Show that $$ \tan(A)=\frac{\sin2A}{1+\cos 2A} $$
I've tried a few methods, and it stumped my teacher.
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asked Mar 15, 2012 at 23:44
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1$\begingroup$ Failure to prove is not surprising. It is quite false. For example, take $A=\pi/4$ ($45$ degrees). $\endgroup$– André NicolasMar 15, 2012 at 23:48
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$\begingroup$ I think you want $\cos(2A)$ there... $\endgroup$– David MitraMar 15, 2012 at 23:53
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$\begingroup$ This is not a valid identity: try $A=\pi/4$ for example. $\endgroup$– Shane O RourkeMar 15, 2012 at 23:54
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5 Answers
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Proof without words: $\tan(A)=\dfrac{\color{red}{\sin(2A)}}{\color{blue}{1}+\color{green}{\cos(2A)}}$
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3$\begingroup$ I've never seen anyone prove a trig identity like that, but they should! $\endgroup$ Mar 29, 2012 at 23:54
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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.
$$\frac{\sin A}{1 + \cos A} = \frac{ 2 \sin A/2 \cos A/2}{2 \cos^2 A/2} = \tan A/2$$
answered Mar 15, 2012 at 23:51
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$\begingroup$ This is one of my favorite identities. It is the basis for one form of $\operatorname{atan2}(x,y)=2\operatorname{atan}\left(\dfrac{y}{r+x}\right)$ which is useful if you have to compute $r=\sqrt{x^2+y^2}$ anyway. It also plays a significant role in the stereographic projection. (+1) $\endgroup$– robjohn ♦Mar 29, 2012 at 22:40
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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. $$\frac{\sin(A)}{\cos(A)}$$ $$=\tan(A)$$ $$\text{LHS=RHS}$$ $$\displaystyle \boxed{\therefore \dfrac{\sin(2A)}{1+\cos(2A)}=\tan(A)}$$
answered Mar 15, 2014 at 3:31
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$$\sin 2A = 2 \sin A \cos A$$
$$\cos 2A = 2 \cos^2A - 1$$
Substitute these identities and you will get $\tan A$.
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1$\begingroup$ I am downvoting this answer. BTW, you may want to look at the TeX edit I have made by clicking on the time stamp above my name. $\endgroup$– user21436Mar 29, 2012 at 22:28
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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 $$\begin{array}{lll} \frac{\sin 2A}{1+\cos 2A}&=& \frac{\sin 2A}{1+\cos 2A}\cdot\frac{\frac{1}{\cos 2A}}{\frac{1}{\cos 2A}}\\ &=& \frac{\tan 2A}{\sec 2A +1}\\ &=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)(\sec 2A +1)}\\ &=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)\sec 2A +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{\sec^2 A +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{(\tan^2 A+1) +(1-\tan^2 A)}\\ &=& \frac{2\tan A}{2}\\ &=& \tan A\\ \end{array}$$ Lessons learned: As just a quick scan of some of the other answers will indicate, a clever substitution can shorten your workload considerably.
