Step to prove that $cos (n arccos (x))$ is a polynomial of $n$-th degreeProve that $cos (n arccos (x))$ is a polynomial of $n$-th degreeQuestion about chebyshev polynomialProve that $sin(2arcsin x + arccos x) = sqrt1-x^2$Find the least degree Polynomial whose one of the roots is $ cos(12^circ)$Finding Imaginary Values of a Degree 6 PolynomialRewriting $sin(arccos(y))$ and $arcsin(cos(x))$Show that, $arccosleft(fracacright)-arccosleft(fracbcright)=…$Evaluating $arccos(cosfrac15pi 4)$Prove $ frac12 (arccos(x) - arccos(-x)) = -arcsin(x)$How to prove $sin(x)arccos(x) + xcos(x) > 0$$cos(x)cos(mx)$ is a polynomial in $sin(x)^2$ with degree $le (m-1)/2$
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Step to prove that $cos (n arccos (x))$ is a polynomial of $n$-th degree
Prove that $cos (n arccos (x))$ is a polynomial of $n$-th degreeQuestion about chebyshev polynomialProve that $sin(2arcsin x + arccos x) = sqrt1-x^2$Find the least degree Polynomial whose one of the roots is $ cos(12^circ)$Finding Imaginary Values of a Degree 6 PolynomialRewriting $sin(arccos(y))$ and $arcsin(cos(x))$Show that, $arccosleft(fracacright)-arccosleft(fracbcright)=…$Evaluating $arccos(cosfrac15pi 4)$Prove $ frac12 (arccos(x) - arccos(-x)) = -arcsin(x)$How to prove $sin(x)arccos(x) + xcos(x) > 0$$cos(x)cos(mx)$ is a polynomial in $sin(x)^2$ with degree $le (m-1)/2$
$begingroup$
I am confronted to the same problem stated in that question, namely to prove that cos(𝑛arccos(𝑥)) is a polynomial of 𝑛-th degree.
However to begin with I don't understand how
$$
cos[n arccos(x) + arccos(x)] = cos[n arccos(x)] cos[arccos(x)] - sin[n arccos(x)] sin[arccos(x)]
$$
trigonometry polynomials chebyshev-polynomials
asked Apr 4 at 19:04
ecjbecjb
28718
$endgroup$
add a comment |
$begingroup$
I am confronted to the same problem stated in that question, namely to prove that cos(𝑛arccos(𝑥)) is a polynomial of 𝑛-th degree.
However to begin with I don't understand how
$$
cos[n arccos(x) + arccos(x)] = cos[n arccos(x)] cos[arccos(x)] - sin[n arccos(x)] sin[arccos(x)]
$$
trigonometry polynomials chebyshev-polynomials
asked Apr 4 at 19:04
ecjbecjb
28718
$endgroup$
add a comment |
$begingroup$
I am confronted to the same problem stated in that question, namely to prove that cos(𝑛arccos(𝑥)) is a polynomial of 𝑛-th degree.
However to begin with I don't understand how
$$
cos[n arccos(x) + arccos(x)] = cos[n arccos(x)] cos[arccos(x)] - sin[n arccos(x)] sin[arccos(x)]
$$
trigonometry polynomials chebyshev-polynomials
asked Apr 4 at 19:04
ecjbecjb
28718
$endgroup$
I am confronted to the same problem stated in that question, namely to prove that cos(𝑛arccos(𝑥)) is a polynomial of 𝑛-th degree.
However to begin with I don't understand how
$$
cos[n arccos(x) + arccos(x)] = cos[n arccos(x)] cos[arccos(x)] - sin[n arccos(x)] sin[arccos(x)]
$$
trigonometry polynomials chebyshev-polynomials
trigonometry polynomials chebyshev-polynomials
asked Apr 4 at 19:04
ecjbecjb
28718
asked Apr 4 at 19:04
ecjbecjb
28718
edited Apr 4 at 19:25
ecjb
asked Apr 4 at 19:04
ecjbecjb
28718
asked Apr 4 at 19:04
ecjbecjb
28718
asked Apr 4 at 19:04
ecjbecjb
28718
28718
add a comment |
add a comment |
1 Answer
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$begingroup$
Use $x=narccos(x)$ and $y=arccos(x)$. Then $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$. This a simple trigonometric identity. You can find a possible proof at http://www.math.ubc.ca/~feldman/m100/trigId.pdf, but just search for it and you get many variants.
answered Apr 4 at 19:13
AndreiAndrei
13.6k21230
$endgroup$
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1 Answer
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1 Answer
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votes
$begingroup$
Use $x=narccos(x)$ and $y=arccos(x)$. Then $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$. This a simple trigonometric identity. You can find a possible proof at http://www.math.ubc.ca/~feldman/m100/trigId.pdf, but just search for it and you get many variants.
answered Apr 4 at 19:13
AndreiAndrei
13.6k21230
$endgroup$
add a comment |
$begingroup$
Use $x=narccos(x)$ and $y=arccos(x)$. Then $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$. This a simple trigonometric identity. You can find a possible proof at http://www.math.ubc.ca/~feldman/m100/trigId.pdf, but just search for it and you get many variants.
answered Apr 4 at 19:13
AndreiAndrei
13.6k21230
$endgroup$
add a comment |
$begingroup$
Use $x=narccos(x)$ and $y=arccos(x)$. Then $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$. This a simple trigonometric identity. You can find a possible proof at http://www.math.ubc.ca/~feldman/m100/trigId.pdf, but just search for it and you get many variants.
answered Apr 4 at 19:13
AndreiAndrei
13.6k21230
$endgroup$
Use $x=narccos(x)$ and $y=arccos(x)$. Then $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$. This a simple trigonometric identity. You can find a possible proof at http://www.math.ubc.ca/~feldman/m100/trigId.pdf, but just search for it and you get many variants.
answered Apr 4 at 19:13
AndreiAndrei
13.6k21230
answered Apr 4 at 19:13
AndreiAndrei
13.6k21230
answered Apr 4 at 19:13
AndreiAndrei
13.6k21230
answered Apr 4 at 19:13
AndreiAndrei
13.6k21230
13.6k21230
add a comment |
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