True or false: tan(pi) = 0. Explain your answer.

Step-by-step solution

1. 1

   Translate the angle. Since $\pi$ radians $=180^{\circ}$, the question asks for the tangent of a half-turn — the point $(-1,0)$ on the unit circle, on the negative $x$-axis.

2. 2

   Use the quotient identity. By definition,

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

   so the value of $\tan\pi$ is decided entirely by the coordinates of that point: $\sin\pi$ is the $y$-coordinate and $\cos\pi$ is the $x$-coordinate.

3. 3

   Read off the two coordinates. At $\theta=\pi$,

   $\sin(\pi)=0,\qquad \cos(\pi)=-1.$

   The sine is zero because the point lies exactly on the horizontal axis; the cosine is $-1$ because the point is one unit to the left of the origin.

4. 4

   Divide, and check the denominator is not zero. Substituting,

   $\tan(\pi)=\frac{0}{-1}=0.$

   The statement is true. The negative denominator does not make the answer negative — zero divided by any nonzero number is zero, with no sign attached.

5. 5

   Contrast with the case where tangent fails. Tangent is undefined precisely where $\cos\theta=0$, that is at $\theta=\tfrac{\pi}{2}+k\pi$; there the quotient would be a nonzero number over zero. Its zeros are where $\sin\theta=0$, namely $\theta=k\pi$ — so $\tan0=\tan\pi=\tan2\pi=0$, and $\tan$ has period $\pi$ rather than $2\pi$.