To rewrite $\sin^6(x)$ in terms of the first power of cosine, we can use power-reducing formulas. These formulas relate higher powers of sine and cosine to lower powers. The key formulas are:

$\sin^2(x) = \frac{1 - \cos(2x)}{2}$

To break down $\sin^6(x)$, we write it as: $\sin^6(x) = (\sin^2(x))^3$

Now, substitute $\sin^2(x)$ using the power-reducing formula: $\sin^2(x) = \frac{1 - \cos(2x)}{2}$

Thus: $\sin^6(x) = \left(\frac{1 - \cos(2x)}{2}\right)^3$

Next, expand this cube: $\sin^6(x) = \frac{(1 - \cos(2x))^3}{8}$

Now, expand $(1 - \cos(2x))^3$ using the binomial theorem: $(1 - \cos(2x))^3 = 1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x)$

We can replace $\cos^2(2x)$ with another power-reducing formula: $\cos^2(2x) = \frac{1 + \cos(4x)}{2}$

Substituting this: $(1 - \cos(2x))^3 = 1 - 3\cos(2x) + 3\left(\frac{1 + \cos(4x)}{2}\right) - \cos^3(2x)$

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

Finally, divide everything by 8 to fully express $\sin^6(x)$: $\sin^6(x) = \frac{1}{8} \left(1 - 3\cos(2x) + \frac{3}{2}(1 + \cos(4x)) - \cos^3(2x)\right)$

This simplifies $\sin^6(x)$ in terms of cosine.

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