Show that tan(30 degrees) = root 3 divided by 3, and explain why the answer is not root 3.

Step-by-step solution

1. 1

   Start from the definition of tangent. Tangent is not an independent value to memorise; it is a ratio:

   $\tan\theta=\frac{\sin\theta}{\cos\theta}$

   So the exact value of $\tan30^\circ$ follows from the exact sine and cosine.

2. 2

   Recall the 30-60-90 triangle values. In a right triangle with angles $30^\circ,60^\circ,90^\circ$ the sides are in ratio $1:\sqrt3:2$, which gives

   $\sin30^\circ=\frac{1}{2},\qquad \cos30^\circ=\frac{\sqrt3}{2}$

3. 3

   Form the quotient and cancel. Dividing fractions means multiplying by the reciprocal:

   $\tan30^\circ=\frac{\tfrac12}{\tfrac{\sqrt3}{2}}=\frac{1}{2}\cdot\frac{2}{\sqrt3}=\frac{1}{\sqrt3}$

   The two halves cancel, leaving $\tfrac{1}{\sqrt3}$.

4. 4

   Rationalise the denominator. $\tfrac{1}{\sqrt3}$ is correct but conventionally rewritten without a radical below the line:

   $\frac{1}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=\frac{\sqrt3}{3}$

   So $\tan30^\circ=\tfrac{\sqrt3}{3}$, and $\tfrac{1}{\sqrt3}$ and $\tfrac{\sqrt3}{3}$ are the same number, not two different answers.

5. 5

   Explain the common confusion with root 3. $\sqrt3$ is $\tan60^\circ$, because at $60^\circ$ the roles of the legs swap: $\tan60^\circ=\frac{\sin60^\circ}{\cos60^\circ}=\frac{\sqrt3/2}{1/2}=\sqrt3$. The two are reciprocals: $\tan30^\circ\cdot\tan60^\circ=1$.

6. 6

   Check the size, which settles it instantly. $\tfrac{\sqrt3}{3}\approx0.5774$, which is less than $\tan45^\circ=1$ — correct, since $30^\circ<45^\circ$. But $\sqrt3\approx1.732$ is greater than $1$, so it could never be the tangent of an angle below $45^\circ$ ✓.