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How do you simplify the following? $$\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝1$$

What I've tried:

$$\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝1$$ $$\sin^2(x)+\cos^2(x)=1$$

$$\implies\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}＝\sin^2(x)+\cos^2(x)$$

$$\implies \frac {\cos^4(x)}{\cos^2(y)}－\cos^2(x)＝\sin^2(x)－\frac {\sin^4(x)}{\sin^2(y)}\\~\\\implies \frac {\cos^4(x)－\cos^2(x)\cos^2(y)}{\cos^2(y)}＝\frac {\sin^2(x)\sin^2(y)－\sin^4(x)}{\sin^2(y)}\\~\\\implies \frac {\cos^2(x)}{\cos^2(y)}\left(\cos^2(x)－\cos^2(y)\right)＝\frac {\sin^2(x)}{\sin^2(y)}\left(\sin^2(y)－\sin^2(x)\right)$$

Jonas Hardt
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Moustapha_M_I
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    You're almost there. Note that: $\cos^2x+\sin^2x=\cos^2y+\sin^2y$. – Koro Jan 14 '23 at 19:53
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    A typographical point: Use \sin and \cos to render the trig functions. Then they'll appear as normal font and you won't need the parentheses for the functions' arguments. I'll get you started in the title. – Robert Shore Jan 14 '23 at 20:00
  • A generalization : https://math.stackexchange.com/questions/639223/if-frac-sin4-xa-frac-cos4-xb-frac1ab-then-show-that-frac – lab bhattacharjee Jan 15 '23 at 04:18
  • A minor typographical point: Some of your "+", "-", and "=" symbols are non-ASCII characters, and it slightly messes up the ability of MathJax to format the formulas neatly. You get cleaner output by using all ASCII input. – David K Jan 15 '23 at 04:20

1 Answers1

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\begin{eqnarray}\frac {\cos^4(x)}{\cos^2(y)}＋\frac {\sin^4(x)}{\sin^2(y)}&=&1\\ \cos^4x\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\ (1-\sin^2x)^2\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\ (1-2\sin^2x+\sin^4x)\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\ (1-2\sin^2x)\sin^2y+\sin^4x\sin^2y+\sin^4x\cos^2y&=&\sin^2y\cos^2y\\ (1-2\sin^2x)\sin^2y+\sin^4x&=&\sin^2y(1-\sin^2y)\\ \sin^4x&=&\sin^2y(1-\sin^2y)-(1-2\sin^2x)\sin^2y\\ \sin^4x&=&(1-\sin^2y-1+2\sin^2x)\sin^2y\\ \sin^4x&=&2\sin^2x\sin^2y-\sin^4y\\ \sin^4x+\sin^4y&=&2\sin^2x\sin^2y\\ \frac{\sin^2x}{\sin^2y}+\frac{\sin^2y}{\sin^2x}&=&2 \end{eqnarray}

Let $u=\frac{\sin^2x}{\sin^2y}$, then $u+\frac{1}{u}=2$. So $(u-1)^2=0$.

Thus $u=1$ which gives $$\sin^2x=\sin^2y$$

Note that the original expression is undefined when either of $\cos y$ or $\sin y$ is zero whereas the 'equivalent' equation is defined for all $x,y$. If the original equation is set to $a$ instead of $1$ and we allow $a\to1$ we will see why.

The following desmos graph compares the graph for $a=1.2$ (in red) compared to the graph of $\sin^2x=\sin^2y$ (in black).

Here is a link to a graph in motion where $a$ varies in value from 1 to 2.

graph of equation

Desmos dynamic graph

John Wayland Bales
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