How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$

Question

How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$

I tried to factor and I got $2\cos^4(x)+(-2\sin^2(x)-3)(\cos^4(x)+\sin^4(x))$ but that doesn't lead me to my goal.

I also tried to write all the cosines in terms of sines to have: $-3\sin^6(x)-9\sin^2(x)-2-3\sin^4(x)$

But I don't see how to continue

Any hint is welcome! thnxx

Answer 1

HINT:

Use $\displaystyle \cos^2x=1-\sin^2x$

in $\cos^6x=(\cos^2x)^3,\cos^4x=(\cos^2x)^2$ in the Left Hand Side

to eliminate $\cos x$ as expected in the Right Hand Side

Answer 2

Apply $a^2+b^2=(a+b)^2-2ab$ in $$\sin^4x+\cos^4x=(\sin^2x)^2+(\cos^2x)^2$$

and $a^3+b^3=(a+b)^3-3ab(a+b)$ in $$\sin^6x+\cos^6x=(\sin^2x)^3+(\cos^2x)^3$$

Finally the re-arrange the terms

Answer 3

According to the equation:

$$2(cos^6(x) - sin^6(x)) - 3(cos^4(x) + sin^4(x)) = -4sin^6(x) - 1$$

this means that:

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

And since we are dividing by a negative where $x\in \mathbb{Q}$ I assume, this means that:

$$3(cos^4(x) + sin^4(x)) - 1 > 2(cos^6(x) - sin^6(x))$$

Meaning that:

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

Therefore when we do some of the simplifications, we find that:

$$-4sin^6(x) - 1 < 2 - 6(cos^4(x) + sin^4(x)) + 4cos^6(x) + 8sin^6(x)$$

$$\implies sin^6(x) < \frac{3 - 6(cos^4(x) + sin^4(x)) + 4cos^6(x) + 8sin^6(x)}{-4}$$

$$\therefore \frac{3(cos^4(x) + sin^4(x)) - 1}{2} - cos^6(x) + 2sin^6(x) = \frac{3 - 6(cos^4(x) + sin^4(x)) + 4cos^6(x) + 8sin^6(x)}{-4}$$

So there's your proof. Since there is no contradiction and we basically come to a fact something like $-4n = \frac{-4n}{-4}$ then we cannot go any further with this, meaning that every single equation in this answer is true! So there ya go. Hope I helped! :)