# 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.

• FYI, $(A + B)^2 = A^2 + B^2$ is not true in general. Jan 24, 2022 at 1:48
• It should be $$A^2 + B^2 = (A + B)^2 - 2AB.$$ Jan 24, 2022 at 1:49
• I guess you should try to use the fact that $sin^2x + cos^2x = 1$ and that $sin2x = 2sinxcosx$. That is $1=(sin^2x + cos^2x)^2 = ...$ Jan 24, 2022 at 1:50
• Also, $(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^2(2x)}{2}$. Jan 24, 2022 at 1:53
• My question, therefore, is how does one transform $\sin^4 x + \cos^4x$ into $\sin^4x + \cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\;?$ $\quad$ Let $A=\sin^2x\,$ and $\,B=\cos^2x.$ Then \begin{align}\text{RHS}&=(A+B)^2-2AB\\ &=(A^2+2AB+B^2)-2AB\\ &=A^2+B^2\\ &=\text{LHS}.\end{align} Jan 24, 2022 at 4:06

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

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

$$\sin(2x) = ~$$ ?

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

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

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

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

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 $$\displaystyle \sin(2x) = \frac{2}{3}.$$

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

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

$$2 sc = \sin 2x = \pm \frac23$$