Calculating large sums without Calculator (with sin)

[pivu0]

Homework Statement

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

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

Homework Equations

I don't know any

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?

 

[jhicks]

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.

 

[pivu0]

Sorry Its sin². The second sum ends with +2006!, not with n

 

[jhicks]

Sorry Its sin²

Recall that cos(x)=-cos(180-x). Can you perhaps express the sin^2 in another way? Remember your trig identities.

The second sum ends with +2006!, not with n

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?

 

[pivu0]

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?

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.

 

[jhicks]

I just want to know how to calculate these sums without a calculator.

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 120 and 16! 136, so it seems that 136-120 is 16. So 2006!-2005!

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?

And with the Cos part, I don't really get what you mean.

Use the double angle trig identity for sin^2 and you'll see what I mean.

 

[pivu0]

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.

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!

 

[jhicks]

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

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.

Ok, so 2006! last two numbers are 00. But the problem is, I also have to add 1! up to 2005!

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.

 

[pivu0]

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.

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.

 

[jhicks]

So the awnser to the sum of sin² is 45 right?

You're on the right track but remember to sum from 1 to 359.