What is the value of \(\tan(2820°)\)?
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
What is the value of \(\tan(2820°)\)?
\(-\sqrt{3}\)
\(-\frac{\sqrt{3}}{3}\)
\(\frac{\sqrt{3}}{3}\)
\(\sqrt{3}\)
1. INFER the key strategy
- Large angles like \(2820°\) are unwieldy, but trigonometric functions are periodic
- Key insight: Tangent repeats every \(180°\) (unlike sine and cosine which repeat every \(360°\))
- Strategy: Reduce \(2820°\) to an equivalent angle between \(0°\) and \(180°\)
2. SIMPLIFY using modular arithmetic
- Divide the given angle by the period: \(2820 ÷ 180 = 15.666...\)
- The integer part tells us how many complete periods: 15 complete periods
- Calculate: \(15 × 180 = 2700°\)
- Find remainder: \(2820 - 2700 = 120°\)
- Therefore: \(\mathrm{tan}(2820°) = \mathrm{tan}(120°)\)
3. INFER the quadrant and sign
- \(120°\) is between \(90°\) and \(180°\), so it's in quadrant II
- In quadrant II, tangent values are negative
- Reference angle: \(180° - 120° = 60°\)
4. INFER the final value
- \(\mathrm{tan}(120°) = -\mathrm{tan}(60°)\)
- From unit circle: \(\mathrm{tan}(60°) = \sqrt{3}\)
- Therefore: \(\mathrm{tan}(120°) = -\sqrt{3}\)
Answer: A
Why Students Usually Falter on This Problem
Most Common Error Path:
Conceptual confusion about periods: Many students incorrectly use \(360°\) as the period for all trig functions because they're more familiar with sine and cosine.
If they use \(360°\): \(2820 ÷ 360 = 7.833...\), so \(7 × 360 = 2520\), remainder \(= 300°\)
Then \(\mathrm{tan}(300°) = \mathrm{tan}(-60°) = -\frac{\sqrt{3}}{3}\), leading them to select Choice B \(-\frac{\sqrt{3}}{3}\)
Second Most Common Error:
Weak INFER skill regarding signs in quadrants: Students correctly find that \(\mathrm{tan}(2820°) = \mathrm{tan}(120°)\) but forget that tangent is negative in quadrant II.
They calculate \(\mathrm{tan}(120°) = \mathrm{tan}(60°) = \sqrt{3}\) (forgetting the negative sign), leading them to select Choice D \(\sqrt{3}\)
The Bottom Line:
This problem tests whether students truly understand the period of tangent \(180°\) vs \(360°\)) and can correctly apply quadrant sign rules. The large angle is just a distraction - the real challenge is applying fundamental periodic properties accurately.
\(-\sqrt{3}\)
\(-\frac{\sqrt{3}}{3}\)
\(\frac{\sqrt{3}}{3}\)
\(\sqrt{3}\)