Mixing
closed form expression for $sin 10^o$?
$begingroup$
As we know that $sin 15^o$, $sin 30^o$,$sin 45^o$ have simple closed form expressions as these are multiples of 3, but i have never seen any simple closed form expression for $sin 10^o$ or simply sine for any non-multiple of 3, if there exists a closed form expression, do help me,
PS. I know $sin 10^o$ is solution of $8x^3-6x+1=0$ but i can't solve it as its too tedious.
Why is being multiple of 3 such a great thing for an angle???
trigonometry closed-form angle
asked Jan 29 at 8:17
mathaholicmathaholic
706
$endgroup$
|
show 7 more comments
$begingroup$
As we know that $sin 15^o$, $sin 30^o$,$sin 45^o$ have simple closed form expressions as these are multiples of 3, but i have never seen any simple closed form expression for $sin 10^o$ or simply sine for any non-multiple of 3, if there exists a closed form expression, do help me,
PS. I know $sin 10^o$ is solution of $8x^3-6x+1=0$ but i can't solve it as its too tedious.
Why is being multiple of 3 such a great thing for an angle???
trigonometry closed-form angle
asked Jan 29 at 8:17
mathaholicmathaholic
706
$endgroup$
1
$begingroup$
It's exactly the case for which you can not solve it without trigonometry.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:29
$begingroup$
@MichaelRozenberg but there does exist a cubic formula for solution of cubic equations
$endgroup$
– mathaholic
Jan 29 at 8:31
$begingroup$
It gives complex numbers and the trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:32
$begingroup$
The cubic equation is not THAT difficult to solve: en.wikipedia.org/wiki/…
$endgroup$
– Matti P.
Jan 29 at 8:33
$begingroup$
@Matti P But in our case the discriminant is negative and we get a trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:34
|
show 7 more comments
$begingroup$
As we know that $sin 15^o$, $sin 30^o$,$sin 45^o$ have simple closed form expressions as these are multiples of 3, but i have never seen any simple closed form expression for $sin 10^o$ or simply sine for any non-multiple of 3, if there exists a closed form expression, do help me,
PS. I know $sin 10^o$ is solution of $8x^3-6x+1=0$ but i can't solve it as its too tedious.
Why is being multiple of 3 such a great thing for an angle???
trigonometry closed-form angle
asked Jan 29 at 8:17
mathaholicmathaholic
706
$endgroup$
As we know that $sin 15^o$, $sin 30^o$,$sin 45^o$ have simple closed form expressions as these are multiples of 3, but i have never seen any simple closed form expression for $sin 10^o$ or simply sine for any non-multiple of 3, if there exists a closed form expression, do help me,
PS. I know $sin 10^o$ is solution of $8x^3-6x+1=0$ but i can't solve it as its too tedious.
Why is being multiple of 3 such a great thing for an angle???
trigonometry closed-form angle
trigonometry closed-form angle
asked Jan 29 at 8:17
mathaholicmathaholic
706
asked Jan 29 at 8:17
mathaholicmathaholic
706
asked Jan 29 at 8:17
mathaholicmathaholic
706
asked Jan 29 at 8:17
mathaholicmathaholic
706
asked Jan 29 at 8:17
mathaholicmathaholic
706
706
1
$begingroup$
It's exactly the case for which you can not solve it without trigonometry.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:29
$begingroup$
@MichaelRozenberg but there does exist a cubic formula for solution of cubic equations
$endgroup$
– mathaholic
Jan 29 at 8:31
$begingroup$
It gives complex numbers and the trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:32
$begingroup$
The cubic equation is not THAT difficult to solve: en.wikipedia.org/wiki/…
$endgroup$
– Matti P.
Jan 29 at 8:33
$begingroup$
@Matti P But in our case the discriminant is negative and we get a trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:34
|
show 7 more comments
1
$begingroup$
It's exactly the case for which you can not solve it without trigonometry.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:29
$begingroup$
@MichaelRozenberg but there does exist a cubic formula for solution of cubic equations
$endgroup$
– mathaholic
Jan 29 at 8:31
$begingroup$
It gives complex numbers and the trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:32
$begingroup$
The cubic equation is not THAT difficult to solve: en.wikipedia.org/wiki/…
$endgroup$
– Matti P.
Jan 29 at 8:33
$begingroup$
@Matti P But in our case the discriminant is negative and we get a trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:34
1
1
$begingroup$
It's exactly the case for which you can not solve it without trigonometry.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:29
$begingroup$
It's exactly the case for which you can not solve it without trigonometry.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:29
$begingroup$
@MichaelRozenberg but there does exist a cubic formula for solution of cubic equations
$endgroup$
– mathaholic
Jan 29 at 8:31
$begingroup$
@MichaelRozenberg but there does exist a cubic formula for solution of cubic equations
$endgroup$
– mathaholic
Jan 29 at 8:31
$begingroup$
It gives complex numbers and the trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:32
$begingroup$
It gives complex numbers and the trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:32
$begingroup$
The cubic equation is not THAT difficult to solve: en.wikipedia.org/wiki/…
$endgroup$
– Matti P.
Jan 29 at 8:33
$begingroup$
The cubic equation is not THAT difficult to solve: en.wikipedia.org/wiki/…
$endgroup$
– Matti P.
Jan 29 at 8:33
$begingroup$
@Matti P But in our case the discriminant is negative and we get a trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:34
$begingroup$
@Matti P But in our case the discriminant is negative and we get a trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:34
|
show 7 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Being a multiple of $3°$ isn't a great thing, as degrees are an arbitrary unit. It matters much more to be a small fraction of a full turn, as this leads to polynomial equations of a low degree.
E.g.
$$sin 3x=-4sin^3x+3sin x=0$$ leads to the well-known $$sinfracpi3=frac{sqrt3}2.$$
Some other fractions lead to closed-form expressions, but most others not, as explained by the Abel-Ruffini theorem. In particular, angle trisection (dividing by three) involves a cubic equation which is in general not solvable (the above case being an exception).
It is interesting to note that there are analytical formulas to solve cubics, but for some values of the coefficients (the so-called casus irreductibilis), the solution requires… trigonometric functions, with angle trisection, and you are circling in rounds.
answered Jan 29 at 8:31
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
Let's refer to Casus irreducibilis. Consider an equation
$$ax^2+bx^2+cx+d=0.$$
Then if you note
$$D=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
which is equal to $D=4cdot 8 cdot 216-27cdot64 = 5184>0$ in our case, you find that we're in casus irreducibilis case. Which means that you can't write the roots using only integer numbers and roots (square or cubic).
Conclusion: you can't write $sin 10^circ$ using integer numbers and roots.
answered Jan 29 at 8:41
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
I don't think that it's tedious.
Let $x=cosalpha.$
Thus, we need to solve $$4cos^3alpha-3cosalpha=-frac{1}{2}$$ or$$cos3alpha=-frac{1}{2},$$which gives $$3alpha=120^{circ}+360^{circ}k,$$ where $kinmathbb Z$ or
$$3alpha=-120^{circ}+360^{circ}k,$$ which for $k=0$ in the first sequence and for $k=1$ and $k=2$ in the second sequence gives:
$$x_1=cos40^{circ},$$
$$x_2=cos80^{circ}=sin10^{circ}$$ and
$$x_3=cos200^{circ}=-cos20^{circ}.$$
answered Jan 29 at 8:31
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Being a multiple of $3°$ isn't a great thing, as degrees are an arbitrary unit. It matters much more to be a small fraction of a full turn, as this leads to polynomial equations of a low degree.
E.g.
$$sin 3x=-4sin^3x+3sin x=0$$ leads to the well-known $$sinfracpi3=frac{sqrt3}2.$$
Some other fractions lead to closed-form expressions, but most others not, as explained by the Abel-Ruffini theorem. In particular, angle trisection (dividing by three) involves a cubic equation which is in general not solvable (the above case being an exception).
It is interesting to note that there are analytical formulas to solve cubics, but for some values of the coefficients (the so-called casus irreductibilis), the solution requires… trigonometric functions, with angle trisection, and you are circling in rounds.
answered Jan 29 at 8:31
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
Being a multiple of $3°$ isn't a great thing, as degrees are an arbitrary unit. It matters much more to be a small fraction of a full turn, as this leads to polynomial equations of a low degree.
E.g.
$$sin 3x=-4sin^3x+3sin x=0$$ leads to the well-known $$sinfracpi3=frac{sqrt3}2.$$
Some other fractions lead to closed-form expressions, but most others not, as explained by the Abel-Ruffini theorem. In particular, angle trisection (dividing by three) involves a cubic equation which is in general not solvable (the above case being an exception).
It is interesting to note that there are analytical formulas to solve cubics, but for some values of the coefficients (the so-called casus irreductibilis), the solution requires… trigonometric functions, with angle trisection, and you are circling in rounds.
answered Jan 29 at 8:31
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
Being a multiple of $3°$ isn't a great thing, as degrees are an arbitrary unit. It matters much more to be a small fraction of a full turn, as this leads to polynomial equations of a low degree.
E.g.
$$sin 3x=-4sin^3x+3sin x=0$$ leads to the well-known $$sinfracpi3=frac{sqrt3}2.$$
Some other fractions lead to closed-form expressions, but most others not, as explained by the Abel-Ruffini theorem. In particular, angle trisection (dividing by three) involves a cubic equation which is in general not solvable (the above case being an exception).
It is interesting to note that there are analytical formulas to solve cubics, but for some values of the coefficients (the so-called casus irreductibilis), the solution requires… trigonometric functions, with angle trisection, and you are circling in rounds.
answered Jan 29 at 8:31
Yves DaoustYves Daoust
131k676229
$endgroup$
Being a multiple of $3°$ isn't a great thing, as degrees are an arbitrary unit. It matters much more to be a small fraction of a full turn, as this leads to polynomial equations of a low degree.
E.g.
$$sin 3x=-4sin^3x+3sin x=0$$ leads to the well-known $$sinfracpi3=frac{sqrt3}2.$$
Some other fractions lead to closed-form expressions, but most others not, as explained by the Abel-Ruffini theorem. In particular, angle trisection (dividing by three) involves a cubic equation which is in general not solvable (the above case being an exception).
It is interesting to note that there are analytical formulas to solve cubics, but for some values of the coefficients (the so-called casus irreductibilis), the solution requires… trigonometric functions, with angle trisection, and you are circling in rounds.
answered Jan 29 at 8:31
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 8:31
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 8:31
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 8:31
Yves DaoustYves Daoust
131k676229
131k676229
add a comment |
add a comment |
$begingroup$
Let's refer to Casus irreducibilis. Consider an equation
$$ax^2+bx^2+cx+d=0.$$
Then if you note
$$D=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
which is equal to $D=4cdot 8 cdot 216-27cdot64 = 5184>0$ in our case, you find that we're in casus irreducibilis case. Which means that you can't write the roots using only integer numbers and roots (square or cubic).
Conclusion: you can't write $sin 10^circ$ using integer numbers and roots.
answered Jan 29 at 8:41
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
Let's refer to Casus irreducibilis. Consider an equation
$$ax^2+bx^2+cx+d=0.$$
Then if you note
$$D=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
which is equal to $D=4cdot 8 cdot 216-27cdot64 = 5184>0$ in our case, you find that we're in casus irreducibilis case. Which means that you can't write the roots using only integer numbers and roots (square or cubic).
Conclusion: you can't write $sin 10^circ$ using integer numbers and roots.
answered Jan 29 at 8:41
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
Let's refer to Casus irreducibilis. Consider an equation
$$ax^2+bx^2+cx+d=0.$$
Then if you note
$$D=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
which is equal to $D=4cdot 8 cdot 216-27cdot64 = 5184>0$ in our case, you find that we're in casus irreducibilis case. Which means that you can't write the roots using only integer numbers and roots (square or cubic).
Conclusion: you can't write $sin 10^circ$ using integer numbers and roots.
answered Jan 29 at 8:41
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
Let's refer to Casus irreducibilis. Consider an equation
$$ax^2+bx^2+cx+d=0.$$
Then if you note
$$D=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
which is equal to $D=4cdot 8 cdot 216-27cdot64 = 5184>0$ in our case, you find that we're in casus irreducibilis case. Which means that you can't write the roots using only integer numbers and roots (square or cubic).
Conclusion: you can't write $sin 10^circ$ using integer numbers and roots.
answered Jan 29 at 8:41
mathcounterexamples.netmathcounterexamples.net
27k22158
edited Jan 29 at 8:48
answered Jan 29 at 8:41
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 29 at 8:41
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 29 at 8:41
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
add a comment |
add a comment |
$begingroup$
I don't think that it's tedious.
Let $x=cosalpha.$
Thus, we need to solve $$4cos^3alpha-3cosalpha=-frac{1}{2}$$ or$$cos3alpha=-frac{1}{2},$$which gives $$3alpha=120^{circ}+360^{circ}k,$$ where $kinmathbb Z$ or
$$3alpha=-120^{circ}+360^{circ}k,$$ which for $k=0$ in the first sequence and for $k=1$ and $k=2$ in the second sequence gives:
$$x_1=cos40^{circ},$$
$$x_2=cos80^{circ}=sin10^{circ}$$ and
$$x_3=cos200^{circ}=-cos20^{circ}.$$
answered Jan 29 at 8:31
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
I don't think that it's tedious.
Let $x=cosalpha.$
Thus, we need to solve $$4cos^3alpha-3cosalpha=-frac{1}{2}$$ or$$cos3alpha=-frac{1}{2},$$which gives $$3alpha=120^{circ}+360^{circ}k,$$ where $kinmathbb Z$ or
$$3alpha=-120^{circ}+360^{circ}k,$$ which for $k=0$ in the first sequence and for $k=1$ and $k=2$ in the second sequence gives:
$$x_1=cos40^{circ},$$
$$x_2=cos80^{circ}=sin10^{circ}$$ and
$$x_3=cos200^{circ}=-cos20^{circ}.$$
answered Jan 29 at 8:31
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
I don't think that it's tedious.
Let $x=cosalpha.$
Thus, we need to solve $$4cos^3alpha-3cosalpha=-frac{1}{2}$$ or$$cos3alpha=-frac{1}{2},$$which gives $$3alpha=120^{circ}+360^{circ}k,$$ where $kinmathbb Z$ or
$$3alpha=-120^{circ}+360^{circ}k,$$ which for $k=0$ in the first sequence and for $k=1$ and $k=2$ in the second sequence gives:
$$x_1=cos40^{circ},$$
$$x_2=cos80^{circ}=sin10^{circ}$$ and
$$x_3=cos200^{circ}=-cos20^{circ}.$$
answered Jan 29 at 8:31
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
I don't think that it's tedious.
Let $x=cosalpha.$
Thus, we need to solve $$4cos^3alpha-3cosalpha=-frac{1}{2}$$ or$$cos3alpha=-frac{1}{2},$$which gives $$3alpha=120^{circ}+360^{circ}k,$$ where $kinmathbb Z$ or
$$3alpha=-120^{circ}+360^{circ}k,$$ which for $k=0$ in the first sequence and for $k=1$ and $k=2$ in the second sequence gives:
$$x_1=cos40^{circ},$$
$$x_2=cos80^{circ}=sin10^{circ}$$ and
$$x_3=cos200^{circ}=-cos20^{circ}.$$
answered Jan 29 at 8:31
Michael RozenbergMichael Rozenberg
109k1896201
edited Jan 29 at 11:10
answered Jan 29 at 8:31
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 29 at 8:31
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 29 at 8:31
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
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1
$begingroup$
It's exactly the case for which you can not solve it without trigonometry.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:29
$begingroup$
@MichaelRozenberg but there does exist a cubic formula for solution of cubic equations
$endgroup$
– mathaholic
Jan 29 at 8:31
$begingroup$
It gives complex numbers and the trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:32
$begingroup$
The cubic equation is not THAT difficult to solve: en.wikipedia.org/wiki/…
$endgroup$
– Matti P.
Jan 29 at 8:33
$begingroup$
@Matti P But in our case the discriminant is negative and we get a trigonometry again.
$endgroup$
– Michael Rozenberg
Jan 29 at 8:34