Mixing
Computing $sum_{k=0}^n{2nchoose 2k}(-1)^ksin^{2k}theta cos^{2n-2k}theta$ using Euler's formula
$begingroup$
Compute the following sum by using Euler's formula, $
e^{i theta} = cos theta + i sin theta$,
$$cos^{2n}theta-{2nchoose 2}cos^{2n-2}theta sin^2theta +...+(-1)^{n-1}{2nchoose 2n-2}cos^2theta sin^{2n-2}theta +(-1)^nsin^{2n}theta$$
I have tried to rewrite the expression as:
$$sum_{k=0}^n{2nchoose 2k}(-1)^ksin^{2k}theta cos^{2n-2k}theta$$
But I have no certain idea about how to continue. Could you give me some hints? Thanks!
trigonometry complex-numbers
asked Jan 22 at 18:14
GibbsGibbs
130111
$endgroup$
add a comment |
$begingroup$
Compute the following sum by using Euler's formula, $
e^{i theta} = cos theta + i sin theta$,
$$cos^{2n}theta-{2nchoose 2}cos^{2n-2}theta sin^2theta +...+(-1)^{n-1}{2nchoose 2n-2}cos^2theta sin^{2n-2}theta +(-1)^nsin^{2n}theta$$
I have tried to rewrite the expression as:
$$sum_{k=0}^n{2nchoose 2k}(-1)^ksin^{2k}theta cos^{2n-2k}theta$$
But I have no certain idea about how to continue. Could you give me some hints? Thanks!
trigonometry complex-numbers
asked Jan 22 at 18:14
GibbsGibbs
130111
$endgroup$
add a comment |
$begingroup$
Compute the following sum by using Euler's formula, $
e^{i theta} = cos theta + i sin theta$,
$$cos^{2n}theta-{2nchoose 2}cos^{2n-2}theta sin^2theta +...+(-1)^{n-1}{2nchoose 2n-2}cos^2theta sin^{2n-2}theta +(-1)^nsin^{2n}theta$$
I have tried to rewrite the expression as:
$$sum_{k=0}^n{2nchoose 2k}(-1)^ksin^{2k}theta cos^{2n-2k}theta$$
But I have no certain idea about how to continue. Could you give me some hints? Thanks!
trigonometry complex-numbers
asked Jan 22 at 18:14
GibbsGibbs
130111
$endgroup$
Compute the following sum by using Euler's formula, $
e^{i theta} = cos theta + i sin theta$,
$$cos^{2n}theta-{2nchoose 2}cos^{2n-2}theta sin^2theta +...+(-1)^{n-1}{2nchoose 2n-2}cos^2theta sin^{2n-2}theta +(-1)^nsin^{2n}theta$$
I have tried to rewrite the expression as:
$$sum_{k=0}^n{2nchoose 2k}(-1)^ksin^{2k}theta cos^{2n-2k}theta$$
But I have no certain idea about how to continue. Could you give me some hints? Thanks!
trigonometry complex-numbers
trigonometry complex-numbers
asked Jan 22 at 18:14
GibbsGibbs
130111
asked Jan 22 at 18:14
GibbsGibbs
130111
asked Jan 22 at 18:14
GibbsGibbs
130111
asked Jan 22 at 18:14
GibbsGibbs
130111
asked Jan 22 at 18:14
GibbsGibbs
130111
130111
add a comment |
add a comment |
1 Answer
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$(-1)^ksin^{2k}t=(isin t)^{2k}$
$$(cos t+isin t)^{2n}+(cos t-isin t)^{2n}=?$$
answered Jan 22 at 18:18
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
Well, that's $(e^{it})^{2n}+(e^{-it})^{2n}$... but I don't see clearly what I am supposed to do.
$endgroup$
– Gibbs
Jan 22 at 18:53
$begingroup$
@Gibbs, math.stackexchange.com/questions/1432568/…
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– lab bhattacharjee
Jan 22 at 18:58
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
$(-1)^ksin^{2k}t=(isin t)^{2k}$
$$(cos t+isin t)^{2n}+(cos t-isin t)^{2n}=?$$
answered Jan 22 at 18:18
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
Well, that's $(e^{it})^{2n}+(e^{-it})^{2n}$... but I don't see clearly what I am supposed to do.
$endgroup$
– Gibbs
Jan 22 at 18:53
$begingroup$
@Gibbs, math.stackexchange.com/questions/1432568/…
$endgroup$
– lab bhattacharjee
Jan 22 at 18:58
add a comment |
$begingroup$
$(-1)^ksin^{2k}t=(isin t)^{2k}$
$$(cos t+isin t)^{2n}+(cos t-isin t)^{2n}=?$$
answered Jan 22 at 18:18
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
Well, that's $(e^{it})^{2n}+(e^{-it})^{2n}$... but I don't see clearly what I am supposed to do.
$endgroup$
– Gibbs
Jan 22 at 18:53
$begingroup$
@Gibbs, math.stackexchange.com/questions/1432568/…
$endgroup$
– lab bhattacharjee
Jan 22 at 18:58
add a comment |
$begingroup$
$(-1)^ksin^{2k}t=(isin t)^{2k}$
$$(cos t+isin t)^{2n}+(cos t-isin t)^{2n}=?$$
answered Jan 22 at 18:18
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$(-1)^ksin^{2k}t=(isin t)^{2k}$
$$(cos t+isin t)^{2n}+(cos t-isin t)^{2n}=?$$
answered Jan 22 at 18:18
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 22 at 18:18
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 22 at 18:18
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 22 at 18:18
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
$begingroup$
Well, that's $(e^{it})^{2n}+(e^{-it})^{2n}$... but I don't see clearly what I am supposed to do.
$endgroup$
– Gibbs
Jan 22 at 18:53
$begingroup$
@Gibbs, math.stackexchange.com/questions/1432568/…
$endgroup$
– lab bhattacharjee
Jan 22 at 18:58
add a comment |
$begingroup$
Well, that's $(e^{it})^{2n}+(e^{-it})^{2n}$... but I don't see clearly what I am supposed to do.
$endgroup$
– Gibbs
Jan 22 at 18:53
$begingroup$
@Gibbs, math.stackexchange.com/questions/1432568/…
$endgroup$
– lab bhattacharjee
Jan 22 at 18:58
$begingroup$
Well, that's $(e^{it})^{2n}+(e^{-it})^{2n}$... but I don't see clearly what I am supposed to do.
$endgroup$
– Gibbs
Jan 22 at 18:53
$begingroup$
Well, that's $(e^{it})^{2n}+(e^{-it})^{2n}$... but I don't see clearly what I am supposed to do.
$endgroup$
– Gibbs
Jan 22 at 18:53
$begingroup$
@Gibbs, math.stackexchange.com/questions/1432568/…
$endgroup$
– lab bhattacharjee
Jan 22 at 18:58
$begingroup$
@Gibbs, math.stackexchange.com/questions/1432568/…
$endgroup$
– lab bhattacharjee
Jan 22 at 18:58
add a comment |
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