How exactly do I find the exact value of $\tan(-15^{\circ})$?
Here, I tried to simplify it to something like this: $$\frac{-1+\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}$$
But, the value seems too large. Any help on simplifying this?
$\endgroup$ 33 Answers
$\begingroup$$$\tan (-15)=-\tan (\frac {30}{2}) $$
$$\tan (30)=\frac {2\tan (15)}{1-\tan^2 (15)} $$ $$=\frac {1}{\sqrt {3}} $$
let $x=\tan (15) >0$ and $a=\tan (30) $. then
$$2x=(1-x^2)a $$ or $$ax^2+2x-a=0$$ and $$x=\frac {-1+\sqrt {1+a^2}}{a} $$
$$=(-1+\frac {2}{\sqrt {3}} )\sqrt {3} $$ $$=2-\sqrt {3} $$ finally $$\tan (-15)=\sqrt {3}-2$$
$\endgroup$ 0 $\begingroup$You need to know a few things to do this. First of all, you need the fact that the tangent function is odd, or that $$\tan(-x)=-\tan(x)$$ You also need the half-angle formula: $$\tan(\frac{\theta}{2})=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$ Now you can say that $$\tan(-15)=-\tan(15)$$ $$\tan(-15)=-\tan(\frac{30}{2})$$ $$\tan(-15)=-\sqrt{\frac{1-\cos30}{1+\cos30}}$$ The cosine of $30$ is well-known to be $\frac{\sqrt{3}}{2}$, so now we can say that $$\tan(-15)=-\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}}$$ Now we simplify: $$\tan(-15)=-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}$$ And that's your exact value (though it may still be simplifiable).
$\endgroup$ 2 $\begingroup$You can multiply numerator and denominator by the conjugate of the denominator, here $1-\frac {\sqrt 3}3$. That makes the denominator $1^2-(\frac {\sqrt 3}3)^2=\frac 23$ which helps.
Added: $$\left(-1+\frac{\sqrt{3}}{3}\right) \left(1-\frac{\sqrt{3}}{3}\right) =-1 \cdot 1 +\frac {\sqrt 3}3\cdot 1 +(-1)(-\frac{\sqrt 3}3)-\frac {\sqrt 3}3\cdot \frac {\sqrt 3}3 \\=-1 +\frac{2\sqrt{3}}{3}-\left(\frac{\sqrt{3}}{3}\right)^2\\ =-1+\frac{2\sqrt 3}3-\frac 13\\ =\frac 23(\sqrt 3-2)$$
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