Skills Beta

Trigonometry and the Unit Circle

Radians measure angles by arc length on the unit circle, where the point at angle θ is (cos θ, sin θ) .

Practice 1: Implementing Mathematical Processes

Question set for this topic

Part 1 · Hook

Why this matters

Calculus measures angles in radians, and the six trig functions show up in every unit: as rates of change, as areas, as models for anything that repeats. Most trig mistakes on the exam are precalculus slips: a wrong sign in a quadrant, a degree-mode calculator, sin2⁡x\sin^2 x misread. This page rebuilds trig from one picture, the unit circle.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Which expression is 1x\frac{1}{x} written with an exponent?

  1. x−1x^{-1}
  2. x1x^{1}
  3. −x-x
  4. x1/2x^{1/2}
Show the answer

A negative exponent is a reciprocal.

  • Correct: x−1x^{-1}:
  • x1x^{1}:
  • −x-x:
  • x1/2x^{1/2}:

2. The graph of y=f(x−2)y = f(x - 2) is the graph of y=f(x)y = f(x) moved

  1. 2 units right
  2. 2 units left
  3. 2 units up
  4. 2 units down
Show the answer

Replacing xx by x−2x - 2 shifts right by 2.

  • Correct: 2 units right:
  • 2 units left:
  • 2 units up:
  • 2 units down:

Part 4 · See it

See it first

The unit circle with points at the angles 0, pi/6, pi/4, pi/3 and pi/2 and their coordinates (1, 0), (square root 3 over 2, 1/2), (square root 2 over 2, square root 2 over 2), (1/2, square root 3 over 2) and (0, 1); also (-1, 0) at pi and (0, -1) at 3 pi/2. Each quadrant is labeled with the signs of sine and cosine.
The unit circle: the point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). The special angles in the first quadrant and the signs in each quadrant give every exact value you need. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Measure an angle by the arc it cuts on a circle of radius 1radians: a half turn is π\pi, a full turn 2π2\pi
  2. Walk an angle θ\theta around the unit circle from (1,0)(1, 0)the point reached is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), and tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}
  3. The point lies on x2+y2=1x^2 + y^2 = 1sin2⁡θ+cos2⁡θ=1\sin^2\theta + \cos^2\theta = 1, the Pythagorean identity
  4. A reference angle in the first quadrant plus the quadrant signsgives the exact value at any special angle

Part 6 · Key ideas

Key ideas

  • Radians: π\pi radians =180∘= 180^{\circ}. To convert degrees to radians, multiply by π180\frac{\pi}{180}. Calculus always uses radians, including on the calculator.
  • On the unit circle, the point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta): cosine is the xx-coordinate, sine the yy-coordinate.
  • Know the first-quadrant values at π6,π4,π3\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}: 12,22,32\frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}. Other quadrants use a reference angle and a sign.
  • sec⁡=1cos⁡\sec = \frac{1}{\cos}, csc⁡=1sin⁡\csc = \frac{1}{\sin}, cot⁡=cos⁡sin⁡\cot = \frac{\cos}{\sin}. sin2⁡x\sin^2 x means (sin⁡x)2(\sin x)^2.
  • Pythagorean identity: sin2⁡x+cos2⁡x=1\sin^2 x + \cos^2 x = 1; also 1+tan2⁡x=sec2⁡x1 + \tan^2 x = \sec^2 x and sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x.

Part 7 · Misconception

A common mistake

The wrong idea: sin⁡(2x)=2sin⁡x\sin(2x) = 2\sin x, and sin2⁡x=sin⁡(x2)\sin^2 x = \sin(x^2).

What actually happens: Trig functions are not multiplied through: sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x\cos x. And sin2⁡x\sin^2 x means (sin⁡x)2(\sin x)^2. Test with x=π2x = \frac{\pi}{2}: sin⁡π=0\sin\pi = 0, but 2sin⁡π2=22\sin\frac{\pi}{2} = 2.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. Find cos⁡(2π3)\cos\left(\frac{2\pi}{3}\right) exactly.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Reference angle π−2π3=π3\pi - \frac{2\pi}{3} = \frac{\pi}{3}, with cos⁡π3=12\cos\frac{\pi}{3} = \frac{1}{2}. In the second quadrant cosine is negative: −12-\frac{1}{2}.

  • Answer: −12\frac{-1}{2} ≈ -0.500

2. Find sin⁡(7π6)\sin\left(\frac{7\pi}{6}\right) exactly.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

7π6=π+π6\frac{7\pi}{6} = \pi + \frac{\pi}{6}: reference angle π6\frac{\pi}{6}, third quadrant, so sin⁡7π6=−12\sin\frac{7\pi}{6} = -\frac{1}{2}.

  • Answer: −12\frac{-1}{2} ≈ -0.500

3. Find tan⁡(3π4)\tan\left(\frac{3\pi}{4}\right) exactly.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

tan⁡3π4\tan\frac{3\pi}{4}=sin⁡(3π/4)cos⁡(3π/4)= \frac{\sin(3\pi/4)}{\cos(3\pi/4)}=2/2−2/2= \frac{\sqrt{2}/2}{-\sqrt{2}/2}=−1= -1.

  • Answer: -1

4. Which values of xx in [0,2π)[0, 2\pi) solve 2sin⁡x=−32\sin x = -\sqrt{3}?

  1. 4π3\frac{4\pi}{3} and 5π3\frac{5\pi}{3}
  2. π3\frac{\pi}{3} and 2π3\frac{2\pi}{3}
  3. 7π6\frac{7\pi}{6} and 11π6\frac{11\pi}{6}
  4. 2π3\frac{2\pi}{3} and 4π3\frac{4\pi}{3}
Show the answer

sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}. The reference angle with sine 32\frac{\sqrt{3}}{2} is π3\frac{\pi}{3}; sine is negative in quadrants III and IV: π+π3=4π3\pi + \frac{\pi}{3} = \frac{4\pi}{3} and 2π−π3=5π32\pi - \frac{\pi}{3} = \frac{5\pi}{3}.

  • Correct: 4π3\frac{4\pi}{3} and 5π3\frac{5\pi}{3}: Quadrants III and IV with reference angle π3\frac{\pi}{3}.
  • π3\frac{\pi}{3} and 2π3\frac{2\pi}{3}: Those give sin⁡x=+32\sin x = +\frac{\sqrt{3}}{2}.
  • 7π6\frac{7\pi}{6} and 11π6\frac{11\pi}{6}: Those give sin⁡x=−12\sin x = -\frac{1}{2}; the reference angle should be π3\frac{\pi}{3}, not π6\frac{\pi}{6}.
  • 2π3\frac{2\pi}{3} and 4π3\frac{4\pi}{3}: Those give cos⁡x=−12\cos x = -\frac{1}{2}; sine at 2π3\frac{2\pi}{3} is positive.

5. Which expression equals 1−cos2⁡x1 - \cos^2 x for every xx?

  1. sin2⁡x\sin^2 x
  2. tan2⁡x\tan^2 x
  3. −sin2⁡x-\sin^2 x
  4. sin⁡x\sin x
Show the answer

From the Pythagorean identity sin2⁡x+cos2⁡x=1\sin^2 x + \cos^2 x = 1, subtract cos2⁡x\cos^2 x from both sides.

  • Correct: sin2⁡x\sin^2 x: sin2⁡x=1−cos2⁡x\sin^2 x = 1 - \cos^2 x.
  • tan2⁡x\tan^2 x: tan2⁡x=sec2⁡x−1\tan^2 x = \sec^2 x - 1, a different identity.
  • −sin2⁡x-\sin^2 x: The sign is positive: 1−cos2⁡x=sin2⁡x≥01 - \cos^2 x = \sin^2 x \ge 0.
  • sin⁡x\sin x: The identity involves squares: sin2⁡x\sin^2 x, not sin⁡x\sin x.

6. Which expression equals sin⁡(2x)\sin(2x) for every xx?

  1. 2sin⁡xcos⁡x2\sin x \cos x
  2. 2sin⁡x2\sin x
  3. sin2⁡x+cos2⁡x\sin^2 x + \cos^2 x
  4. sin⁡xcos⁡x\sin x \cos x
Show the answer

The double-angle identity: sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x. Check at x=π4x = \frac{\pi}{4}: sin⁡π2=1\sin\frac{\pi}{2} = 1 and 2⋅22⋅22=12 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 1.

  • Correct: 2sin⁡xcos⁡x2\sin x \cos x: The double-angle identity.
  • 2sin⁡x2\sin x: At x=π2x = \frac{\pi}{2}: sin⁡π=0\sin\pi = 0, but 2sin⁡π2=22\sin\frac{\pi}{2} = 2.
  • sin2⁡x+cos2⁡x\sin^2 x + \cos^2 x: That always equals 1, but sin⁡(2x)\sin(2x) varies.
  • sin⁡xcos⁡x\sin x \cos x: Missing the factor 2: at x=π4x = \frac{\pi}{4} it gives 12\frac{1}{2}, not 1.

7. Use a calculator to find cos⁡(1)\cos(1), where the angle is in radians. Give your answer to three decimal places.

Calculator allowed. Give the answer correct to three decimal places.

Show the answer

In radian mode, cos⁡(1)≈0.540\cos(1) \approx 0.540. Calculus formulas assume radians, so check the mode before every trig calculation.

  • Answer: 0.540

Part 9 · Summary

Summary

Radians measure angles by arc length on the unit circle, where the point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). Exact values come from the first-quadrant special angles plus the quadrant signs. Tangent, secant, cosecant and cotangent are ratios of sine and cosine, and sin2⁡x+cos2⁡x=1\sin^2 x + \cos^2 x = 1 links them. Keep calculators in radian mode.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections