Why does $$\frac{p(\cos(x) - p)}{1 - 2p\cos(x)+p^2} = \frac{1}{2} \cdot \left( \frac{1 - p^2}{1 - 2p\cos(x) + p^2} - 1 \right)$$ I'm unsure as to the steps to take to show that it is equivalent.
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4Write a common denominator on the right side. – Feb 22 '14 at 21:36
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http://math.stackexchange.com/questions/685705/what-steps-are-taken-to-make-this-complex-expression-equal-this – lab bhattacharjee Feb 23 '14 at 04:17
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Start from the RHS by adding the terms in the bracket. $$\frac{1-p^2}{1-2p\cos x+p^2}-\frac{1-2p\cos x+p^2}{1-2p\cos x+p^2}$$ Simplify. $$\frac{1-p^2-(1-2p\cos x+p^2)}{1-2p\cos x+p^2}=\frac{2p\cos x-2p^2}{1-2p\cos x+p^2}$$ Factor out $2p$. $$\frac{2p(\cos x-p)}{1-2p\cos x+p^2}$$ Multiply by half and you'll get the LHS.
Zhoe
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