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How would I prove the following two trigonometric identity.

$$\cot A\sin 2A=1+\cos 2A$$

This is my work so far

$$\frac{\cos A}{\sin A}(2\sin A \cos A)=1+\cos 2A$$

I am not sure what I would do next to make them equal.

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0% accept rate is very low. Why?! Please consider accepting and also upvoting some of the useful answers. Click here to know how. –  user2468 Jul 26 '12 at 17:25
    
I would but it wont let me click the checkmark. –  Fernando Martinez Jul 26 '12 at 17:27
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4 Answers

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\begin{eqnarray*} cotAsin2A=\frac{cosA}{sinA}(2sinAcosA)=cosA(2cosA)=2cos^2A=1+cos2A \end{eqnarray*}

This should be everything.

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After cancellation, you've got $2\cos^2A$ on the left. One of your double angle identities will get you the rest of the way.

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Can you give a hint of which one is it 2cos^2-1? –  Fernando Martinez Jul 26 '12 at 17:24
    
Precisely that one. That will let you rewrite $2\cos^2 A$ in terms of $\cos(2A)$--namely, $2\cos^2 A=1+\cos(2A)$, as desired. –  Cameron Buie Jul 26 '12 at 17:34
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$$ \frac{\cos A}{\sin A}(2\sin A \cos A) = 2\cos^2 A \\ $$ Then use the identity $2\cos^2 A = \cos(2A) + 1$ from this link here.

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Oh I see I have a website with the double angle indemnities yet they can be written in more ways than just 2cos^2A-1=cos(2A). –  Fernando Martinez Jul 26 '12 at 17:42
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$$\cot (a) \cdot \sin(2a) = 1 + 1 - 2\sin^2(a)$$

$$=\frac{\cos (a)}{\sin (a)} \cdot 2\sin(a)\cos(a) = 2(1-\sin^2(a))$$

$$=2\cos^2(a) = 2\cos^2(a)$$

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