Trig Challenge: Proving $\cos^3 y+\sin^3 y=\cos x+\sin x$

[anemone]

Show that if $\dfrac{\cos x}{\cos y}+\dfrac{\sin x}{\sin y}=-1$, then $\dfrac{\cos^3 y}{\cos x}+\dfrac{\sin^3 y}{\sin x}=1$.

 

[Albert1]

Show that if $\dfrac{\cos x}{\cos y}+\dfrac{\sin x}{\sin y}=-1$, then $\dfrac{\cos^3 y}{\cos x}+\dfrac{\sin^3 y}{\sin x}=1$.

Spoiler

$\dfrac{\cos x}{\cos y}+\dfrac{\sin x}{\sin y}=-1---(1)$,
then $\dfrac{\cos^3 y}{\cos x}+\dfrac{\sin^3 y}{\sin x}=1---(2)$
let:$a=\dfrac{cos\, y}{cos\, x}$
$b=\dfrac{sin\, y}{sin\, x}$
if (2) is true then we must have :$\cos^2\,y=\dfrac {1-b}{a-b}$
and (1)+(2)=$a\,cos^2\,y+b\,sin^2\,y+\dfrac{1}{a}+\dfrac {1}{b}=0$
=$cos^2\,y(a-b)+b+\dfrac{1}{a}+\dfrac{1}{b}=0$
$1-b+b+\dfrac{1}{a}+\dfrac{1}{b}=0$
$\dfrac{1}{a}+\dfrac{1}{b}=-1$
this is given in (1)

 

[anemone]

Thanks Albert for your solution!:)

Here is another solution of other I wanted to share:

Spoiler

Let $a=\dfrac{\cos y}{\cos x}$ and $b=\dfrac{\sin y}{\sin x}$.

Since $\dfrac{1}{a}+\dfrac{1}{b}=-1$, we have that $-(a+b)=ab$ or $-(a+b)(a-b)=ab(a-b)\implies (b^2-a^2)=ab(a-b)$. Now,

$\begin{align*}1&=\cos^2 y+\sin^2 y\\&=a^2\cos^2 x+b^2\sin^2 x\\&=a^2+(b^2-a^2)\sin^2 x\\&=a^2+ab(a-b)\sin^2 x\end{align*}$

Hence,

$\begin{align*}\dfrac{\cos^3 y}{\cos x}+\dfrac{\sin^3 y}{\sin x}&=a^3\cos^2 x+b^3\sin^2 x\\&=a(a^2\cos^2 x+b^2\sin^2 x)+(b-a)b^2\sin^2 x\\&=a(1)-(a-b)b^2\sin^2 x\\&=a-\dfrac{(1-a^2)(b)}{a}\\&=\dfrac{a^2+a^2b-b}{a}\\&=\dfrac{a^2+a^2b-(-a-ab)}{a}\\&=a+ab+1+b\\&=0+1\\&=1\end{align*}$