Where's my misstep in this trigonometric problem?

Question

$$\begin{align} \tan\alpha+\cot\alpha=\sqrt{ 6 } \\ \tan^6\alpha+\cot^6\alpha= \ ? \end{align}$$

The given answer is $52$, but I got $214$ instead:

\begin{align} \tan\alpha+\cot\alpha=\frac{1}{\sin\alpha \cos\alpha}&=\sqrt{ 6 } \\ \\ \tan^6\alpha+\cot^6\alpha=\frac{\sin^6\alpha}{\cos^6\alpha}+\frac{\cos^6\alpha}{\sin^6\alpha}&= \\ \\ \frac{\sin^{12}\alpha+\cos^{12}\alpha}{\sin^6\alpha \cos^6\alpha}&= \\ \\ \frac{(\sin^3\alpha)^4+(\cos^3\alpha)^4}{\sin^6\alpha \cos^6\alpha}&= \\ \\ \frac{1-2\sin^6\alpha \cos^6\alpha}{\sin^6\alpha \cos^6\alpha}&= \\ \\ \frac{1}{\sin^6\alpha \cos^6\alpha}-2&= \\ \\ (\sqrt{ 6 })^6-2=216-2&=214 \end{align}

I know my method is not the most efficient, but my question is: what is wrong with my solution?

Answer 1

What is Correct :

Let $\tan \alpha = x$ , then $\cot \alpha = 1/x$

$$x+1/x=\sqrt{6}$$

Squaring , we get :

$x^2+2+1/x^2=6$

$x^2+1/x^2=4$

Cubing , we get :

$x^6+3x^4/x^2+3x^2/x^4+1/x^6=64$

$x^6+3(x^2+1/x^2)+1/x^6=64$

$x^6+12+1/x^6=64$

$x^6+1/x^6=52$

$$\tan^6 \alpha + \cot^6 \alpha = 52$$

Where you went wrong :

This is totally suspicious :

Answer 2

Try evaluating $(\sin^3\alpha)^4+(\cos^3\alpha)^4$ for $x=\frac\pi4.$ Then do the same for $1-2\sin^6\alpha \cos^6\alpha.$

$$ \sin\frac\pi4=\cos\frac\pi4=\frac{\sqrt2}{2}. $$ $$ \sin^6\frac\pi4=\cos^6\frac\pi4=\frac18. $$ $$\left(\sin^3\frac\pi4\right)^4=\left(\cos^3\frac\pi4\right)^4=\frac{1}{64}.$$

Answer 3

Now that we can see what happened with that failed step, we can provide further details about what went wrong. An identity was misapplied. If $\sin^3\alpha=\sin\beta, \cos^3\alpha\ne\cos\beta$. The identity you attempted to use can be proved by completing the square: $$\sin^4\alpha+\cos^4\alpha=\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\cos^4\alpha-2\sin^2\alpha\cos^2\alpha=$$$$(\sin^2\alpha+\cos^2\alpha)^2-2\sin^2\alpha\cos^2\alpha=1-2\sin^2\alpha\cos^2\alpha$$

We can use the same method to correct that step. If we add and subtract $2\sin^6\alpha\cos^6\alpha$, it becomes $$\dfrac{\sin^{12}\alpha+\cos^{12}\alpha}{\sin^6\alpha\cos^6\alpha}=\dfrac{(\sin^6\alpha+\cos^6\alpha)^2}{\sin^6\alpha\cos^6\alpha}-2$$ If you wished to proceed from here, it appears you can arrive at an answer by completing the cube of $\sin^2\alpha+\cos^2\alpha$.

Answer 4

Thanks for everyone's help on this. It seems like I mistakenly tried to apply this Trigonometric Identity: $$\sin^4\alpha+\cos^4\alpha=1-2\sin^2\alpha\cos^2\alpha$$

Like this, which is supposedly wrong:

$$\begin {align} \sin^{12}\alpha + \cos^{12}\alpha&=\\(\sin^3\alpha)^4+(\cos^3\alpha)^4&= 1-2\sin^6\alpha \cos^6\alpha \end{align}$$