Rewrite cos^4(2t) without using any powers.

(1/4)⋅(3+4cos4t+cos8t)
(1/8)⋅(3+4sin4t+sin8t)
(1/8)⋅(3+4cos4t+cos8t)
(1/8)⋅(3−4cos4t+cos8t)

1 answer

To rewrite \( \cos^4(2t) \) without using any powers, we can utilize the double angle and power-reducing formulas.

First, we can use the following power-reducing identity:

\[ \cos^2(x) = \frac{1 + \cos(2x)}{2} \]

Applying this identity to \( \cos^2(2t) \):

\[ \cos^2(2t) = \frac{1 + \cos(4t)}{2} \]

Next, we will square \( \cos^2(2t) \) to find \( \cos^4(2t) \):

\[ \cos^4(2t) = \left( \cos^2(2t) \right)^2 = \left( \frac{1 + \cos(4t)}{2} \right)^2 \]

This expands to:

\[ \cos^4(2t) = \frac{(1 + \cos(4t))^2}{4} = \frac{1 + 2\cos(4t) + \cos^2(4t)}{4} \]

Now, we need to rewrite \( \cos^2(4t) \) using the power-reducing formula again:

\[ \cos^2(4t) = \frac{1 + \cos(8t)}{2} \]

Substituting this back into our expression for \( \cos^4(2t) \):

\[ \cos^4(2t) = \frac{1 + 2\cos(4t) + \frac{1 + \cos(8t)}{2}}{4} \]

Now simplifying,

\[ = \frac{1 + 2\cos(4t) + \frac{1}{2} + \frac{\cos(8t)}{2}}{4} \]

\[ = \frac{\frac{2}{2} + \frac{1}{2} + 2\cos(4t) + \frac{\cos(8t)}{2}}{4} \]

\[ = \frac{\frac{3}{2} + 2\cos(4t) + \frac{\cos(8t)}{2}}{4} \]

Now, multiplying through by \( \frac{1}{4} \) gives:

\[ = \frac{1}{8} \left( 3 + 4\cos(4t) + \cos(8t) \right) \]

Thus, we have rewritten \( \cos^4(2t) \) without using any powers:

\[ \cos^4(2t) = \frac{1}{8} \left( 3 + 4\cos(4t) + \cos(8t) \right) \]

The correct answer is:

\[ \boxed{\frac{1}{8} (3 + 4\cos(4t) + \cos(8t))} \]