Find the value of sin(20 degrees) to four decimal places.

Step-by-step solution

1. 1

   Check whether the angle is special. The exact-value angles in a first course are $0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$ and their reflections. $20^\circ$ is none of these, so there is no clean radical expression to quote — a calculator is genuinely needed rather than a shortcut being missed.

2. 2

   Bracket the answer before computing. Sine increases on $[0^\circ,90^\circ]$, and $\sin0^\circ=0$ while $\sin30^\circ=0.5$. Since $0^\circ<20^\circ<30^\circ$, the value must lie strictly between $0$ and $0.5$, and closer to $0.5$ than to $0$ because $20$ is two-thirds of the way to $30$.

3. 3

   Sharpen the estimate with the small-angle approximation. In radians $20^\circ=\tfrac{\pi}{9}=0.349066$, and for small $\theta$, $\sin\theta\approx\theta-\tfrac{\theta^3}{6}=0.349066-0.007089=0.341977$. So expect about $0.342$.

4. 4

   Evaluate to four decimal places. With the calculator in degree mode:

   $\sin20^\circ=0.3420201\ldots\approx0.3420$

   This matches the series estimate to four decimals ✓ and sits inside the predicted $(0,0.5)$ window ✓.

5. 5

   Guard against the classic mode error. In radian mode the same keystrokes return $\sin(20\ \text{rad})=0.9129$ — a completely different number. The bracket from step 2 catches this instantly, which is why estimating first is worth the few seconds.

6. 6

   Note why no exact radical form exists. $20^\circ$ is one third of $60^\circ$, and trisecting a constructible angle is not generally possible with square roots: $\sin20^\circ$ is a root of $8t^3-6t+\sqrt3=0$ (from $\sin3\theta=3\sin\theta-4\sin^3\theta$ with $\sin60^\circ=\tfrac{\sqrt3}{2}$, multiplied through by $-2$), while the companion identity $\cos3\theta=4\cos^3\theta-3\cos\theta$ with $\cos60^\circ=\tfrac12$ gives $\cos20^\circ$ the cubic $8t^3-6t-1=0$ — and that one has rational coefficients and is irreducible over the rationals, which is precisely what rules out any square-root expression. Verifying both: $\cos20^\circ=0.9396926$ gives $8(0.9396926)^3-6(0.9396926)-1=0.000000$ ✓, and $\sin20^\circ=0.3420201$ gives $8(0.3420201)^3-6(0.3420201)+\sqrt3=0.000000$ ✓.