Maths Olympiad Problem

[Gotcha]

The following is a problem I got in a Maths Olympiad, I had to solve it without a calculator, although I couldn't solve it:

sin 1 + sin 2 + sin 3 + ... + sin 90

If anyone could show me how to solve this I would really appreciate it.

[matt grime]

write sin(n) = sin(1)cos(n-1)+cos(1)sin(n-1), and use the fact that sin(x) = cos(90-x) and you're done.

[3.14lwy]

I don't know will this method work ,

but i will try to find

(sin 1 + sin 2 + sin 3 + ... + sin 90)(cos 1)

= (sin 1)(cos 1) + (sin 2)(cos 1) + (sin 3)(cos 1) + ... + (sin 90)(cos 1)


or (sin 1 + sin 2 + sin 3 + ... + sin 90)(sin 1)

this is just a suggection , may not work. :smile:

[TenaliRaman]

Or simply use euler's rule
and u r done

-- AI

[aeroegnr]

There is another way to do this, but you have to know how to use complex numbers, series, etc. It gives you a general formula.

Basically, you can say that this is a sum from 1 to N of sin(n). This is hard in itself to evaluate, so you use Euler's formula where e^(i*x)=cos(x)+i*sin(x). You then use the sum from 1 to N of e^(i*n*x) which is a geometric series. The nth sum is then equal to (1-e^iN*x)/(1-e^i*x).

If you can get this far, figure out a way to rewrite the (1-e^iN*x)/(1-e^i*x) so that it looks like a different function. If you have no idea what I'm talking about (I don't know what level math olympiad is at), I don't know method without using complex numbers/series/functions.

[uart]

The following is a problem I got in a Maths Olympiad, I had to solve it without a calculator, although I couldn't solve it:

sin 1 + sin 2 + sin 3 + ... + sin 90

If anyone could show me how to solve this I would really appreciate it.

Are you sure they didn't ask you merely to approximate it rather than actually solve it exactly?

I've got a feeling that they are expecting you to realize that it is a close approximation to,
180/\pi \, \int_{0}^{\pi/2}\sin(x) \,dx = 180/\pi

or numerically about 57.

[uart]

Actually that integral (above) is an ok approximation but not all that good because it's just an upper sum. It would be a lot closer if you averaged the upper and lower sums as in

Integral ~= 1/2 { (sin 0 + sin 1 + ... sin 89) + (sin 1 + sin 2 + ..sin 90) }

= 1/2 { S -1 + S} = S - 1/2

So a better approximation to the given Sum (S) would be,

S \simeq 180/\pi \, \int_{0}^{\pi/2}\sin(x) \,dx + 1/2 = 180/\pi +1/2

[devious_]

write sin(n) = sin(1)cos(n-1)+cos(1)sin(n-1), and use the fact that sin(x) = cos(90-x) and you're done.
I don't really understand how you arrived at that, and how it'd help solve the problem. Can you please elaborate?

[Gotcha]

All of this stuff is great, thanks guys!
This still doesn't solve the problem without the use of a calculator ...
except if you can calculate the sines of numbers in your head!

[matt grime]

For other people who might not be aware, maths olympiad questions are multiple choice.

Perhaps if you posted the options we might be able to help you decide which decimal answer was the most likely.

I'd presumed that the possible answers were in exact forms.

As for devious:

sin(n)=sin(n-1+1) = sin(n-1)cos(1) + sin(1)cos(n-1)


Let S be the original sum, then

S= sin(1)(1+cos(1)+cos(2)+cos(3)+...+cos(89)) + cos(1)(sin(1)+sin(2)+..+sin(89))

S=sin(1)S + cos(1)S - cos(1)

since the sums of the cos's is the same as the sum of the sin's.
from which we can find S.

[Gotcha]

This was a 3rd round maths olympiad, and it wasn't multiple choice.

[matt grime]

Ah, sorry. thought it was one of the first two.

DId it explicitly ask you to work out the decimal expansion of the sum, or evaluate it? surely an exact solution involving sin(1) and cos(1) is far better than any decimal expansion? This is the Olympiad, so I'd presume it uses proper mathematical standards ie an answer of 2pi is written and not "6.28 to 2 dp", and hence that the solutions offered here which are 'exact' constitute the one that they wanted.

[aeroegnr]

I'm looking at this and I think that the way to handle this is the way I above mentioned.

If you follow the steps above, you get the value (1-e^iN*x)/(1-e^i*x).

This can then be factored into (you take e^(iN*x/2) out of the top and e^(i*x/2) out of the bottom)

e^( iN*x/2 ) * ( e^(-iN*x/2) - e^(iN*x/2) ) / ( e^(ix/2) * ( e^(-ix/2) - e^(ix/2) )

Recognize that e^(-iN*x/2) - e^(iN*x/2) is = -2i*sin(Nx/2) and that e^(-ix/2) - e^(ix/2) = -2i*sin(x/2)

Also, this was a combined expansion of cos(nx)+isin(nx) so the value for sin(nx) will be the imaginary part of the above.

You get:

sum sin(nx)= sin(Nx/2)/sin(x/2) * sin((N-1)x/2)
Plugging in N=90 and x=1, you get the value 56.7943 (using degrees not radians), the same as mentioned above, and you now have a general formula for all problems of this type.

[TenaliRaman]

I don't think the value needs to evaluated ....
if one gets sin(pi/35) as an answer , thats an answer enough .......
(if u catch my drift...)

-- AI

[devious_]

S=sin(1)S + cos(1)S - cos(1)
from which we can find S.
I don't really get this one. :shy:
Shouldn't it be: S= sin(1)S + cos(1)S + sin(1)?

[Salt]

I don't really get this one. :shy:
Shouldn't it be: S= sin(1)S + cos(1)S + sin(1)?

cos(1)[sin(1)+sin(2)+..+sin(89)]
=cos(1)[sin(1)+sin(2)+..+sin(89)+ sin(90) - sin(90)]
=cos(1)[S - sin(90)]
=cos(1)[S - 1]
=cos(1)S - cos(1)

if I'm right
:tongue2:

[devious_]

cos(1)[sin(1)+sin(2)+..+sin(89)]
=cos(1)[sin(1)+sin(2)+..+sin(89)+ sin(90) - sin(90)]
=cos(1)[S - sin(90)]
=cos(1)[S - 1]
=cos(1)S - cos(1)

if I'm right
:tongue2:
You are. I messed up some stuff when I typed it out. :approve: