4
$\begingroup$

Suppose the following equation $$ A+D\sin^{2}x=B\sin x+C\cos x, $$ where $A,B,C,D\in\mathbb{R}$ are the real constants. Initially, I tried to find its solution from a simple substitution \begin{align*} A-B\sin x+D\sin^{2}x & =\pm C\sqrt{1-\sin^{2}x}, \end{align*} that after $t=\sin x$ leads to the following quartic equation $$ (A^{2}-C^{2})-2ABt+(B^{2}+2AD+C^{2})t^{2}-2BDt^{3}+D^{2}t^{4}=0. $$ The Weierstrass substitution, where $t=\tan\frac{x}{2}$, and $$ \sin x=2\sin\frac{x}{2}\cos\frac{x}{2}=\frac{2\sin\frac{x}{2}}{\cos\frac{x}{2}}\cos^{2}\frac{x}{2}=\frac{2\tan\frac{x}{2}}{\frac{1}{\cos^{2}\frac{x}{2}}}=\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\frac{2t}{1+t^{2}}, $$ and $$ \cos x=\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}=\left(1-\frac{\sin^{2}\frac{x}{2}}{\cos^{2}\frac{x}{2}}\right)\cos^{2}\frac{x}{2}=\frac{1-\tan^{2}\frac{x}{2}}{\frac{1}{\cos^{2}\frac{x}{2}}}=\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\frac{1-t^{2}}{1+t^{2}}, $$ leads to $$ A+D\frac{4t^{2}}{(1+t^{2})^{2}}=B\frac{2t}{1+t^{2}}+C\frac{1-t^{2}}{1+t^{2}}, $$ 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.

$\endgroup$
  • $\begingroup$ $\begin{align*} A-B\sin x+D\sin^{2}x & = \pm C\sqrt{1-\sin^{2}x} \end{align*} \not = C\sqrt{1-\sin^{2}x}$ $\endgroup$ – S.H.W Aug 5 '18 at 15:24
  • $\begingroup$ @ S.H.W: thanks, corrected $\endgroup$ – justik Aug 5 '18 at 17:55
  • 1
    $\begingroup$ Note that $A+B\sin^2 x$ can be written as $P + Q\cos 2x$ 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. $\endgroup$ – Blue Aug 5 '18 at 21:01
  • $\begingroup$ @ Blue: Thanks for the explanation, you are right... $\endgroup$ – justik Aug 6 '18 at 10:23
1
$\begingroup$

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 $D\ne0$), 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 $A=A\cos^2x+A\sin^2x$, 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 $$ A+\frac{4Dt^2}{(1+t^2)^2}=\frac{2Bt}{1+t^2}+\frac{C(1-t^2)}{1+t^2} $$ 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.

$\endgroup$
  • $\begingroup$ @ 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 :-( $\endgroup$ – justik Aug 5 '18 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.