Question #bdbc2

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Question #bdbc2

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Answer 1 · 2016-05-23T00:34:30

See below.

Explanation:

To prove that: ${cos}^{4} (x) = \frac{3}{8} + \frac{1}{2} cos (2 x) + \frac{1}{8} cos (4 x)$ $cos^4(x) = 3/8+1/2cos(2x)+1/8cos(4x)$

First we have that:

${cos}^{4} (x) = {cos}^{2} (x) {cos}^{2} (x)$ $cos^4(x)=cos^2(x)cos^2(x)$

We can now make use of trig identity:

${cos}^{2} (x) = \frac{1}{2} cos (2 x) + \frac{1}{2}$ $cos^2(x) = 1/2cos(2x) +1/2$

So we can say that:

${cos}^{4} (x) = (\frac{1}{2} cos (2 x) + \frac{1}{2}) (\frac{1}{2} cos (2 x) + \frac{1}{2})$ $cos^4(x) = (1/2cos(2x)+1/2)(1/2cos(2x)+1/2)$

Expanding these brackets gives us:

$\frac{1}{4} {cos}^{2} (2 x) + \frac{1}{2} cos (2 x) + \frac{1}{4}$ $1/4cos^2(2x)+1/2cos(2x)+1/4$

Now using the identity again we can say that

${cos}^{2} (2 x) = \frac{1}{2} cos (4 x) + \frac{1}{4}$ $cos^2(2x) = 1/2cos(4x) +1/4$

Substituting this into our expression gives:

${cos}^{4} (x) = \frac{1}{4} (\frac{1}{2} cos (4 x) + \frac{1}{2}) + \frac{1}{2} cos (2 x) + \frac{1}{4})$ $cos^4(x) = 1/4(1/2cos(4x)+1/2)+1/2cos(2x)+1/4)$

$= \frac{1}{8} cos (4 x) + \frac{1}{8} + \frac{1}{2} cos (2 x) + \frac{1}{4}$ $=1/8cos(4x)+1/8 +1/2cos(2x) +1/4$

$= \frac{1}{8} cos (4 x) + \frac{1}{2} cos (2 x) + \frac{3}{8}$ $=1/8cos(4x)+1/2cos(2x)+3/8$

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Question #bdbc2

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