nerdsforever
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Re: 2009 3U Maths Marathon
solve: sin3x + sin2x = sinx
0 << x << 360
solve: sin3x + sin2x = sinx
0 << x << 360
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nerdsforever
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Re: 2009 3U Maths Marathon
lol, x = 0, 60, 180, 300, 360.
lol, x = 0, 60, 180, 300, 360.
youngminii
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Re: 2009 3U Maths Marathon
I deleted my post. Can't be bothered to think about the question..
Wait was I wrong in the first place? o.o
LOL whoops?
I deleted my post. Can't be bothered to think about the question..
Wait was I wrong in the first place? o.o
nerdsforever
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Re: 2009 3U Maths Marathon
1) An arithmetic series is 6.5, 9, 11.5, 14, 16.5 + .........
Find the sum of the first 500 terms.
1) An arithmetic series is 6.5, 9, 11.5, 14, 16.5 + .........
Find the sum of the first 500 terms.
Re: 2009 3U Maths Marathon
=1/2*500(2(6.5)+(599)(2.5))
=377625
Solve abs((x+1)/(x-2)) >2
Sum=1/2*n*(2a+(n-1)d)
=1/2*500(2(6.5)+(599)(2.5))
=377625
Solve abs((x+1)/(x-2)) >2
gurmies
Drover
Re: 2009 3U Maths Marathon
 = x_{1}-\overline{x}+x_{2}-\overline{x}+...+x_{n}-\overline{x} \,\, so \,\, there \,\, are \,\, n \,\, \overline{x}s \,\, as \,\, there \,\, are \,\, n \,\, terms.)
}{n})
 = 0)
1 ≤ x ≤ 5, x ≠ 2
Define the arithmetic mean of a set of numbers x1, x2, ...., xn as:
Show that:
 = 0)
Last edited:
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Might've have been quicker to do:
Question (relatively straightforward):
Recall that an object exhibits simple harmonic motion if:
Show that if a displacement x is a function consisting of a sum of sines and cosines of equal periods then it is simple harmonic, i.e. prove the following expression describes simple harmonic motion:
 + \sum_{j=1}^r \cos(nt+\alpha_j) )
Question (relatively straightforward):
Recall that an object exhibits simple harmonic motion if:
Show that if a displacement x is a function consisting of a sum of sines and cosines of equal periods then it is simple harmonic, i.e. prove the following expression describes simple harmonic motion:
Last edited:
Dragonmaster262
Unorthodox top student
Re: 2009 3U Maths Marathon
Air Jordan
Member
Re: 2009 3U Maths Marathon
is it 3.5?
is it 3.5?
In answer to Trebla's question:
[maths]x=\sum_{i=1}^{k}\sin(nt+\alpha _i)+\sum_{j=1}^{r}\cos(nt+\alpha_j)[/maths]
[maths]\frac{dx}{dt}=n\sum_{i=1}^{k}\cos(nt+\alpha _i)-n\sum_{j=1}^{r}\sin(nt+\alpha_j)[/maths]
[maths]\frac{d^2x}{dt^2}=-n^2(\sum_{i=1}^{k}\sin(nt+\alpha _i)+\sum_{j=1}^{r}\cos(nt+\alpha_j))[/maths]
[maths]=-n^2x[/maths]
[maths]x=\sum_{i=1}^{k}\sin(nt+\alpha _i)+\sum_{j=1}^{r}\cos(nt+\alpha_j)[/maths]
[maths]\frac{dx}{dt}=n\sum_{i=1}^{k}\cos(nt+\alpha _i)-n\sum_{j=1}^{r}\sin(nt+\alpha_j)[/maths]
[maths]\frac{d^2x}{dt^2}=-n^2(\sum_{i=1}^{k}\sin(nt+\alpha _i)+\sum_{j=1}^{r}\cos(nt+\alpha_j))[/maths]
[maths]=-n^2x[/maths]
studentcheese
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Nice and simple:
cos(@ + a) = cos@cosa - sin@sina
Derive the result for sin(@-a) using the above result.
cos(@ + a) = cos@cosa - sin@sina
Derive the result for sin(@-a) using the above result.
duckcowhybrid
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Replace sin(@-a) with cos (90-(@-a)) and solve.
gurmies
Drover
I think since it said use the result above, they want the addition. What I mean is:
90 - (@ - a) = 90 + (a-@)