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Calculating large sums without Calculator (with sin)

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    sin²(1°)+sin²(3°)+sin²(5°)+.......sin²(359°)= ?

    And : 1!+2!+3!+4!+.....+2006!, asked are he last two numbers of this sum.

    2. Relevant equations
    I don't know any

    3. The attempt at a solution
    I don't know how to calculate Sin with your head, and 2006! is way to hard to calculate.
    Is there a simple formula for these questions?
    Last edited: Oct 22, 2008
  2. jcsd
  3. Oct 22, 2008 #2
    For the first one, you switch suddenly from sin^2 to sin. Which is it?

    For the second, think about what happens to the last 2 digits in n! as n becomes large. As it turns out, the summation is quite simple.
  4. Oct 22, 2008 #3
    Sorry Its sin². The second sum ends with +2006!, not with n
  5. Oct 22, 2008 #4
    Recall that cos(x)=-cos(180-x). Can you perhaps express the sin^2 in another way? Remember your trig identities.

    I was just speaking in general since the result is not dependent on most of the numbers in the summation. Compute 15!, 16! etc. What are the last 2 digits?
    Last edited: Oct 22, 2008
  6. Oct 22, 2008 #5
    15! is 120 and 16! 136, so it seems that 136-120 is 16. So 2006!-2005! is 2006. However I still don't get how to get the last two numbers of the sum of all !'s until 2006.

    And with the Cos part, I don't really get what you mean.
    I just want to know how to calculate these sums without a calculator.
  7. Oct 22, 2008 #6
    I'm giving you hints, I'm not going to tell you how to do it. The point is to make YOU think!

    15! is definitely NOT 120. Think about this also: If you multiplied 5782389105 by 4 and looked only at the last 2 digits, is it really any different than looking at 05*4?

    Use the double angle trig identity for sin^2 and you'll see what I mean.
    Last edited: Oct 22, 2008
  8. Oct 22, 2008 #7

    I'm not English so I don't know what you mean with "Use the double angle trig identity for sin^2".

    Ok, so 2006! last two numbers are 00. But the problem is, I also have to add 1! up to 2005!
  9. Oct 22, 2008 #8
    sin^2(x) = (1-cos(2x))/2. Combined with the identity cos(x) = -cos(180-x), for example, sin^2(1)+sin^2(89) = (1-cos(2))/2+(1-cos(178)/2 = (1 - cos(2) + cos(2) + 1)/2 = 1.

    You will find that for much smaller factorials that this is true. 15! also has the last 2 digits of 00, for example. You can guarantee that the smallest number n such that the last 2 digits of n! are 0 that all numbers larger than n also have this property. Your task is to find the smallest n where this is true then you instantly know that all factorials larger than that do not contribute to the smallest 2 digits.

    KEEP IN MIND: Any digits other than the 2 smallest in any factorial do not concern you! For example, when you want to find the last 2 digits of 7427128 multiplied by 781276, you need only consider what is 28 multiplied by 76. This is an important idea in something called modular arithmetic.
    Last edited: Oct 22, 2008
  10. Oct 22, 2008 #9
    So the awnser to the sum of sin² is 45 right?
    the smallest n! for 00 is 11, so I just have to add the first 10!
    Thank you.
  11. Oct 22, 2008 #10
    You're on the right track but remember to sum from 1 to 359.
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