Finding $\sin 2x$ from transforming $\sin^4 x+ \cos^4 x = \frac{7}{9}$ using trigonometric identities While studying for the first exams of the year, the following question found on one of Kognity's questionbank was extremely challenging for myself, a pre calculus student. The current topic is that of trigonometric equations, mostly making use of basic trigonometric identities (see image below for a screenshot of my formula booklet). Even after seeing the resolution, I just did not understand what magic was used to transform the sine and cosine.
The question below stated: Given that X is in quadrant III and $\sin^4 x + \cos^4 x = \frac{7}{9}$, work out the value of $\sin 2x$.
The question in its original format.
At first, one may notice that: $$\sin^4 x + \cos ^4 x=\left(\sin^2 x +\cos^2x\right)^2=1$$
because of the fundamental trigonometrical identity ($\sin^2x+\cos^2x=1)$. However to my frustration, this was as far as I got. Substituting by 1 did not yield any result, and I was forced to admit defeat and to submit the question without a resolution. Fortunately, I was at least pleased that I would have access to the resolution. Yet upon seeing the resolution, my confusion only doubled.
Resolution of the question.
By making use of black magic, the resolution transformed $$(\sin^4 x + \cos^4x)$$ into $$(\sin^4 x + \cos^4x)=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$$ Analysing and transforming the terms with the identities, one reaches that these two terms could be rewritten as: $$1-\sin^22x$$
The rest of the question was very straightforward, simple algebra to find an expression for X and noticing that $\sin2x=2\sin x\cos x$ were both in the third quadrant, therefore a negative multiplying a negative resulting in a positive result. My question, therefore, is how does one transform $$(\sin^4 x + \cos^4x)$$ into $$(\sin^4 x + \cos^4x)=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$$ It seems illogical since the second term appeared out of nowhere.
 A: $$
\begin{array}{c}
\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x=\frac{1}{9} \\
1-\frac{\sin ^{2}(2 x)}{2}=\frac{7}{9} \\
\sin ^{2}(2 x)=\frac{4}{9} \\
\sin (2 x)=\frac{2}{3}\quad \textrm{ (as x is the quadrant III)}
\end{array}
$$
A: Hint
Since you did most of the work...
$$s^4+c^4 =1-2 s^2c^2= \frac79$$
$$ 2 s^2c^2=  2/9;\; s^2c^2=  1/9;\;$$
Multiply by 4, take sqrt
$$ 2 sc = \sin 2x = \pm \frac23 $$
Negative sign is discarded to obey given quadrant condition.
A: 
$\displaystyle \sin^4(x) + \cos^4(x) = \frac{7}{9}.$


$\sin(2x) = ~$ ?

$\underline{\text{Preliminary Results}}$
$$\cos(2x) = 2\cos^2(x) - 1 \implies 2\cos^2(x) = \cos(2x) + 1.\tag{R-1}$$
$$\cos^4(x) = \frac{1}{8} \times [\cos(4x) + 4\cos(2x) + 3]. \tag{R-2}$$
$$\sin^4(x) = \frac{1}{8} \times [\cos(4x) - 4\cos(2x) + 3]. \tag{R-3}$$
See the Addendum for a proof of (R-2) and (R-3).

Using (R-2) and (R-3),
$\displaystyle \frac{7}{9} = \frac{1}{4} \times 
[\cos(4x) + 3] \implies $
$\displaystyle \frac{28}{9} = \cos(4x) + 3 \implies $
$\displaystyle \frac{1}{9} = \cos(4x) = 
\cos^2(2x) - \sin^2(2x) = 1 - 2\sin^2(2x) \implies $
$\displaystyle \frac{8}{9} = 1 - \frac{1}{9} = 2\sin^2(2x) \implies $
$\displaystyle \sin^2(2x) = \frac{4}{9} \implies $
$\displaystyle \sin(2x) = \pm \frac{2}{3}.$
Given that $x$ is in quadrant $3$, you have that
$180^\circ < x < 270^\circ$.
This implies that $360^\circ < 2x < 540^\circ.$
This implies that $\displaystyle \sin(2x) = \frac{2}{3}.$

Addendum
Proof of (R-2) and (R-3).
Proof of (R-2)
Using (R-1), 
$\cos(4x) = 2\cos^2(2x) - 1 $
$\displaystyle = 2 \left[2\cos^2(x) - 1\right]^2 - 1$
$ = 2[4\cos^4(x) - 4\cos^2(x) + 1] - 1$
$= 8\cos^4(x) - 8\cos^2(x) + 1$
$= 8\cos^4(x) - 4[\cos(2x) + 1] + 1$
$= 8\cos^4(x) - 4\cos(2x) - 3.$
This implies that
$\cos(4x) + 4\cos(2x) + 3 = 8\cos^4(x).$
Proof of (R-3)
$\sin^4(x) = [\sin^2(x)]^2 = [1 - \cos^2(x)]^2$
$\displaystyle = 1 - 2\cos^2(x) + \cos^4(x)$
$\displaystyle = 1 + \cos^4(x) - [\cos(2x) + 1]$
[using R-2]
$\displaystyle = \left\{ ~\frac{1}{8} \times [\cos(4x) + 4\cos(2x) + 3] ~\right\} - \cos(2x).$
