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I really am not able to find any identity which results in this expression, any help regarding how we obtain the right-hand side of the equation $$\cos2x + 1 = 2 \cos^2 x$$ would be really appreciated. :)

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Hint: $\cos(2x)=\cos^2(x)-\sin^2(x)\quad$ and $\quad\sin^2(x)+\cos^2(x)=1$


Observation: Both of the identities above follow from $$ \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) $$ The first by setting $a=b=x$, the second by setting $a=-b=x$.

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Using the identities $$\cos{(a+b)}=\cos{(a)} \cos{(b)}-\sin{(a)} \sin{(b)}$$ and $$\sin^2{(x)}+\cos^2{(x)}=1 \Rightarrow \sin^2{(x)}=1-\cos^2{(x)}$$ we have the following: $$\cos{(2x)}+1=\cos{(x+x)}+1=\cos{(x)} \cos{(x)}-\sin{(x)} \sin{(x)}+1= \\ \cos^2{(x)}-\sin^2{(x)}+1=\cos^2{(x)}-(1-\cos^2{(x)})+1=2\cos^2{(x)}-1+1=2 \cos^2{(x)}$$

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    $\begingroup$ (+1) Your answer prompted me to make the observation that both of the identities in my answer follow from the formula for the cosine of a sum. :-) $\endgroup$ – robjohn Sep 1 '14 at 15:54

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