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.
Trigonometry identity help: $\frac{2\cos\ 2x}{\cos\ x}$ and $2 \cos^2\ x-1$
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trigonometry
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0Doesn't look true: [see this](http://www.wolframalpha.com/input/?i=%282cos%282x%29%29%2Fcos%28x%29+-%282cos^2%28x%29-1%29) – 2012-09-02
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0$2\cos^2x-1$ 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.) – 2012-09-02
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1$\frac{2\cos2x}{\cos x}$ is an expression; it is not an identity. Where's the equal sign? – 2012-09-02
1 Answers
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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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0Doesn't seem to be an answer to the question. – 2012-09-02
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0@GerryMyerson Apperantly $2\cos 2x/\cos x\neq 2\cos^2 x-1$. – 2012-09-02
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0Indeed. So perhaps you ought to call attention to that in your answer, since that seems to be the point. – 2012-09-02
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0@GerryMyerson thanks for the comment. – 2012-09-02