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Reference Angle

Reference Angle – definition and example, Algebra 2 math glossary – Mathovore

Key takeaway: A reference angle is the sharp angle that an angle’s terminal side makes with the x-axis, and it fixes the size of the trig values of the original angle.

Grade 11 – Algebra 2F-TF.A.2Trigonometry

Definition of Reference Angle

Definition

A reference angle tells you how far the terminal side of an angle sits from the nearest part of the x-axis. It is always positive and always between 0° and 90°, or between 0 and \(\frac{\pi}{2}\) in radians.

Why bother? Angles that share one reference angle give sine, cosine and tangent values of equal size and possibly different signs. Learn the values for the acute angles once, find the partner for any other angle, then attach the sign from the quadrant. A big angle such as 500° is first reduced by full turns.

Formula and notation

For an angle \(\theta\) between \(0^{\circ}\) and \(360^{\circ}\), the partner \(\theta_r\) equals

\[\theta,\quad 180^{\circ}-\theta,\quad \theta-180^{\circ},\quad 360^{\circ}-\theta\]

in quadrants I, II, III and IV in that order. In radians use \(\pi\) and \(2\pi\) in place of 180° and 360°.

140°40°500° − 360° = 140°180° − 140° = 40°partner angle 40°
The terminal side of 140° sits 40° above the negative x-axis.

Key facts

  • The measurement is always taken from the x-axis, never from the y-axis.
  • Quadrant signs decide whether sine, cosine and tangent come out positive or negative.
  • Angles with the same partner have trig values equal in absolute value.
  • Reduce angles outside 0° to 360° by adding or subtracting full turns first.

Reference Angle example

Problem

Find the reference angle of \(500^{\circ}\) and use it to rewrite \(\sin 500^{\circ}\) as the sine of an acute angle.

  1. Remove one full turn: \(500^{\circ}-360^{\circ}=140^{\circ}\).

  2. 140° lies in quadrant II, so the partner is \(180^{\circ}-140^{\circ}=40^{\circ}\).

  3. Sine is positive in quadrant II, so \(\sin 500^{\circ}=\sin 40^{\circ}\approx 0.643\).

The reference angle is 40°, and \(\sin 500^{\circ}=\sin 40^{\circ}\).

Common mistake

Watch out

Wen says the reference angle of \(140^{\circ}\) is \(140^{\circ}-90^{\circ}=50^{\circ}\), measuring from the y-axis.

Why it is wrong

It is always found against the x-axis. The terminal side of 140° is \(180^{\circ}-140^{\circ}=40^{\circ}\) from the negative x-axis, so the answer is 40°.

Related terms and lessons

Browse the glossary: A–Z list · Algebra 2 vocabulary · Trigonometry

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