Math Detours · Grade 10 · Sharpen · 11

SOHCAHTOA grows up: the unit circle

Exact values, angles past 90°, and why sin² + cos² = 1 is just Pythagoras.

Name ____________________Date ____________

Tonight's idea

In a right triangle, \(\sin\theta\) is opposite over hypotenuse, \(\cos\theta\) adjacent over hypotenuse, \(\tan\theta\) opposite over adjacent. Tonight: a definition that also works past 90°.

Two triangles give every exact value you need. Half an equilateral triangle of side 2 is a 30-60-90 triangle with sides \(1, \sqrt3, 2\). Half a unit square is a 45-45-90 triangle with sides \(1, 1, \sqrt2\).

√312 30°60° 11√2 45°45°
Read exact values off these: \(\sin 30° = \tfrac12\), \(\sin 45° = \tfrac{\sqrt2}{2}\).

The unit circle. Draw a circle of radius 1 centred at the origin. Start at \((1, 0)\) and turn anticlockwise through an angle \(\theta\). Wherever you land, define

\(\cos\theta =\) the \(x\)-coordinate of the point, \(\;\sin\theta =\) the \(y\)-coordinate, \(\;\tan\theta = \dfrac{\sin\theta}{\cos\theta}\).

For angles under 90° this agrees with SOHCAHTOA. Signs by quadrant: in I all three are positive, in II only sine, in III only tangent, in IV only cosine ("All Students Take Calculus"). The reference angle is the acute angle between the point's radius and the \(x\)-axis: it fixes the size of sine and cosine, and the quadrant fixes the signs.

(−√3⁄2, ½)(1, 0) 150°30° IIIIIIIV
Turn 150° from \((1, 0)\). The point is \((\cos 150°, \sin 150°) = (-\tfrac{\sqrt3}{2}, \tfrac12)\). The shaded 30° is the reference angle.

Angles are also measured in radians, the arc length turned on this circle, so \(90° = \tfrac{\pi}{2}\), \(180° = \pi\), \(360° = 2\pi\).

Pythagoras in disguise. The point \((\cos\theta, \sin\theta)\) lies on the circle \(x^2 + y^2 = 1\). That is the whole proof of

\(\sin^2\theta + \cos^2\theta = 1\) for every angle \(\theta\).

Drag the point round the circle at mathdetours.com/demos/unitcircle.html.

Watch one: \(\sin 150°\), \(\cos 150°\) and \(\tan 225°\)

150° is 30° short of 180°, so the reference angle is 30° and the point is in quadrant II. Apply the signs: \(\sin 150° = \tfrac12\), \(\cos 150° = -\tfrac{\sqrt3}{2}\).

225° is \(180° + 45°\), quadrant III, reference angle 45°. Both coordinates are negative: \((-\tfrac{\sqrt2}{2}, -\tfrac{\sqrt2}{2})\). So \(\tan 225° = 1\).

Your turn

  1. Without a calculator, give exact values: \(\sin 60°\), \(\cos 45°\), \(\tan 30°\), \(\sin 90°\), \(\cos 180°\). Write \(\tan 30°\) with no square root in the denominator.
  2. Use the unit circle, not a calculator. Find every angle \(\theta\) with \(0° \le \theta \lt 360°\) such that (a) \(\cos\theta = -\tfrac12\), (b) \(\sin\theta = -\tfrac{\sqrt2}{2}\), (c) \(\tan\theta = -1\). Each part has two answers: give the reference angle, then the quadrants, then the angles.
  3. \(\sin\theta = \tfrac35\) and \(90° \lt \theta \lt 180°\). Find \(\cos\theta\) and \(\tan\theta\) exactly, with the correct signs. Explain how you knew the signs.
  4. Prove \(\sin^2\theta + \cos^2\theta = 1\) in one sentence using the unit circle. Then divide every term by \(\cos^2\theta\) to get a second identity, in terms of \(\tan\theta\) and \(1/\cos\theta\).

Stretch: a twelve-sided π

Any triangle with sides \(a\) and \(b\) meeting at angle \(C\) has area \(\tfrac12 ab\sin C\). (Why? Drop a perpendicular from the end of side \(a\) onto side \(b\).) A regular 12-sided polygon is inscribed in a circle of radius 1. Cut it into 12 triangles from the centre and find its exact area. What does that tell you about \(\pi\)?

Answers

  1. \(\sin 60° = \tfrac{\sqrt3}{2}\), \(\cos 45° = \tfrac{\sqrt2}{2}\), \(\tan 30° = \tfrac{1}{\sqrt3} = \tfrac{\sqrt3}{3}\), \(\sin 90° = 1\), \(\cos 180° = -1\). The last two come from the points \((0,1)\) and \((-1,0)\) on the unit circle.
  2. (a) Reference angle 60° (since \(\cos 60° = \tfrac12\)), cosine negative in quadrants II and III: \(120°\) and \(240°\). (b) Reference 45°, sine negative in III and IV: \(225°\) and \(315°\). (c) Reference 45°, tangent negative in II and IV: \(135°\) and \(315°\).
  3. From \(\sin^2\theta + \cos^2\theta = 1\): \(\cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}\), so \(\cos\theta = \pm\tfrac45\). Quadrant II has negative \(x\), so \(\cos\theta = -\tfrac45\) and \(\tan\theta = \tfrac{3/5}{-4/5} = -\tfrac34\). (The 3-4-5 triangle, reflected into the second quadrant.)
  4. The point \((\cos\theta, \sin\theta)\) is on the circle \(x^2 + y^2 = 1\), so \(\cos^2\theta + \sin^2\theta = 1\). Dividing by \(\cos^2\theta\): \(\tan^2\theta + 1 = \dfrac{1}{\cos^2\theta}\) (valid whenever \(\cos\theta \ne 0\)). The quantity \(1/\cos\theta\) is called \(\sec\theta\), so this is usually written \(1 + \tan^2\theta = \sec^2\theta\).

Stretch. The perpendicular has length \(a\sin C\) and is the height on base \(b\), so area \(= \tfrac12 ab\sin C\). Each of the 12 triangles has two sides of length 1 meeting at \(360°/12 = 30°\), so area \(\tfrac12 \sin 30° = \tfrac14\). Total: \(12 \times \tfrac14 = 3\), exactly. The circle has area \(\pi\) and the polygon sits inside it, so \(\pi \gt 3\). Archimedes pushed this to 96 sides and got \(\pi\) between \(3\tfrac{10}{71}\) and \(3\tfrac17\).

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