So I have a small problem here where I have to prove the following :
$$\cos^4x - \sin^4x - \cos^2x + \sin^2x = 0 $$
I know that the 2nd part is always $1$, so I need to prove that the first part also equals $1$. So how should I prove it ?
Edit : Sorry, the equation itself was wrong. I've edited it.
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$\begingroup$Simple: it's not even true. Unless $x$ is a very specific value.
Your other statement, that's a different story. You can write down
$$\cos^4 x + \sin^4 x = \cos^4 x + \sin^4 x-2\sin^2 x \cos^2x + 2\sin^2x \cos^2 x =$$ $$= (\cos^2x+\sin^2x)^2-2\sin^2 x \cos^2 x = $$ $$=1-2\sin^2 x \cos^2 x=\cos^2 x + \sin^2 x \underbrace{(1-2\cos^2 x)}_{\neq \pm 1}$$ So not even this is true. Nothing's true here.
EDIT:
Now it's obvious.
$$\cos^4 x - \sin^4 x -\cos^2 x +\sin^2 x = $$ $$(\cos^2 x -\sin^2 x )\underbrace{(\cos^2 x + \sin^2 x )}_{1}-\cos^2 x +\sin^2 x = 0$$
$\endgroup$ 1 $\begingroup$One more thing that is not true: '2nd part' is not equal to $1$. $$\cos^4x+\sin^4x-\cos^2x+\sin^2x=\cos^4x+\sin^4x-\bbox[5px,border:2px solid #F0A]{(\cos^2x-\sin^2x)}$$
$\endgroup$ 4 $\begingroup$This is true for any x if you take $cos^2x-sin^2x$ common and simplify $$(cos^2x-sin^2x)(cos^2x+sin^2x-1)$$ where $cos^2x+sin^2x=1$ therfore the expression is zero.
$\endgroup$ $\begingroup$Your statement is false in general. Try to plot
$$ y=\cos^4(x)+\sin^4(x) $$
and you will see that it is far from constant!
$\endgroup$ $\begingroup$Notice, your statement isn't true. Now we know that:
$$\cos^2(x)+\sin^2(x)=1$$
So, we can see that:
$$\cos^4(x)+\sin^4(x)-\cos^2(x)-\sin^2(x)=\cos^4(x)+\sin^4(x)-1$$
Now, prove that $\cos^4(x)+\sin^4(x)=\frac{\cos(4x)+3}{4}$:
$$\cos^4(x)+\sin^4(x)=\frac{\cos(4x)+3}{4}\Longleftrightarrow$$$$4\left(\cos^4(x)+\sin^4(x)\right)=\cos(4x)+3\Longleftrightarrow$$
Use $\sin^4(x)=\left(\sin^2(x)\right)^2=\left(1-\cos^2(x)\right)^2$:
$$4\left(\cos^4(x)+\left(1-\cos^2(x)\right)^2\right)=\cos(4x)+3\Longleftrightarrow$$
Use $\left(1-\cos^2(x)\right)^2=1-2\cos^2(x)+\cos^4(x)$:
$$4\left(\cos^4(x)+1-2\cos^2(x)+\cos^4(x)\right)=\cos(4x)+3\Longleftrightarrow$$$$4\left(1-2\cos^2(x)+2\cos^4(x)\right)=\cos(4x)+3\Longleftrightarrow$$$$4-8\cos^2(x)+8\cos^4(x)=\cos(4x)+3\Longleftrightarrow$$
Use $\cos(4x)=\cos(2(2x))=2\cos^2(2x)-1$:
$$4-8\cos^2(x)+8\cos^4(x)=2\cos^2(2x)-1+3\Longleftrightarrow$$$$4-8\cos^2(x)+8\cos^4(x)=2\cos^2(2x)+2\Longleftrightarrow$$
Use $\cos(2x)=2\cos^2(x)-1$:
$$4-8\cos^2(x)+8\cos^4(x)=2\left(2\cos^2(x)-1\right)^2+2\Longleftrightarrow$$$$4-8\cos^2(x)+8\cos^4(x)=4-8\cos^2(x)+8\cos^4(x)$$
So:
$$\color{red}{\cos^4(x)+\sin^4(x)-\cos^2(x)-\sin^2(x)=\frac{\cos(4x)+3}{4}-1=\frac{\cos(4x)-1}{4}\ne0}$$
$\endgroup$ $\begingroup$The edited identity is correct.
Use $a^2 - b^2 = (a+b)(a-b)$ to rewrite $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(cos^2 x + \sin^2 x)$.
The second factor is one ($\sin^2 x + \cos^2 x = 1$), reducing the original expression to:
$\cos^2x - \sin^2x - \cos^2x + \sin^2x = 0$ as required.
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