# Find exact value of $\tan (\frac{\pi}{12})$ given that $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$

Find exact value of $$\tan (\frac{\pi}{12})$$ given that $$\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$$

I am asking myself how do I manipulate $$\tan (\frac{\pi}{2})$$ such that I can form an equation with $$\sin(\frac{\pi}{12})$$ in it so that I can substitute the value to find the exact value but I couldn't find any co function identities to transform $$\tan (\frac{\pi}{12})$$ into an equation.

The closest one could be $$\cot (\frac{\pi}{2} - \theta) = \tan \theta$$

But $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$, I do not have the exact value of $$\cos \theta$$ to use this identity.

So how do actually manipulate $$\tan (\frac{\pi}{12})$$ into an equation ?

• You know that $\tan(x)=\sin(x)/\cos(x)$ and $\cos(x)^2+\sin^2(x)=1$. From the latter you can solve for $\cos(x)$ in terms of $\sin(x)$, and plug it into the formula for $\tan(x)$. Sep 26 at 16:27
• Actually, $\sin\left(\frac\pi{12}\right)=\frac{\sqrt3-1}{2\sqrt2}$, rather than $\frac{\sqrt3-1}{2\sqrt{12}}$. Sep 26 at 16:33

As per the suggestion of Santos: $$\sin(\pi/12)=\frac{\sqrt{3}-1}{2\sqrt{2}}\implies \tan(\pi/12)=\frac{\sqrt{3}-1}{\sqrt{(2\sqrt{2})^2-(\sqrt{3}-1)^2}}=\frac{\sqrt{3}-1}{\sqrt{4+2\sqrt{3}}}$$ Next let $$\sqrt{4+2\sqrt{3}}=\sqrt{x}+\sqrt{y} \implies x+y=4,xy=3\implies x=3,y=1.$$ Then $$\tan(\pi/12)=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{3-1}=2-\sqrt{3}.$$
$$\frac{1}{2}=\sin \frac{\pi}{6}=\sin (2 \times \frac{\pi}{12})=2\times\sin \frac{\pi}{12} \times \cos \frac{\pi}{12}=2\times \frac{\sqrt3 -1}{2\sqrt{2}}\times \cos \frac{\pi}{12}$$ $$\cos \frac{\pi}{12}=\frac{\sqrt2}{2\times (\sqrt3 -1)}$$ $$\tan\frac{\pi}{12}=\frac{\frac{\sqrt3 -1}{2\sqrt{2}}}{\frac{\sqrt2}{2\times (\sqrt3 -1)}}=2-\sqrt3$$
• I have a different $\cos \frac{\pi}{12}$ value as you using $\cos^2 \theta = 1 - \sin^2 \theta$, $\sqrt{\cos^2 \frac{\pi}{12}} = \sqrt {1- (\frac{\sqrt{3}-1}{2\sqrt{2}})^2} = \sqrt{1- \frac{2-\sqrt{3}}{4}} =\frac{ \sqrt{2+\sqrt{3}}}{2}$ Sep 26 at 16:57
• @user307640 The two values are equivalent. $\frac{\sqrt{2}}{2(\sqrt{3} - 1)} = \frac{\sqrt{2}(\sqrt{3} + 1)}{2(3 - 1)} = \frac{\sqrt{6} + \sqrt{2}}{4}$ and $\frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{\sqrt{8 + 4\sqrt{3}}}{4} = \frac{\sqrt{6 + 4\sqrt{3} + 2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$. Sep 26 at 17:26