Prove $32 \sin^6 x = -15 \cos (2x) + 6 \cos (4x) - \cos (6x) + 10$?

Question

I already tried to do it, but my answer is incorrect! This is how I do it.

assuming $$2 \cos x = z + \frac1z\space\space\text{ and }\space\space 2j \sin x = z - \frac1z$$ $$(2j \sin x)^6 \cdot \frac12 = \left(z - \frac1z\right)^6 \cdot \frac12$$ $$32j \sin^6 x \cdot \frac12 = \left(z - \frac1z\right)^6 \cdot \frac12$$ I expanded $(z - \frac1z)^6$ using pascal's triangle, however this is what I got: $$z^6 + \frac1{z^6} - 6 \left(z^4 + \frac1{z^4}\right) + 15 \left(z^2 + \frac1{z^2}\right) - 20$$ $$2 \cos {6x} - 12 \cos {4x} + 30 \cos {2x} - 20$$ multiply it with $\frac12$, final answer is: $$\cos {6x} - 6 \cos {4x} + 15 \cos {2x} - 10$$

The "+" and "-" signs seems to be reversed compared to the actual answer, however I don't know what I did wrong! Can anyone help me?

Sorry for bad question format, I'm new to using StackExchange. Thank you for your help!

Answer 1

$2\sin^2x = 1- \cos2x$

So,

$8\sin^6x = (2\sin^2x)^3 = (1-\cos2x)^3 = 1 - 3\cos2x+3\cos^2(2x)-\cos^3(2x)$

Now, $\cos^2A = \frac{1+\cos(2A)}{2}$ and $\cos^3(A) = \frac{\cos(3A) + 3\cos(A)}{4}$

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

Multipying both sides by $4$

$32\sin^6x = 4 - 12\cos(2x) + 6 + 6\cos(4x) - \cos(6x) - 3\cos(2x)$

$$32\sin^6x = -15\cos(2x)+6\cos(4x)- \cos(6x)+10$$