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My math book has a trigonometry identity $\frac{2\cos2x}{\cos x}$ and they simplify it to $2\cos^2x-1$ but do not show the steps. I have tried many time to simplify but can never get the same answer.

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Doesn't look true: see this –  tomasz Sep 2 '12 at 1:23
    
$2\cos^2x-1$ is identically equal to $\cos(2x)$, which is definitely NOT identically equal to $2\cos(2x)/\cos x$. Someone has a typo. (Or worse than a typo ... I'm not even seeing what the book might have intended.) –  Blue Sep 2 '12 at 1:26
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$\frac{2\cos2x}{\cos x}$ is an expression; it is not an identity. Where's the equal sign? –  Joel Reyes Noche Sep 2 '12 at 5:04

1 Answer 1

Use the fact that $\cos2 x = \left(2\cos^2 (x) - 1\right)$ to get:

$$\frac{2\cos(2x)}{\cos(x)} = \frac{2\left(2 \cos^2(x)-1\right)}{\cos(x)} = 4\cos(x)-\frac{2}{\cos(x)}$$

EDIT : The solution implies that $$\frac{2\cos(2x)}{\cos(x)}\neq 2 \cos^2(x)-1.$$

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Doesn't seem to be an answer to the question. –  Gerry Myerson Sep 2 '12 at 5:50
    
@GerryMyerson Apperantly $2\cos 2x/\cos x\neq 2\cos^2 x-1$. –  Seyhmus Güngören Sep 2 '12 at 9:23
    
Indeed. So perhaps you ought to call attention to that in your answer, since that seems to be the point. –  Gerry Myerson Sep 2 '12 at 11:36
    
@GerryMyerson thanks for the comment. –  Seyhmus Güngören Sep 2 '12 at 11:43

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