# Prove that $\tan(75^\circ) = 2 + \sqrt{3}$

My (very simple) question to a friend was how do I prove the following using basic trig principles:

$\tan75^\circ = 2 + \sqrt{3}$

He gave this proof (via a text message!)

$1. \tan75^\circ$

$2. = \tan(60^\circ + (30/2)^\circ)$

$3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \tan60^\circ \tan(30/2)^\circ)$

$4. \tan (30/2)^\circ = \dfrac{(1 - \cos30^\circ)}{ \sin30^\circ}$

Can this be explained more succinctly as I'm new to trigonometry and a little lost after (2.) ?

EDIT

Using the answers given I'm almost there:

1. $\tan75^\circ$
2. $\tan(45^\circ + 30^\circ)$
3. $\sin(45^\circ + 30^\circ) / \cos(45^\circ + 30^\circ)$
4. $(\sin30^\circ.\cos45^\circ + \sin45^\circ.\cos30^\circ) / (\cos30^\circ.\cos45^\circ - \sin45^\circ.\sin30^\circ)$
5. $\dfrac{(1/2\sqrt{2}) + (3/2\sqrt{2})}{(3/2\sqrt{2}) - (1/2\sqrt{2})}$
6. $\dfrac{(1 + \sqrt{3})}{(\sqrt{3}) - 1}$
7. multiply throughout by $(\sqrt{3}) + 1)$

Another alternative approach:

1. $\tan75^\circ$
2. $\tan(45^\circ + 30^\circ)$
3. $\dfrac{\tan45^\circ + \tan30^\circ}{1-\tan45^\circ.\tan30^\circ}$
4. $\dfrac{1 + 1/\sqrt{3}}{1-1/\sqrt{3}}$
5. at point 6 in above alternative
• Hint 75 = 45 + 30 Commented Apr 13, 2013 at 20:58
• @DavidH so there's an easier way? Commented Apr 13, 2013 at 21:00
• My way certainly requires fewer steps. I'd recommend using these values with Mettin's answer. Commented Apr 13, 2013 at 21:04
• From step 6 you just need to rationalize the denominator. Multiple the top and bottom by sqrt(3) + 1. Commented Apr 13, 2013 at 21:45
• @Inceptio thanks - half the effort was writing the TeX commands (my first time) Commented Apr 15, 2013 at 6:53

A proof without words (but it uses some geometry). Is that OK?

• This is very pleasing! Commented Apr 16, 2013 at 23:46
• how did you know that the top triangle was going to be equilateral? (...that certainly helped) Commented Apr 17, 2013 at 14:11
• @whytheq The upper triangle is isoceles, not equilateral. I drew the lower triangle first, extended the vertical side, and marked on it the uppermost vertex to ensure that the upper triangle is isoceles. I know this upper triangle has an exterior angle of $30^\circ$ (because of the way I constructed the lower triangle), and so the two interior opposite angles (which I know have equal measure by the isoceles triangle property) must be $15^\circ$ each. Arrgh... too many words... (Of course, I worked out what I wanted first, and the above explanation is backfill to justify the diagram). Commented Apr 17, 2013 at 14:26

The formula you want to see is: $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ for any degrees $x$ and $y$.

On the other hand, proving this tangent equality from the formulas $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$ and $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ will be a good exercise for a beginner.

You can rather use $\tan (75)=\tan(45+30)$ and plug into the formula by Metin. Cause: Your $15^\circ$ is not so trivial.

• @Inceptio Out of curiosity, what's the equation editor command for the degree symbol? Commented Apr 13, 2013 at 21:09
• It is '\circ'. For $x^\circ$ you can write x^\circ between two \$ :) Commented Apr 13, 2013 at 21:12
• ....something like "15^\circ" Commented Apr 13, 2013 at 21:13
• @David H You shall use detexify.kirelabs.org/classify.html Commented Apr 13, 2013 at 21:15
• Wow, definitely bookmarking that page. Thanks! Commented Apr 13, 2013 at 21:20

$$\begin{array}{l} \tan (75^{\circ})=\frac{\sin (75^{\circ})}{\cos (75^{\circ})}=\frac{\cos (15^{\circ})}{\sin (15^{\circ})}=\frac{\cos (30^{\circ} / 2)}{\sin (30^{\circ} / 2)}=\sqrt{\frac{1+\cos (30^{\circ})}{1-\cos (30^{\circ})}} \\ =\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}=\sqrt{\frac{(2+\sqrt{3})^{2}}{(2-\sqrt{3})(2+\sqrt{3})}}=2+\sqrt{3} \end{array}$$