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Normal of trigonometry points

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data
    x = (-4[tex]\pi[/tex]sin[tex]\pi[/tex]t, 4[tex]\pi[/tex]cos[tex]\pi[/tex]t, 1)
    what is the length of x?

    2. Relevant equations

    3. The attempt at a solution
    Well, I do realize that length of something is calculated by sqrt(x^2 + y^2) etc.
    However, when I plug in the numbers above, I get sqrt(16[tex]\pi[/tex]^3 + 1), but the answer is sqrt[1 + (4[tex]\pi[/tex])^2)]. If anyone can show me step by step (trig is my weak point), I'd really appreciate it. I'm guessing (sin^2[tex]\pi[/tex]t + cos^2[tex]\pi[/tex]t) = 1, instead of [tex]\pi[/tex]? If so, why?
  2. jcsd
  3. Nov 5, 2009 #2


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    Science Advisor

    You should have learned long ago that [itex]sin^2(x)+ cos^2(x)= 1[/itex] for all x, even if [itex]x= \pi t[/itex]!
  4. Nov 5, 2009 #3


    Staff: Mentor

    Yes, sin2A + cos2A = 1 for any A. This is the only trig in this problem. The rest is algebra, which also seems to be a weak point.

    How did you get sqrt(16[itex]\pi[/itex]3 + 1)? Show what you did to get this and we can set you straight.
  5. Nov 5, 2009 #4


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    I do hope you are NOT arguing that you can take a "[itex]\pi[/itex] out of the function and saying that:

    "[itex]sin^2(\pi t)+ cos^2(\pi t)= (\pi sin^2(t))^2+ (\pi cos^2(t))^2[/itex][itex]= \pi^2[/itex]".

    sin(at) is NOT equal to a sin(t) and cos(at) is NOT equal to a cos(t)!
  6. Nov 5, 2009 #5
    lol yeah, been away from math for a long time so did forget it. don't worry about 16pi^3, it was done quickly and obv wrong.
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