Evaluate arctan(0.003), giving the answer in radians and in degrees.

Step-by-step solution

1. 1

   State what is being asked. $\arctan(0.003)$ is the angle $\theta$ in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ with $\tan\theta=0.003$. Arctangent accepts every real input, so no domain check is needed — unlike arcsin or arccos.

2. 2

   Apply the small-angle rule to predict the answer. The Taylor series is

   $\arctan t=t-\frac{t^3}{3}+\frac{t^5}{5}-\cdots$

   For $t=0.003$ the second term is $\tfrac{(0.003)^3}{3}=9\times10^{-9}$ — utterly negligible. So $\arctan(0.003)\approx0.003$ radians, accurate to eight decimal places.

3. 3

   Give the radian value.

   $\arctan(0.003)=0.0029999910\ \text{radians}$

   Exactly as the series predicts: $0.003-0.000000009=0.002999991$ ✓. The answer is very slightly less than the input, because $\tan\theta>\theta$ for small positive $\theta$.

4. 4

   Convert to degrees. Multiply by $\dfrac{180}{\pi}=57.29578$:

   $0.002999991\times57.29578=0.171887^\circ\approx0.1719^\circ$

   That is about $10.3$ arcminutes — a very shallow angle, the kind that appears in gradients and alignment tolerances.

5. 5

   Note where this matters in practice. A tangent of $0.003$ is a slope of $3$ in $1000$, so the small-angle rule says: for shallow gradients, the angle in radians equals the slope. Multiplying by $57.3$ converts directly to degrees, which is why surveyors and engineers can treat gradient and angle as interchangeable below a few degrees.

6. 6

   Verify by going backwards. $\tan(0.002999991)=0.00300000$ ✓, recovering the input to eight decimal places, and $\tan(0.171887^\circ)=0.003000$ ✓.