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Answer 1 · 2016-04-12T13:36:44

$2 {(cos (4 α) + 1)}^{2}$ $2(cos(4alpha)+1)^2$

We should initially rewrite $cos (8 α)$ $cos(8alpha)$ using the following cosine double-angle identity:

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

We can apply this to $cos (8 α)$ $cos(8alpha)$ as follows:

$cos (8 α) = 2 {cos}^{2} (4 α) - 1$ $cos(8alpha)=2cos^2(4alpha)-1$

Plugging this into the original expression, we see that it equals

$3 + 4 cos (4 α) + 2 {cos}^{2} (4 α) - 1$ $3+4cos(4alpha)+2cos^2(4alpha)-1$

$= 2 + 4 cos (4 α) + 2 {cos}^{2} (4 α)$ $=2+4cos(4alpha)+2cos^2(4alpha)$

Factor a $2$ $2$ from each term and rearrange order.

$= 2 ({cos}^{2} (4 α) + 2 cos (4 α) + 1)$ $=2(cos^2(4alpha)+2cos(4alpha)+1)$

Finally note that $c^{2} + 2 c + 1 = {(c + 1)}^{2}$ $c^2+2c+1 = (c+1)^2$ , so we can write

$= 2 {(cos (4 α) + 1)}^{2}$ $=2(cos(4alpha)+1)^2$