees

  27. freckles
    • one year ago
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    right so you can replace cos(pi/4) is sqrt(2)/2

  28. freckles
    • one year ago
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    well pi/4 is 45 deg

  29. freckles
    • one year ago
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    \[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}}\]

  30. anonymous
    • one year ago
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    yeah I didn't know if I needed the degrees or radians so I put both lol

  31. freckles
    • one year ago
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    you don't need either you just need to find cos(45) or cos(pi/4)

  32. freckles
    • one year ago
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    cos(45 deg) is equivalent to finding cos(pi/4)

  33. anonymous
    • one year ago
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    do I need to get rid of the radical or leave that there ?

  34. freckles
    • one year ago
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    I would get rid of the compound fraction action inside the radical and then see if there are any perfect squares afterwards

  35. anonymous
    • one year ago
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    so would I multiply top and bottom by the opposite of the denominator ?

  36. freckles
    • one year ago
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    \[\sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \cdot \sqrt{\frac{2}{2}} \\ \sec(\frac{\pi}{8}) = \sqrt{ \frac{2 (2)}{(1+\frac{\sqrt{2}}{2})(2)}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{4}{2+\sqrt{2}}}\] notice the numerator inside the squareroot is a perfect square

  37. freckles
    • one year ago
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    \[\sec(\frac{\pi}{8})=\frac{\sqrt{4}}{\sqrt{2+\sqrt{2}}}\]

  38. anonymous
    • one year ago
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    wouldn't the rad 2 or rad 2 equal 1 ?

  39. freckles
    • one year ago
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    yes sqrt(2/2)=1 that is why I multiplied it to get rid of the compound fraction action inside the square root

  40. anonymous
    • one year ago
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    I'm confused on the multiplying of 2 in the denominator and numerator

  41. freckles
    • one year ago
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    let's look inside the square root we have inside the square root: \[\frac{2}{1+\frac{\sqrt{2}}{2}}\] this has a mini-fraction the sqrt(2)/2 the sqrt(2)/2 has a 2 in the denominator I multiply top and bottom inside the square root by 2 so I have inside the square root: \[\frac{2}{1+\frac{\sqrt{2}}{2}} \cdot \frac{2}{2} \\ \frac{2(2)}{(1+\frac{\sqrt{2}}{2})(2)} \\ \frac{4}{1(2)+\frac{\sqrt{2}}{2}(2)} \\ \frac{4}{2+ \frac{2}{2} \sqrt{2}} \\ \\ \frac{4}{2+(1) \sqrt{2}} \\ \frac{4}{2+ \sqrt{2}}\]

  42. anonymous
    • one year ago
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    why did you transfer the radical to the two ?

  43. freckles
    • one year ago
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    what 2 ?

  44. freckles
    • one year ago
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    are you asking me about this: \[\frac{\sqrt{2}}{2}(2)=\frac{2}{2} \sqrt{2}\]

  45. anonymous
    • one year ago
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    this part

  46. freckles
    • one year ago
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    multiplication is commuative

  47. freckles
    • one year ago
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    \[\frac{\sqrt{2}}{2}(2) =\frac{\sqrt{2}}{2} \cdot \frac{2}{1} = \frac{\sqrt{2} \cdot 2 }{2 \cdot 1 } \\ \text{ you change order \in multiplication } \\ =\frac{2 \sqrt{2}}{2 \cdot 1} \\ \text{ you can reseperate the fraction } \\ =\frac{2}{2} \frac{\sqrt{2}}{1} \\ =1 \frac{\sqrt{2}}{1} \\ 1 \cdot \sqrt{2} \ = \sqrt{2}\]

  48. anonymous
    • one year ago
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    oooh okay , that makes sense

  49. freckles
    • one year ago
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    all I did was use the fact that a*b is the same as b*a

  50. anonymous
    • one year ago
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    okay, that makes a lot of sense

  51. anonymous
    • one year ago
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    for my final answer , I got \[-\sqrt{2}+2\] is that right ?

  52. freckles
    • one year ago
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    \[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \\ \] \[\sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \cdot \sqrt{\frac{2}{2}} \\ \sec(\frac{\pi}{8}) = \sqrt{ \frac{2 (2)}{(1+\frac{\sqrt{2}}{2})(2)}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{4}{2+\sqrt{2}}}\] \[\sec(\frac{\pi}{8})=\frac{\sqrt{4}}{\sqrt{2+\sqrt{2}}}\] \[\sec(\frac{\pi}{8})=\frac{2}{\sqrt{2+\sqrt{2}}}\]

  53. anonymous
    • one year ago
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    wouldn't the rad 2 cancel out since it's two radicals ?

  54. anonymous
    • one year ago
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    so it'd be \[\frac{ 2 }{ \sqrt{2}+2 }\]

  55. freckles
    • one year ago
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    no \[\sqrt{a+\sqrt{b}} \neq \sqrt{a}+b \\ \text{ take } a=4 \text{ and } b=16 \\ \text{ so } \sqrt{a}=2 \text{ and } \sqrt{b}=4 \\ \sqrt{a+\sqrt{b}}=\sqrt{4+4}=\sqrt{8} \\ \text{ while } \sqrt{a}+b=2+16=18 \\ \text{ and } 18 \text{ is definitely not equal to } \sqrt{8} \\ \text{ so } \sqrt{a+\sqrt{b}} \text{ is not } \sqrt{a}+b\]

  56. anonymous
    • one year ago
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    so it stays like that ?

  57. freckles
    • one year ago
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    it stays like \[\frac{2}{\sqrt{2+\sqrt{2}}}\] yes you could rationalize the bottom but it will still look ugly

  58. freckles
    • one year ago
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    so I would leave it as is

  59. anonymous
    • one year ago
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    I never learned that lol . I have to get ready for work now though , thank you for the help !

  60. freckles
    • one year ago
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    good luck

  61. anonymous
    • one year ago
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    If I post up a new question with this activity I have to do , do you think you could figure it out ? I tried but I couldn't get it . I got stuck

  62. anonymous
    • one year ago
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    The activities that are worth points in this packet are super confusing

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