What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?
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What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?
I can use De Moivre's formula for the left part:
$[cos(pi/12)+isin(pi/12)]^16 = cos(4pi/3) + isin(4pi/3) = -dfracsqrt32 + dfraci2$
but I'm stuck at the right part. Thanks in advance.
complex-analysis analysis complex-numbers
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up vote
2
down vote
favorite
What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?
I can use De Moivre's formula for the left part:
$[cos(pi/12)+isin(pi/12)]^16 = cos(4pi/3) + isin(4pi/3) = -dfracsqrt32 + dfraci2$
but I'm stuck at the right part. Thanks in advance.
complex-analysis analysis complex-numbers
Use math.stackexchange.com/questions/3510/â¦
â lab bhattacharjee
Aug 26 at 11:45
2
By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
â Henning Makholm
Aug 26 at 11:53
1
I can use De Moivre's formula for the left part
Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.
â dxiv
Aug 27 at 5:25
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?
I can use De Moivre's formula for the left part:
$[cos(pi/12)+isin(pi/12)]^16 = cos(4pi/3) + isin(4pi/3) = -dfracsqrt32 + dfraci2$
but I'm stuck at the right part. Thanks in advance.
complex-analysis analysis complex-numbers
What is $[cos(pi/12)+isin(pi/12)]^16+[cos(pi/12)-isin(pi/12)]^16$?
I can use De Moivre's formula for the left part:
$[cos(pi/12)+isin(pi/12)]^16 = cos(4pi/3) + isin(4pi/3) = -dfracsqrt32 + dfraci2$
but I'm stuck at the right part. Thanks in advance.
complex-analysis analysis complex-numbers
asked Aug 26 at 11:44
Surzilla
1087
1087
Use math.stackexchange.com/questions/3510/â¦
â lab bhattacharjee
Aug 26 at 11:45
2
By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
â Henning Makholm
Aug 26 at 11:53
1
I can use De Moivre's formula for the left part
Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.
â dxiv
Aug 27 at 5:25
add a comment |Â
Use math.stackexchange.com/questions/3510/â¦
â lab bhattacharjee
Aug 26 at 11:45
2
By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
â Henning Makholm
Aug 26 at 11:53
1
I can use De Moivre's formula for the left part
Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.
â dxiv
Aug 27 at 5:25
Use math.stackexchange.com/questions/3510/â¦
â lab bhattacharjee
Aug 26 at 11:45
Use math.stackexchange.com/questions/3510/â¦
â lab bhattacharjee
Aug 26 at 11:45
2
2
By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
â Henning Makholm
Aug 26 at 11:53
By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
â Henning Makholm
Aug 26 at 11:53
1
1
I can use De Moivre's formula for the left part
Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.â dxiv
Aug 27 at 5:25
I can use De Moivre's formula for the left part
Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.â dxiv
Aug 27 at 5:25
add a comment |Â
4 Answers
4
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4
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accepted
Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$
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2
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$$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$
as $(cos x+isin x)(cos x-isin x)=1$
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up vote
2
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$(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$
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0
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It's much simpler with the exponential notation:
beginalign
(cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
endalign
@dxiv: Fixed. Thanks for pointing it!
â Bernard
Aug 27 at 9:20
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$
add a comment |Â
up vote
4
down vote
accepted
Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$
Note that $$cos(pi/12)-isin(pi/12)=cos(-pi/12)+isin(-pi/12)$$
answered Aug 26 at 11:45
Suzet
2,421527
2,421527
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add a comment |Â
up vote
2
down vote
$$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$
as $(cos x+isin x)(cos x-isin x)=1$
add a comment |Â
up vote
2
down vote
$$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$
as $(cos x+isin x)(cos x-isin x)=1$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$
as $(cos x+isin x)(cos x-isin x)=1$
$$(cos y-isin y)^n=dfrac1(cos y+isin y)^n=dfrac1cos(ny)+isin(ny)=cos(ny)-isin(ny)$$
as $(cos x+isin x)(cos x-isin x)=1$
edited Aug 26 at 11:50
Lord Shark the Unknown
88.2k955115
88.2k955115
answered Aug 26 at 11:47
lab bhattacharjee
216k14153265
216k14153265
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add a comment |Â
up vote
2
down vote
$(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$
add a comment |Â
up vote
2
down vote
$(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$
$(cos y-isin y)^n=(cos(-y)+isin(-y))^n=cos(n(-y))+isin(n(-y))=cos(ny)-isin(ny)$
answered Aug 26 at 11:59
drhab
88.4k541120
88.4k541120
add a comment |Â
add a comment |Â
up vote
0
down vote
It's much simpler with the exponential notation:
beginalign
(cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
endalign
@dxiv: Fixed. Thanks for pointing it!
â Bernard
Aug 27 at 9:20
add a comment |Â
up vote
0
down vote
It's much simpler with the exponential notation:
beginalign
(cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
endalign
@dxiv: Fixed. Thanks for pointing it!
â Bernard
Aug 27 at 9:20
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It's much simpler with the exponential notation:
beginalign
(cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
endalign
It's much simpler with the exponential notation:
beginalign
(cospi/12+isinpi/12) ^16&+(cospi/12-isinpi/12 )^16\&=mathrm e^tfrac4ipi3+mathrm e^tfrac-4ipi3=2cosfrac4pi3=2cdotfrac12
endalign
edited Aug 27 at 9:19
answered Aug 26 at 12:30
Bernard
111k635102
111k635102
@dxiv: Fixed. Thanks for pointing it!
â Bernard
Aug 27 at 9:20
add a comment |Â
@dxiv: Fixed. Thanks for pointing it!
â Bernard
Aug 27 at 9:20
@dxiv: Fixed. Thanks for pointing it!
â Bernard
Aug 27 at 9:20
@dxiv: Fixed. Thanks for pointing it!
â Bernard
Aug 27 at 9:20
add a comment |Â
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Use math.stackexchange.com/questions/3510/â¦
â lab bhattacharjee
Aug 26 at 11:45
2
By the way, you're getting the cosine and sine wrong values in the last part of your computation. It should be $-frac12 - fracsqrt32i$.
â Henning Makholm
Aug 26 at 11:53
1
I can use De Moivre's formula for the left part
Call that $z$, then the right part is just $bar z$, so their sum is $z + bar z = 2 operatornameRe(z)$. But you need to fix the left part, first.â dxiv
Aug 27 at 5:25