Finding $\sin 2x$ from transforming $\sin^4 x+ \cos^4 x = \frac{7}{9}$ using trigonometric identities

Question

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.

Answer 1

$$ \begin{array}{c} \left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x=\frac{1}{9} \\ 1-\frac{\sin ^{2}(2 x)}{2}=\frac{7}{9} \\ \sin ^{2}(2 x)=\frac{4}{9} \\ \sin (2 x)=\frac{2}{3}\quad \textrm{ (as x is the quadrant III)} \end{array} $$

Answer 2

$\displaystyle \sin^4(x) + \cos^4(x) = \frac{7}{9}.$

$\sin(2x) = ~$ ?

$\underline{\text{Preliminary Results}}$

$$\cos(2x) = 2\cos^2(x) - 1 \implies 2\cos^2(x) = \cos(2x) + 1.\tag{R-1}$$

$$\cos^4(x) = \frac{1}{8} \times [\cos(4x) + 4\cos(2x) + 3]. \tag{R-2}$$

$$\sin^4(x) = \frac{1}{8} \times [\cos(4x) - 4\cos(2x) + 3]. \tag{R-3}$$

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]
$\displaystyle = \left\{ ~\frac{1}{8} \times [\cos(4x) + 4\cos(2x) + 3] ~\right\} - \cos(2x).$

Answer 3

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.