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Prove that:

$$\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$$

My Work: I guess that I have to use the formula : $$\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}$$

But, I am not being able to use it. Please help me.

Lovecraft
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pi-π
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  • If you have knowledge of $15$ and $22.5$ (half of $45$) then you can take the difference of these two, and use the tangent difference formula. – Sarvesh Ravichandran Iyer Jan 11 '17 at 12:28
  • Or you can start from $\tan 30^\circ$ and bisect it twice. Notice that the relation you wrote, with the due considerations, becomes $\beta \tan^2 \frac A2+2 \tan \frac A2-\beta=0$ if you set $\beta=\tan A$. –  Jan 11 '17 at 12:29
  • What is the value of $Tan 15°$ and $Tan22.5°$? – pi-π Jan 11 '17 at 12:29

5 Answers5

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Hint: We have $$\tan 7.5^\circ =\frac {\sin 7.5^\circ}{\cos 7.5^\circ} =\frac {2\sin^2 7.5^\circ}{2\cos 7.5^\circ \sin 7.5^\circ} =\frac {1-\cos 15^\circ}{\sin 15^\circ} =\frac {1- \cos (45^\circ -30^\circ)}{\sin (45^\circ -30^\circ)} $$

Can you take it from here? Hope it helps.

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$\cos 15^{\circ}=\cos(45^{\circ}-30^{\circ})=\cos45^{\circ}\cos30^{\circ}+\sin30^{\circ}\sin45^{\circ}=\frac{\sqrt6+\sqrt2}{4}$

$$\tan \frac{15^{\circ}}2=\sqrt{\frac{1-\cos15^{\circ}}{1+\cos15^{\circ}}}$$

Roman83
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2

You may prove that identity in a very geometric flavour, i.e. by bisecting twice a $30^\circ$ angle, through the Pythagorean theorem and the bisector theorem. It is best to keep the expressions of the involved lengths as simple as possible during the process. So, let we consider a triangle $BAE$ with $BA=\sqrt{3}, AE=2, EB=1$. Let $AD$ be the bisector of $\widehat{BAE}$ and $AC$ be the bisector of $\widehat{BAD}$. enter image description here

By the bisector theorem, we have: $$ BD = \frac{\sqrt{3}}{2+\sqrt{3}} = 2\sqrt{3}-3. \tag{1}$$ By the Pythagorean theorem, it follows that: $$ AD^2 = 3+(2\sqrt{3}-3)^2 = 24-12\sqrt{3} = 6(\sqrt{3}-1)^2 \tag{2} $$ hence $AD=3\sqrt{2}-\sqrt{6}$. By the bisector theorem again, $$ BC = BD\cdot\frac{AB}{AB+AD},\quad \frac{BC}{BA} = \frac{BD}{BA+AD} = \frac{2\sqrt{3}-3}{\sqrt{3}+3\sqrt{2}-\sqrt{6}}\tag{3} $$ so: $$ \tan(7^\circ 30') = \frac{2-\sqrt{3}}{1+\sqrt{6}-\sqrt{2}}\tag{4}$$ and the claim turns out to be equivalent to: $$ (1+\sqrt{6}-\sqrt{2})(\sqrt{6}-\sqrt{3}+\sqrt{2}-2)=(2-\sqrt{3})\tag{5} $$ that is tedious but straightforward to check.

Jack D'Aurizio
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Hint $$\tan{(15^{\circ})}=2-\sqrt{3}$$ and let$x=\tan{7.5^{\circ}}$.we have $$\dfrac{2x}{1-x^2}=2-\sqrt{3}\Longrightarrow (2-\sqrt{3})x^2+2x-(2-\sqrt{3})=0$$ so we have $$x^2+2(2+\sqrt{3})x-1=0$$ then we have $$(x+2+\sqrt{3})^2=1+(2+\sqrt{3})^2=8+4\sqrt{3}=2(\sqrt{3}+1)^2$$ then we have $$x=\sqrt{2}(\sqrt{3}+1)-2-\sqrt{3}=\sqrt{6}+\sqrt{2}-2-\sqrt{3}$$

math110
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Hint 1:

For $x\in [0,\frac{\pi}{2}]$: $$\tan \frac{x}{2} = \frac{\sin x}{1+\cos x}$$ $$\cos \frac{x}{2} = \sqrt{\frac{1+\cos x}{2}}$$ $$\sin \frac{x}{2} = \sqrt{\frac{1-\cos x}{2}}$$

Hint 2:

$$7^{\circ}30' = 15^{\circ}\cdot \frac{1}{2} =30^{\circ}\cdot \frac{1}{2}\cdot \frac{1}{2} $$

Jaroslaw Matlak
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