Show that $ \tan(A)=\frac{\sin2A}{1+\cos 2A} $
I've tried a few methods, and it stumped my teacher.
Show that $ \tan(A)=\frac{\sin2A}{1+\cos 2A} $
I've tried a few methods, and it stumped my teacher.
Proof without words: $\tan(A)=\dfrac{\color{red}{\sin(2A)}}{\color{blue}{1}+\color{green}{\cos(2A)}}$
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The given equality is false. Set $A = \pi/2$. (Note: this applied to an earlier version of the problem).
Perhaps what you meant was
$ \tan \frac{A}{2} = \frac{\sin A}{1 + \cos A}$
or
$ \tan A = \frac{\sin 2A}{1 + \cos 2A}$
which is true, by using the half/double angle formulas.
$\frac{\sin A}{1 + \cos A} = \frac{ 2 \sin A/2 \cos A/2}{2 \cos^2 A/2} = \tan A/2$
$\sin 2A = 2 \sin A \cos A$
$\cos 2A = 2 \cos^2A - 1$
Substitute these identities and you will get $\tan A$.
We need to prove that: $\frac{\sin(2A)}{1+\cos(2A)}=\tan(A)$ Let's do LHS-RHS to prove it. I will try to make the left side equal the right side. $\frac{\sin(2A)}{1+\cos(2A)}$ Using double angle identites for both sine and cosine: $\frac{2\sin(A)\cos(A)}{1+2\cos^2(A)-1}$ How nice. The $1$ and $-1$ in the denominator cancel out. $\frac{2\sin(A)\cos(A)}{2\cos^2(A)}$ Cancelling out the $\cos(A)$ in the numerator and the denominator yields: $\frac{2\sin(A)}{2\cos(A)}$ We can also cancel out the $2$ in the numerator and the denominator. $\frac{\sin(A)}{\cos(A)}$ $=\tan(A)$ $\text{LHS=RHS}$ $\displaystyle \boxed{\therefore \dfrac{\sin(2A)}{1+\cos(2A)}=\tan(A)}$
First, lets develop a couple of identities. Given that $\sin 2A = 2\sin A\cos A$, and $\cos 2A = \cos^2A - \sin^2 A$ we have
$\begin{array}{lll} \tan 2A &=& \frac{\sin 2A}{\cos 2A}\\ &=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\\ &=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\cdot\frac{\frac{1}{\cos^2 A}}{\frac{1}{\cos^2 A}}\\ &=&\frac{2\tan A}{1-\tan^2A} \end{array}$ Similarly, we have $\begin{array}{lll} \sec 2A &=& \frac{1}{\cos 2A}\\ &=&\frac{1}{\cos^2 A-\sin^2A}\\ &=&\frac{1}{\cos^2 A-\sin^2A}\cdot\frac{\frac{1}{\cos^2 A}}{\frac{1}{\cos^2 A}}\\ &=&\frac{\sec^2 A}{1-\tan^2A} \end{array}$
But sometimes it is just as easy to represent these identities as
$\begin{array}{lll} (1-\tan^2 A)\sec 2A &=& \sec^2 A\\ (1-\tan^2 A)\tan 2A &=& 2\tan A \end{array}$
Applying these identities to the problem at hand we have $\begin{array}{lll} \frac{\sin 2A}{1+\cos 2A}&=& \frac{\sin 2A}{1+\cos 2A}\cdot\frac{\frac{1}{\cos 2A}}{\frac{1}{\cos 2A}}\\ &=& \frac{\tan 2A}{\sec 2A +1}\\ &=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)(\sec 2A +1)}\\ &=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)\sec 2A +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{\sec^2 A +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{(\tan^2 A+1) +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{2}\\ &=& \tan A\\ \end{array}$ discalimer: I like JChau's answer better :)