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Can we solve $A+Dsin^2x=Bsin x+Ccos x$ without having to solve a quartic polynomial?
Clash Royale CLAN TAG#URR8PPP
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Suppose the following equation
$$
A+Dsin^2x=Bsin x+Ccos x,
$$
where $A,B,C,DinmathbbR$ are the real constants. Initially, I tried to find its solution from a simple substitution
beginalign*
A-Bsin x+Dsin^2x & =pm Csqrt1-sin^2x,
endalign*
that after $t=sin x$ leads to the following quartic equation
$$
(A^2-C^2)-2ABt+(B^2+2AD+C^2)t^2-2BDt^3+D^2t^4=0.
$$
The Weierstrass substitution, where $t=tanfracx2$, and
$$
sin x=2sinfracx2cosfracx2=frac2sinfracx2cosfracx2cos^2fracx2=frac2tanfracx2frac1cos^2fracx2=frac2tanfracx21+tan^2fracx2=frac2t1+t^2,
$$
and
$$
cos x=cos^2fracx2-sin^2fracx2=left(1-fracsin^2fracx2cos^2fracx2right)cos^2fracx2=frac1-tan^2fracx2frac1cos^2fracx2=frac1-tan^2fracx21+tan^2fracx2=frac1-t^21+t^2,
$$
leads to
$$
A+Dfrac4t^2(1+t^2)^2=Bfrac2t1+t^2+Cfrac1-t^21+t^2,
$$
which is also the quartic equation for $t$
$$
(A+C)t^4-2Bt^3+2(A+2D)t^2-2Bt+A-C=0,
$$
$$
(A+C)t^4-2Bt(t^2+1)+2(A+2D)t^2+A-C=0.
$$
Is there any better substitution avoiding the transformation of the trigonometric identity to the quartic equation for $t$?
Thanks for your help.
algebra-precalculus trigonometry polynomials quartic-equations
asked 23 hours ago
justik
22528
add a comment |Â
up vote
4
down vote
favorite
Suppose the following equation
$$
A+Dsin^2x=Bsin x+Ccos x,
$$
where $A,B,C,DinmathbbR$ are the real constants. Initially, I tried to find its solution from a simple substitution
beginalign*
A-Bsin x+Dsin^2x & =pm Csqrt1-sin^2x,
endalign*
that after $t=sin x$ leads to the following quartic equation
$$
(A^2-C^2)-2ABt+(B^2+2AD+C^2)t^2-2BDt^3+D^2t^4=0.
$$
The Weierstrass substitution, where $t=tanfracx2$, and
$$
sin x=2sinfracx2cosfracx2=frac2sinfracx2cosfracx2cos^2fracx2=frac2tanfracx2frac1cos^2fracx2=frac2tanfracx21+tan^2fracx2=frac2t1+t^2,
$$
and
$$
cos x=cos^2fracx2-sin^2fracx2=left(1-fracsin^2fracx2cos^2fracx2right)cos^2fracx2=frac1-tan^2fracx2frac1cos^2fracx2=frac1-tan^2fracx21+tan^2fracx2=frac1-t^21+t^2,
$$
leads to
$$
A+Dfrac4t^2(1+t^2)^2=Bfrac2t1+t^2+Cfrac1-t^21+t^2,
$$
which is also the quartic equation for $t$
$$
(A+C)t^4-2Bt^3+2(A+2D)t^2-2Bt+A-C=0,
$$
$$
(A+C)t^4-2Bt(t^2+1)+2(A+2D)t^2+A-C=0.
$$
Is there any better substitution avoiding the transformation of the trigonometric identity to the quartic equation for $t$?
Thanks for your help.
algebra-precalculus trigonometry polynomials quartic-equations
asked 23 hours ago
justik
22528
$beginalign* A-Bsin x+Dsin^2x & = pm Csqrt1-sin^2x endalign* not = Csqrt1-sin^2x$
– S.H.W
18 hours ago
@ S.H.W: thanks, corrected
– justik
15 hours ago
Note that $A+Bsin^2 x$ can be written as $P + Qcos 2x$ for appropriate $P$ and $Q$; also, $B sin x + C cos x$ can be written as $M sin(x + N)$ for appropriate $M$ and $N$. Considering how a "$cos 2x$" wave and a "$sin(x+N)"$ wave look, it seems reasonable to expect/fear that —barring some special relationships among the coefficients— the curves will meet in four asymmetric points.
– Blue
12 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Suppose the following equation
$$
A+Dsin^2x=Bsin x+Ccos x,
$$
where $A,B,C,DinmathbbR$ are the real constants. Initially, I tried to find its solution from a simple substitution
beginalign*
A-Bsin x+Dsin^2x & =pm Csqrt1-sin^2x,
endalign*
that after $t=sin x$ leads to the following quartic equation
$$
(A^2-C^2)-2ABt+(B^2+2AD+C^2)t^2-2BDt^3+D^2t^4=0.
$$
The Weierstrass substitution, where $t=tanfracx2$, and
$$
sin x=2sinfracx2cosfracx2=frac2sinfracx2cosfracx2cos^2fracx2=frac2tanfracx2frac1cos^2fracx2=frac2tanfracx21+tan^2fracx2=frac2t1+t^2,
$$
and
$$
cos x=cos^2fracx2-sin^2fracx2=left(1-fracsin^2fracx2cos^2fracx2right)cos^2fracx2=frac1-tan^2fracx2frac1cos^2fracx2=frac1-tan^2fracx21+tan^2fracx2=frac1-t^21+t^2,
$$
leads to
$$
A+Dfrac4t^2(1+t^2)^2=Bfrac2t1+t^2+Cfrac1-t^21+t^2,
$$
which is also the quartic equation for $t$
$$
(A+C)t^4-2Bt^3+2(A+2D)t^2-2Bt+A-C=0,
$$
$$
(A+C)t^4-2Bt(t^2+1)+2(A+2D)t^2+A-C=0.
$$
Is there any better substitution avoiding the transformation of the trigonometric identity to the quartic equation for $t$?
Thanks for your help.
algebra-precalculus trigonometry polynomials quartic-equations
asked 23 hours ago
justik
22528
Suppose the following equation
$$
A+Dsin^2x=Bsin x+Ccos x,
$$
where $A,B,C,DinmathbbR$ are the real constants. Initially, I tried to find its solution from a simple substitution
beginalign*
A-Bsin x+Dsin^2x & =pm Csqrt1-sin^2x,
endalign*
that after $t=sin x$ leads to the following quartic equation
$$
(A^2-C^2)-2ABt+(B^2+2AD+C^2)t^2-2BDt^3+D^2t^4=0.
$$
The Weierstrass substitution, where $t=tanfracx2$, and
$$
sin x=2sinfracx2cosfracx2=frac2sinfracx2cosfracx2cos^2fracx2=frac2tanfracx2frac1cos^2fracx2=frac2tanfracx21+tan^2fracx2=frac2t1+t^2,
$$
and
$$
cos x=cos^2fracx2-sin^2fracx2=left(1-fracsin^2fracx2cos^2fracx2right)cos^2fracx2=frac1-tan^2fracx2frac1cos^2fracx2=frac1-tan^2fracx21+tan^2fracx2=frac1-t^21+t^2,
$$
leads to
$$
A+Dfrac4t^2(1+t^2)^2=Bfrac2t1+t^2+Cfrac1-t^21+t^2,
$$
which is also the quartic equation for $t$
$$
(A+C)t^4-2Bt^3+2(A+2D)t^2-2Bt+A-C=0,
$$
$$
(A+C)t^4-2Bt(t^2+1)+2(A+2D)t^2+A-C=0.
$$
Is there any better substitution avoiding the transformation of the trigonometric identity to the quartic equation for $t$?
Thanks for your help.
algebra-precalculus trigonometry polynomials quartic-equations
asked 23 hours ago
justik
22528
edited 15 hours ago
asked 23 hours ago
justik
22528
asked 23 hours ago
justik
22528
asked 23 hours ago
justik
22528
22528
$beginalign* A-Bsin x+Dsin^2x & = pm Csqrt1-sin^2x endalign* not = Csqrt1-sin^2x$
– S.H.W
18 hours ago
@ S.H.W: thanks, corrected
– justik
15 hours ago
Note that $A+Bsin^2 x$ can be written as $P + Qcos 2x$ for appropriate $P$ and $Q$; also, $B sin x + C cos x$ can be written as $M sin(x + N)$ for appropriate $M$ and $N$. Considering how a "$cos 2x$" wave and a "$sin(x+N)"$ wave look, it seems reasonable to expect/fear that —barring some special relationships among the coefficients— the curves will meet in four asymmetric points.
– Blue
12 hours ago
add a comment |Â
$beginalign* A-Bsin x+Dsin^2x & = pm Csqrt1-sin^2x endalign* not = Csqrt1-sin^2x$
– S.H.W
18 hours ago
@ S.H.W: thanks, corrected
– justik
15 hours ago
Note that $A+Bsin^2 x$ can be written as $P + Qcos 2x$ for appropriate $P$ and $Q$; also, $B sin x + C cos x$ can be written as $M sin(x + N)$ for appropriate $M$ and $N$. Considering how a "$cos 2x$" wave and a "$sin(x+N)"$ wave look, it seems reasonable to expect/fear that —barring some special relationships among the coefficients— the curves will meet in four asymmetric points.
– Blue
12 hours ago
$beginalign* A-Bsin x+Dsin^2x & = pm Csqrt1-sin^2x endalign* not = Csqrt1-sin^2x$
– S.H.W
18 hours ago
$beginalign* A-Bsin x+Dsin^2x & = pm Csqrt1-sin^2x endalign* not = Csqrt1-sin^2x$
– S.H.W
18 hours ago
@ S.H.W: thanks, corrected
– justik
15 hours ago
@ S.H.W: thanks, corrected
– justik
15 hours ago
Note that $A+Bsin^2 x$ can be written as $P + Qcos 2x$ for appropriate $P$ and $Q$; also, $B sin x + C cos x$ can be written as $M sin(x + N)$ for appropriate $M$ and $N$. Considering how a "$cos 2x$" wave and a "$sin(x+N)"$ wave look, it seems reasonable to expect/fear that —barring some special relationships among the coefficients— the curves will meet in four asymmetric points.
– Blue
12 hours ago
Note that $A+Bsin^2 x$ can be written as $P + Qcos 2x$ for appropriate $P$ and $Q$; also, $B sin x + C cos x$ can be written as $M sin(x + N)$ for appropriate $M$ and $N$. Considering how a "$cos 2x$" wave and a "$sin(x+N)"$ wave look, it seems reasonable to expect/fear that —barring some special relationships among the coefficients— the curves will meet in four asymmetric points.
– Blue
12 hours ago
add a comment |Â
1 Answer
1
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votes
up vote
1
down vote
If you set $X=cos x$ and $Y=sin x$, the equation becomes geometrically finding the intersection of $DY^2-CX-BY-A=0$ (a parabola when $Dne0$), with the circle $X^2+Y^2=1$, which generally has four solutions. So, no: you can't reduce the resolvent equation from degree $4$ unless there is some particular relation between the coefficients.
If $D=0$ the problem becomes intersecting a line (provided one among $B$ and $C$ is nonzero) with a circle and indeed can be reduced to a quadratic equation.
It's no better if you use $A=Acos^2x+Asin^2x$, because the conic becomes
$$
AX^2+(A-D)Y^2+CX+BY=0
$$
which is either an ellipse, for $A(A-D)>0$, or a hyperbola, for $A(A-D)<0$.
Another way to see it is setting $t=tan(x/2)$, which transforms the equation into
$$
A+frac4Dt^2(1+t^2)^2=frac2Bt1+t^2+fracC(1-t^2)1+t^2
$$
which is what you did. As you see, this generally is a quartic:
$$
(A+C)t^4-2Bt^3+(2A+4D)t^2-2Bt-C=0
$$
One gets a symmetry when $B=0$, because in this case the parabola is symmetric with respect to the $X$-axis: as you see, the quartic becomes biquadratic.
answered 15 hours ago
egreg
164k1179187
@ egreg: Thanks for the detailed explanation. I also tried $A^2=A^2cos^2x+A^2sin^2x$ with reordering the terms, but without success. Hence, there is no way to simplify the equation :-(
– justik
14 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If you set $X=cos x$ and $Y=sin x$, the equation becomes geometrically finding the intersection of $DY^2-CX-BY-A=0$ (a parabola when $Dne0$), with the circle $X^2+Y^2=1$, which generally has four solutions. So, no: you can't reduce the resolvent equation from degree $4$ unless there is some particular relation between the coefficients.
If $D=0$ the problem becomes intersecting a line (provided one among $B$ and $C$ is nonzero) with a circle and indeed can be reduced to a quadratic equation.
It's no better if you use $A=Acos^2x+Asin^2x$, because the conic becomes
$$
AX^2+(A-D)Y^2+CX+BY=0
$$
which is either an ellipse, for $A(A-D)>0$, or a hyperbola, for $A(A-D)<0$.
Another way to see it is setting $t=tan(x/2)$, which transforms the equation into
$$
A+frac4Dt^2(1+t^2)^2=frac2Bt1+t^2+fracC(1-t^2)1+t^2
$$
which is what you did. As you see, this generally is a quartic:
$$
(A+C)t^4-2Bt^3+(2A+4D)t^2-2Bt-C=0
$$
One gets a symmetry when $B=0$, because in this case the parabola is symmetric with respect to the $X$-axis: as you see, the quartic becomes biquadratic.
answered 15 hours ago
egreg
164k1179187
@ egreg: Thanks for the detailed explanation. I also tried $A^2=A^2cos^2x+A^2sin^2x$ with reordering the terms, but without success. Hence, there is no way to simplify the equation :-(
– justik
14 hours ago
add a comment |Â
up vote
1
down vote
If you set $X=cos x$ and $Y=sin x$, the equation becomes geometrically finding the intersection of $DY^2-CX-BY-A=0$ (a parabola when $Dne0$), with the circle $X^2+Y^2=1$, which generally has four solutions. So, no: you can't reduce the resolvent equation from degree $4$ unless there is some particular relation between the coefficients.
If $D=0$ the problem becomes intersecting a line (provided one among $B$ and $C$ is nonzero) with a circle and indeed can be reduced to a quadratic equation.
It's no better if you use $A=Acos^2x+Asin^2x$, because the conic becomes
$$
AX^2+(A-D)Y^2+CX+BY=0
$$
which is either an ellipse, for $A(A-D)>0$, or a hyperbola, for $A(A-D)<0$.
Another way to see it is setting $t=tan(x/2)$, which transforms the equation into
$$
A+frac4Dt^2(1+t^2)^2=frac2Bt1+t^2+fracC(1-t^2)1+t^2
$$
which is what you did. As you see, this generally is a quartic:
$$
(A+C)t^4-2Bt^3+(2A+4D)t^2-2Bt-C=0
$$
One gets a symmetry when $B=0$, because in this case the parabola is symmetric with respect to the $X$-axis: as you see, the quartic becomes biquadratic.
answered 15 hours ago
egreg
164k1179187
@ egreg: Thanks for the detailed explanation. I also tried $A^2=A^2cos^2x+A^2sin^2x$ with reordering the terms, but without success. Hence, there is no way to simplify the equation :-(
– justik
14 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you set $X=cos x$ and $Y=sin x$, the equation becomes geometrically finding the intersection of $DY^2-CX-BY-A=0$ (a parabola when $Dne0$), with the circle $X^2+Y^2=1$, which generally has four solutions. So, no: you can't reduce the resolvent equation from degree $4$ unless there is some particular relation between the coefficients.
If $D=0$ the problem becomes intersecting a line (provided one among $B$ and $C$ is nonzero) with a circle and indeed can be reduced to a quadratic equation.
It's no better if you use $A=Acos^2x+Asin^2x$, because the conic becomes
$$
AX^2+(A-D)Y^2+CX+BY=0
$$
which is either an ellipse, for $A(A-D)>0$, or a hyperbola, for $A(A-D)<0$.
Another way to see it is setting $t=tan(x/2)$, which transforms the equation into
$$
A+frac4Dt^2(1+t^2)^2=frac2Bt1+t^2+fracC(1-t^2)1+t^2
$$
which is what you did. As you see, this generally is a quartic:
$$
(A+C)t^4-2Bt^3+(2A+4D)t^2-2Bt-C=0
$$
One gets a symmetry when $B=0$, because in this case the parabola is symmetric with respect to the $X$-axis: as you see, the quartic becomes biquadratic.
answered 15 hours ago
egreg
164k1179187
If you set $X=cos x$ and $Y=sin x$, the equation becomes geometrically finding the intersection of $DY^2-CX-BY-A=0$ (a parabola when $Dne0$), with the circle $X^2+Y^2=1$, which generally has four solutions. So, no: you can't reduce the resolvent equation from degree $4$ unless there is some particular relation between the coefficients.
If $D=0$ the problem becomes intersecting a line (provided one among $B$ and $C$ is nonzero) with a circle and indeed can be reduced to a quadratic equation.
It's no better if you use $A=Acos^2x+Asin^2x$, because the conic becomes
$$
AX^2+(A-D)Y^2+CX+BY=0
$$
which is either an ellipse, for $A(A-D)>0$, or a hyperbola, for $A(A-D)<0$.
Another way to see it is setting $t=tan(x/2)$, which transforms the equation into
$$
A+frac4Dt^2(1+t^2)^2=frac2Bt1+t^2+fracC(1-t^2)1+t^2
$$
which is what you did. As you see, this generally is a quartic:
$$
(A+C)t^4-2Bt^3+(2A+4D)t^2-2Bt-C=0
$$
One gets a symmetry when $B=0$, because in this case the parabola is symmetric with respect to the $X$-axis: as you see, the quartic becomes biquadratic.
answered 15 hours ago
egreg
164k1179187
answered 15 hours ago
egreg
164k1179187
answered 15 hours ago
egreg
164k1179187
answered 15 hours ago
egreg
164k1179187
164k1179187
@ egreg: Thanks for the detailed explanation. I also tried $A^2=A^2cos^2x+A^2sin^2x$ with reordering the terms, but without success. Hence, there is no way to simplify the equation :-(
– justik
14 hours ago
add a comment |Â
@ egreg: Thanks for the detailed explanation. I also tried $A^2=A^2cos^2x+A^2sin^2x$ with reordering the terms, but without success. Hence, there is no way to simplify the equation :-(
– justik
14 hours ago
@ egreg: Thanks for the detailed explanation. I also tried $A^2=A^2cos^2x+A^2sin^2x$ with reordering the terms, but without success. Hence, there is no way to simplify the equation :-(
– justik
14 hours ago
@ egreg: Thanks for the detailed explanation. I also tried $A^2=A^2cos^2x+A^2sin^2x$ with reordering the terms, but without success. Hence, there is no way to simplify the equation :-(
– justik
14 hours ago
add a comment |Â
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$beginalign* A-Bsin x+Dsin^2x & = pm Csqrt1-sin^2x endalign* not = Csqrt1-sin^2x$
– S.H.W
18 hours ago
@ S.H.W: thanks, corrected
– justik
15 hours ago
Note that $A+Bsin^2 x$ can be written as $P + Qcos 2x$ for appropriate $P$ and $Q$; also, $B sin x + C cos x$ can be written as $M sin(x + N)$ for appropriate $M$ and $N$. Considering how a "$cos 2x$" wave and a "$sin(x+N)"$ wave look, it seems reasonable to expect/fear that —barring some special relationships among the coefficients— the curves will meet in four asymmetric points.
– Blue
12 hours ago