Evaluate tan 315 degrees exactly.

Step-by-step solution

1. 1

   Locate the angle and fix the sign. $315^\circ$ lies between $270^\circ$ and $360^\circ$, so it is in the fourth quadrant, where $x>0$ and $y<0$. Since $\tan=\dfrac{\sin}{\cos}=\dfrac{y}{x}$ is a negative over a positive, $\tan315^\circ$ must be negative.

2. 2

   Find the reference angle. The nearest horizontal axis is $360^\circ$:

   $360^\circ-315^\circ=45^\circ$

   so the reference angle is $45^\circ$ and the magnitude matches that of $\tan45^\circ$.

3. 3

   Use the fourth-quadrant identity. For an angle written as $360^\circ-\theta$,

   $\tan\!\left(360^\circ-\theta\right)=-\tan\theta$

   (equivalently $\tan(-\theta)=-\tan\theta$, since tangent is an odd function with period $180^\circ$). With $\theta=45^\circ$:

   $\tan315^\circ=-\tan45^\circ$

4. 4

   Substitute the special-angle value. The $45$–$45$–$90$ triangle has equal legs, so $\tan45^\circ=1$ and

   $\tan315^\circ=-1$

5. 5

   Verify from the coordinates. The unit-circle point at $315^\circ$ is $\left(\tfrac{\sqrt2}{2},-\tfrac{\sqrt2}{2}\right)$, so

   $\tan315^\circ=\frac{-\sqrt2/2}{\sqrt2/2}=-1\ \checkmark$

   Numerically $-1$ to within $10^{-12}$ ✓. Note $315^\circ=\tfrac{7\pi}{4}$ radians, and because tangent has period $180^\circ$, $\tan315^\circ=\tan135^\circ=-1$ as well.