Mixing
Trisection of an angle $ Z $ whose cosine is $ -frac{11}{16} $ with straightedge and a compass
$begingroup$
Suppose there exists an angle $Z$ such that $cos Z = -frac{11}{16}$.
Prove or disprove that such an angle can be trisected with a straightedge and a compass.
Well, we know that an angle is constructible if and only if its cosine is constructible. Therefore, we have to prove that $cos (frac{Z}{3})$ is constructible... However, I don't know what to do next.
abstract-algebra trigonometry geometric-construction
edited Jan 30 at 16:56
user549397
1,6591418
asked Nov 20 '14 at 4:49
LinaLina
311
$endgroup$
add a comment |
$begingroup$
Suppose there exists an angle $Z$ such that $cos Z = -frac{11}{16}$.
Prove or disprove that such an angle can be trisected with a straightedge and a compass.
Well, we know that an angle is constructible if and only if its cosine is constructible. Therefore, we have to prove that $cos (frac{Z}{3})$ is constructible... However, I don't know what to do next.
abstract-algebra trigonometry geometric-construction
edited Jan 30 at 16:56
user549397
1,6591418
asked Nov 20 '14 at 4:49
LinaLina
311
$endgroup$
$begingroup$
Refer "I.N. Herstein", Topics in Algebra, for a good discussion on straight edge and compass
$endgroup$
– Hirak
Nov 20 '14 at 4:54
add a comment |
$begingroup$
Suppose there exists an angle $Z$ such that $cos Z = -frac{11}{16}$.
Prove or disprove that such an angle can be trisected with a straightedge and a compass.
Well, we know that an angle is constructible if and only if its cosine is constructible. Therefore, we have to prove that $cos (frac{Z}{3})$ is constructible... However, I don't know what to do next.
abstract-algebra trigonometry geometric-construction
edited Jan 30 at 16:56
user549397
1,6591418
asked Nov 20 '14 at 4:49
LinaLina
311
$endgroup$
Suppose there exists an angle $Z$ such that $cos Z = -frac{11}{16}$.
Prove or disprove that such an angle can be trisected with a straightedge and a compass.
Well, we know that an angle is constructible if and only if its cosine is constructible. Therefore, we have to prove that $cos (frac{Z}{3})$ is constructible... However, I don't know what to do next.
abstract-algebra trigonometry geometric-construction
abstract-algebra trigonometry geometric-construction
edited Jan 30 at 16:56
user549397
1,6591418
asked Nov 20 '14 at 4:49
LinaLina
311
edited Jan 30 at 16:56
user549397
1,6591418
asked Nov 20 '14 at 4:49
LinaLina
311
edited Jan 30 at 16:56
user549397
1,6591418
edited Jan 30 at 16:56
user549397
1,6591418
edited Jan 30 at 16:56
user549397
1,6591418
1,6591418
asked Nov 20 '14 at 4:49
LinaLina
311
asked Nov 20 '14 at 4:49
LinaLina
311
asked Nov 20 '14 at 4:49
LinaLina
311
311
$begingroup$
Refer "I.N. Herstein", Topics in Algebra, for a good discussion on straight edge and compass
$endgroup$
– Hirak
Nov 20 '14 at 4:54
add a comment |
$begingroup$
Refer "I.N. Herstein", Topics in Algebra, for a good discussion on straight edge and compass
$endgroup$
– Hirak
Nov 20 '14 at 4:54
$begingroup$
Refer "I.N. Herstein", Topics in Algebra, for a good discussion on straight edge and compass
$endgroup$
– Hirak
Nov 20 '14 at 4:54
$begingroup$
Refer "I.N. Herstein", Topics in Algebra, for a good discussion on straight edge and compass
$endgroup$
– Hirak
Nov 20 '14 at 4:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Step 1:
$$cos(3z)=4cos^3z-3cos z$$
Step 2:
$$4c^3-3c+frac{11}{16}=frac1{16}(4c-1)(16c^2+4c-11)$$
answered Nov 20 '14 at 5:04
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
add a comment |
$begingroup$
From the addition theorems, find the cubic equation for $cosfrac13 Z$. Either it has a rational root or it is irreducible. In the latter case $cosfrac13Z$ is not constructible (because $3$ is not a power of $2$).
answered Nov 20 '14 at 4:58
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Step 1:
$$cos(3z)=4cos^3z-3cos z$$
Step 2:
$$4c^3-3c+frac{11}{16}=frac1{16}(4c-1)(16c^2+4c-11)$$
answered Nov 20 '14 at 5:04
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
add a comment |
$begingroup$
Step 1:
$$cos(3z)=4cos^3z-3cos z$$
Step 2:
$$4c^3-3c+frac{11}{16}=frac1{16}(4c-1)(16c^2+4c-11)$$
answered Nov 20 '14 at 5:04
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
add a comment |
$begingroup$
Step 1:
$$cos(3z)=4cos^3z-3cos z$$
Step 2:
$$4c^3-3c+frac{11}{16}=frac1{16}(4c-1)(16c^2+4c-11)$$
answered Nov 20 '14 at 5:04
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
Step 1:
$$cos(3z)=4cos^3z-3cos z$$
Step 2:
$$4c^3-3c+frac{11}{16}=frac1{16}(4c-1)(16c^2+4c-11)$$
answered Nov 20 '14 at 5:04
Mario CarneiroMario Carneiro
18.6k34090
answered Nov 20 '14 at 5:04
Mario CarneiroMario Carneiro
18.6k34090
answered Nov 20 '14 at 5:04
Mario CarneiroMario Carneiro
18.6k34090
answered Nov 20 '14 at 5:04
Mario CarneiroMario Carneiro
18.6k34090
18.6k34090
add a comment |
add a comment |
$begingroup$
From the addition theorems, find the cubic equation for $cosfrac13 Z$. Either it has a rational root or it is irreducible. In the latter case $cosfrac13Z$ is not constructible (because $3$ is not a power of $2$).
answered Nov 20 '14 at 4:58
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
From the addition theorems, find the cubic equation for $cosfrac13 Z$. Either it has a rational root or it is irreducible. In the latter case $cosfrac13Z$ is not constructible (because $3$ is not a power of $2$).
answered Nov 20 '14 at 4:58
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
From the addition theorems, find the cubic equation for $cosfrac13 Z$. Either it has a rational root or it is irreducible. In the latter case $cosfrac13Z$ is not constructible (because $3$ is not a power of $2$).
answered Nov 20 '14 at 4:58
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
From the addition theorems, find the cubic equation for $cosfrac13 Z$. Either it has a rational root or it is irreducible. In the latter case $cosfrac13Z$ is not constructible (because $3$ is not a power of $2$).
answered Nov 20 '14 at 4:58
Hagen von EitzenHagen von Eitzen
283k23273508
answered Nov 20 '14 at 4:58
Hagen von EitzenHagen von Eitzen
283k23273508
answered Nov 20 '14 at 4:58
Hagen von EitzenHagen von Eitzen
283k23273508
answered Nov 20 '14 at 4:58
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
add a comment |
add a comment |
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$begingroup$
Refer "I.N. Herstein", Topics in Algebra, for a good discussion on straight edge and compass
$endgroup$
– Hirak
Nov 20 '14 at 4:54