A Trigonometry problem

Problem:

Prove $\frac{\sin^6(x) + \cos^6(x) - 1}{\sin^4(x) + \cos^4(x) - 1} = \frac{3}{2}$.

Solution:

Let's first divide as if we do not know they are trigonometry functions. We have

$\begin{eqnarray*} & & (\sin^4(x) + \cos^4(x) - 1)(\sin^2(x) + \cos^2(x)) \\ &=& \sin^6(x) + \cos^4(x)\sin^2(x) -\sin^2(x) + \sin^4(x)\cos^2(x) + \cos^6(x) - \cos^2(x) \\ &=& \sin^6(x) + \cos^6(x) + \cos^4(x)\sin^2(x) + \sin^4(x)\cos^2(x) -\sin^2(x) - \cos^2(x) \\ &=& \sin^6(x) + \cos^6(x) + \cos^2(x)\sin^2(x)(\cos^2(x) + \sin^2(x)) -\sin^2(x) - \cos^2(x) \end{eqnarray*}$

Of course, we know they are actually trigonometry functions, therefore we can simplify $\sin^2(x) + \cos^2(x) = 1$ and therefore we have

$\begin{eqnarray*} & & (\sin^4(x) + \cos^4(x) - 1) \\ &=& (\sin^4(x) + \cos^4(x) - 1)(\sin^2(x) + \cos^2(x)) \\ &=& \sin^6(x) + \cos^6(x) + \cos^2(x)\sin^2(x)(\cos^2(x) + \sin^2(x)) -\sin^2(x) - \cos^2(x) \\ &=& \sin^6(x) + \cos^6(x) + \cos^2(x)\sin^2(x) - 1 \end{eqnarray*}$

Rearranging, we have $\begin{eqnarray*} \sin^6(x) + \cos^6(x) - 1 = (\sin^4(x) + \cos^4(x) - 1) - \cos^2(x)\sin^2(x) \end{eqnarray*}$ , note this equation, we will use it later.

Now we see some $\cos^2(x)$ and $\sin^2(x)$, next, let try to simplify the 4th power to the 2nd power

$\begin{eqnarray*} & & 1 \\ &=& (\sin^2(x) + \cos^2(x))^2 \\ &=& \sin^4(x) + 2\sin^2(x)\cos^2(x) + \cos^4(x) \end{eqnarray*}$

Rearranging, we have $\begin{eqnarray*} \sin^4(x) + \cos^4(x) - 1 = - 2\sin^2(x)\cos^2(x) \end{eqnarray*}$ , note this equation, we will use it later.

Putting them altogether, we finally conclude:

$\begin{eqnarray*} & & \frac{\sin^6(x) + \cos^6(x) - 1}{\sin^4(x) + \cos^4(x) - 1} \\ &=& \frac{(\sin^4(x) + \cos^4(x) - 1) - \cos^2(x)\sin^2(x)}{\sin^4(x) + \cos^4(x) - 1} \\ &=& 1 - \frac{\cos^2(x)\sin^2(x)}{\sin^4(x) + \cos^4(x) - 1} \\ &=& 1 - \frac{\cos^2(x)\sin^2(x)}{- 2\sin^2(x)\cos^2(x)} \\ &=& 1 - \frac{1}{- 2} \\ &=& \frac{3}{2} \\ \end{eqnarray*}$

Q.E.D.