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.
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.
\begin{eqnarray*} cotAsin2A=\frac{cosA}{sinA}(2sinAcosA)=cosA(2cosA)=2cos^2A=1+cos2A \end{eqnarray*}
This should be everything.
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.
$$ \frac{\cos A}{\sin A}(2\sin A \cos A) = 2\cos^2 A \\ $$ Then use the identity $2\cos^2 A = \cos(2A) + 1$ from this link here.
$$\cot (a) \cdot \sin(2a) = 1 + 1 - 2\sin^2(a)$$
$$=\frac{\cos (a)}{\sin (a)} \cdot 2\sin(a)\cos(a) = 2(1-\sin^2(a))$$
$$=2\cos^2(a) = 2\cos^2(a)$$