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Chapter 5: Problem 44

Use the power-reducing formulas to rewrite the expression in terms of the first power of the cosine. $$\sin ^{4} 2 x$$

Short Answer

Expert verified
\(\frac{3 - 4\cos(4x) + \cos(8x)}{8}\)

Step by step solution

01

Application of the Power-Reducing Formula

We can re-write \(\sin^{4}2x\) as \((\sin^{2}2x)^{2}\). Then apply the power-reducing formula for \(\sin^2(x)\). This formula states that \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\). Therefore, \((\sin^{2}2x)^{2}\) can be rewritten as \(\left(\frac{1 - \cos(4x)}{2} \right)^2\)
02

Squaring the Expression

Next, let's square the expression from the previous step. That means, let's compute \(\left(\frac{1 - \cos(4x)}{2} \right)^2\), which will give us \(\frac{1 - 2\cos(4x) + \cos^2(4x)}{4}\)
03

Use Power-Reducing Formula for \(\cos^{2}(4x)\)

To further simplify \(\cos^{2}(4x)\) term and express it in first power of cosine, let's apply the power-reducing formula for \(\cos^{2}(x)\) which is \(\cos^{2}(x) = \frac{1 + \cos(2x)}{2}\). Using this, we can rewrite the \(\cos^{2}(4x)\) term to be \(\cos^{2}(4x) = \frac{1 + \cos(8x)}{2}\). Substituting this into our equation from the previous step gives us \(\frac{1 - 2\cos(4x) + \frac{1 + \cos(8x)}{2}}{4}\).
04

Simplifying the Final Expression

Let's simplify the equation \(\frac{1 - 2\cos(4x) + \frac{1 + \cos(8x)}{2}}{4}\) by finding the common denominator and simplifying. This results in \(\frac{2 - 4\cos(4x) + 1 + \cos(8x)}{8} = \frac{3 - 4\cos(4x) + \cos(8x)}{8}\).

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