If $x^2 + y^2 = 1$, then find the value of $(3x-4x^3)^2+(3y-4y^3)^2$ where $x,y\in \mathbb {R}$ .

Question

If $x^2 + y^2 = 1$, then find the value of $(3x-4x^3)^2+(3y-4y^3)^2$ where $x$, $y \in \mathbb {R}$ .

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What I've tried :

I thought $x$, $y$ as $\sin \theta$ and $\cos \theta$ .

But, it doesn't seems to work and I'm unable to see any particular pattern in it .

Answer 1

Your first idea was the right one

Hint. Try to express $\cos(3\theta)$ and $\sin(3\theta)$ as functions of $\cos(\theta)$ and $\sin(\theta)$ respectively, and see what happenes.

Answer 2

You do not need to use $\sin$ or $\cos$.

Just rearrange the expression and use $$1=(x^2+y^2)^2 = x^4+y^4+2x^2y^2$$ $$1= (x^2+y^2)^3 = x^6+y^6 + 3 x^2 y^2 (x^2+y^2) = x^6+y^6 + 3x^2 y^2$$ So you get $$(3x-4x^3)^2+(3y-4y^3)^2 = x^2(3-4x^2)^2 + y^2(3-4y^2)^2$$ Expanding and using $x^2+y^2=1$ gives $$= 9-24(x^4+y^4)+16(x^6+y^6)= 9-24(1-2x^2y^2)+16(1-3x^2y^2)$$ $$= 9-24+16 = \boxed{1}$$

Answer 3

An answer that doesn't use algebraic expansion :

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Let

$$\begin{align} P(x,y)&:=(3x-4x^3)^2+(3y-4y^3)^2\end{align}$$

Making the substitution $u=x^2, \thinspace u\in[0,1]$ and $y^2=1-u$, then we can observe that the polynomial $P(x,y)$ is equivalent to the polynomial

$$ \begin{align}Q(u):=u(3-4u)^2&+(1-u)(4u-1)^2\thinspace .\end{align} $$

Then using the quick property $Q(u)=Q(1-u),\thinspace \forall u\in[0,1]$ we see that : If $u\in\left\{0,1,\frac 14,\frac 34\right\}$, then $Q(u)=1\thinspace .$

Finally we observe that, if $0<\deg \{Q\}≤3\thinspace ,$ then the polynomial $Q(u)-1$ has at most $3$ real roots, which leads to a contradiction .

Therefore, we conclude that :

$Q(u)\equiv 1\thinspace,$ since $\deg\left\{Q\right\}=0\thinspace .$ This means, if $(x,y)\in\mathbb R^{2}$ and $x^2+y^2=1$, then $P(x,y)=1\thinspace .$

Answer 4

Another way following the fine suggestion given by Ivan Kaznacheyeu in the comments

$$x^2+y^2=1 \iff z=x+iy = e^{i\theta}$$

and since in general

$$(-3x+4x^3)+i(3y-4y^3)= (x+i y)^3+3(x-i y)(x^2+y^2-1)$$

by the given condition we have

$$w=(-3x+4x^3)+i(3y-4y^3)= (x+i y)^3=z^3=e^{i3\theta}$$

that is

$$(3x-4x^3)^2+(3y-4y^3)^2=1$$

Answer 5

Another "long" way by

• $x=\frac{2t}{1+t^2}$
• $y=\frac{1-t^2}{1+t^2}$

such that $x^2+y^2=1$ and then

$$(3x-4x^3)^2+(3y-4y^3)^2=\left(\frac{6t}{1+t^2}-\frac{32t^3}{(1+t^2)^3}\right)^2+\left(\frac{3(1-t^2)}{1+t^2}-\frac{4(1-t^2)^3}{(1+t^2)^3}\right)^2=$$

$$=\frac{36t^2}{(1+t^2)^2}+\frac{32^2t^6}{(1+t^2)^6}-\frac{384t^4}{(1+t^2)^4}+\frac{9(1-t^2)^2}{(1+t^2)^2}+\frac{16(1-t^2)^6}{(1+t^2)^6}-\frac{24(1-t^2)^4}{(1+t^2)^4}=$$

$$\small=\frac{36t^2(1+t^2)^4+32^2t^6-384t^4(1+t^2)^2+9(1-t^2)^2(1+t^2)^4+16(1-t^2)^6-24(1-t^2)^4(1+t^2)^2}{(1+t^2)^6}=$$

$$=\frac{t^{12} + 6 t^{10} + 15 t^{8} + 20 t^6 + 15 t^4 + 6 t^2 + 1}{(1+t^2)^6}=\frac{(1+t^2)^6}{(1+t^2)^6}=1$$

Answer 6

Let $x^2=u+{1\over 2},$ $y^2=v+{1\over 2}.$ Then $u+v=0.$ The quantity in question takes the form $P(u)+P(-u),$ where $$P(u)=\left (u+{1\over 2}\right )(1-4u)^2$$ The problem becomes whether the quantity $P(u)+P(-u)$ is constant. It is equivalent to $$P'(u)-P'(-u)=0$$ i.e. the function $P'(u)$ is even. We have $$P'(u)=(4u-1)^2+8\left (u+{1\over 2}\right )(4u-1)\\ =(4u-1)(4u-1+8u+4)=3(16u^2-1)$$ Hence $P'(u)$ is indeed an even polynomial. Once we know that $P(u)+P(-u)$ is independent of $u$ we plug in $u=0$ and get $$P(u)+P(-u)=2P(0)=1$$

Remarks

1. The advantage of this method is that we are dealing with expressions of the second degree. Moreover the method can be applied to other expressions of similar form.
2. If we are too lazy to verify whether $P'$ is even, we can instead verify if $P''$ is odd. Then we are dealing with a linear function. Once we know that $P''$ is odd the polynomial $P'$ is even.