Unit 3 · Topic 3.2

Sine, Cosine, and Tangent

An angle in radians is arc length over radius, so a full turn is 2π .

Practice 2: Multiple Representations

Part 1 · Hook

Why this matters

A bike wheel turns, and a reflector on its rim goes round and round. How high is the reflector after the wheel has turned through some angle? Put the wheel's center at the origin and the answer is a coordinate. That coordinate, read off a circle of radius 1, is the sine of the angle, and it turns sine from a triangle ratio into a function of any angle at all.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A circle has radius rr. How long is the arc cut off by a central angle of 90∘90^\circ?

  1. πr2\frac{\pi r}{2}
  2. 90r90r
  3. πr\pi r
  4. r4\frac{r}{4}
Show the answer

90∘90^\circ is a quarter turn: 90360⋅2πr=πr2\frac{90}{360} \cdot 2\pi r = \frac{\pi r}{2}.

  • Correct: πr2\frac{\pi r}{2}:
  • 90r90r:
  • πr\pi r:
  • r4\frac{r}{4}:

2. In a right triangle with hypotenuse 11, the side opposite the angle θ\theta has length 0.60.6. What is sin⁡θ\sin\theta?

  1. 0.60.6
  2. 0.80.8
  3. 10.6\frac{1}{0.6}
  4. 0.40.4
Show the answer

sin⁡θ\sin\theta=oppositehypotenuse= \frac{\text{opposite}}{\text{hypotenuse}}=0.61= \frac{0.6}{1}=0.6= 0.6. With hypotenuse 1, sine is just the opposite side.

  • Correct: 0.60.6:
  • 0.80.8:
  • 10.6\frac{1}{0.6}:
  • 0.40.4:

Part 4 · See it

See it first

A circle of radius 1 centered at the origin. An angle theta in standard position starts on the positive x-axis (the initial ray) and turns counterclockwise to the terminal ray, which meets the circle at P. The arc from (1, 0) to P, drawn in green, has length theta, because the radius is 1. A dashed vertical segment from P down to the x-axis has length sine of theta; the horizontal distance from the origin is cosine of theta. So P is (cos theta, sin theta). The slope of the terminal ray is sin theta over cos theta, which is tangent of theta.
On the unit circle the point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), the arc from (1,0)(1, 0) has length θ\theta, and tan⁡θ\tan\theta is the slope of the terminal ray. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Measure an angle by the arc it cuts off, divided by the radiusthat is its radian measure: a full turn is 2πrr=2π\frac{2\pi r}{r} = 2\pi radians, so π\pi radians =180∘= 180^\circ
  2. Put the angle in standard position: vertex at the origin, initial ray on the positive xx-axis, turning counterclockwiseits terminal ray meets the unit circle x2+y2=1x^2 + y^2 = 1 at one point
  3. Read that point's coordinatesits xx-coordinate is cos⁡θ\cos\theta and its yy-coordinate is sin⁡θ\sin\theta, for any angle, not only acute ones
  4. Divide the yy-coordinate by the xx-coordinatetan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}, the slope of the terminal ray, undefined where cos⁡θ=0\cos\theta = 0

Part 6 · Key ideas

Key ideas

  • Radians: θ=arc lengthradius\theta = \frac{\text{arc length}}{\text{radius}}. On the unit circle, an angle of θ\theta radians cuts off an arc of length exactly θ\theta.
  • Convert with π\pi radians =180∘= 180^\circ: multiply degrees by π180\frac{\pi}{180}, or radians by 180π\frac{180}{\pi}.
  • Signs follow the quadrant: in Quadrant II, cos⁡θ<0\cos\theta < 0 and sin⁡θ>0\sin\theta > 0; in Quadrant III both are negative; in Quadrant IV, cos⁡θ>0\cos\theta > 0 and sin⁡θ<0\sin\theta < 0.
  • Angles that differ by 2π2\pi share a terminal ray, so they have the same sine, cosine and tangent.

Part 7 · Misconception

A common mistake

The wrong idea: sin⁡θ\sin\theta is the xx-coordinate, because xx comes first.

What actually happens: Cosine comes first: the point is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). Sine is the height, the yy-coordinate. At θ=0\theta = 0 the point is (1,0)(1, 0), so cos⁡0=1\cos 0 = 1 and sin⁡0=0\sin 0 = 0.

Part 8 · Check yourself

Check yourself

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

Diagram

An angle θ\theta on the unit circle

The angle θ\theta, with 0<θ<2π0 < \theta < 2\pi, is in standard position. Its terminal ray meets the unit circle at P(−35,45)P\left(-\frac{3}{5}, \frac{4}{5}\right).
A circle of radius 1 centered at the origin, with the x- and y-axes. An angle theta in standard position turns counterclockwise from the positive x-axis to a terminal ray in the second quadrant, which meets the circle at the point P with coordinates (−3/5, 4/5). A dashed segment drops from P to the x-axis.
LevlPrep original diagram.

1. What is sin⁡θ\sin\theta?

  1. 45\frac{4}{5}
  2. −35-\frac{3}{5}
  3. −43-\frac{4}{3}
  4. 35\frac{3}{5}
Show the answer

sin⁡θ\sin\theta is the yy-coordinate of PP: 45\frac{4}{5}.

  • Correct: 45\frac{4}{5}: The height of PP is 45\frac{4}{5}.
  • −35-\frac{3}{5}: −35-\frac{3}{5} is the xx-coordinate, which is cos⁡θ\cos\theta.
  • −43-\frac{4}{3}: −43-\frac{4}{3} is yx\frac{y}{x}, which is tan⁡θ\tan\theta.
  • 35\frac{3}{5}: This drops the sign of the xx-coordinate and uses it for sine; sine is the yy-coordinate.

2. What is cos⁡θ\cos\theta?

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

Show the answer

cos⁡θ\cos\theta is the xx-coordinate of PP: −35-\frac{3}{5}.

  • Answer: −35\frac{-3}{5} ≈ -0.600

3. What is tan⁡θ\tan\theta?

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

Show the answer

tan⁡θ\tan\theta=sin⁡θcos⁡θ= \frac{\sin\theta}{\cos\theta}=4/5−3/5= \frac{4/5}{-3/5}=−43= -\frac{4}{3}, the slope of the ray from the origin to PP.

  • Answer: −43\frac{-4}{3} ≈ -1.333

4. Which is true about θ\theta?

  1. π2<θ<π\frac{\pi}{2} < \theta < \pi
  2. 0<θ<π20 < \theta < \frac{\pi}{2}
  3. π<θ<3π2\pi < \theta < \frac{3\pi}{2}
  4. 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi
Show the answer

PP has a negative xx-coordinate and a positive yy-coordinate, so it is in Quadrant II, between π2\frac{\pi}{2} and π\pi.

  • Correct: π2<θ<π\frac{\pi}{2} < \theta < \pi: Quadrant II runs from π2\frac{\pi}{2} to π\pi.
  • 0<θ<π20 < \theta < \frac{\pi}{2}: In Quadrant I both coordinates are positive, but PP has x<0x < 0.
  • π<θ<3π2\pi < \theta < \frac{3\pi}{2}: In Quadrant III both coordinates are negative, but PP has y>0y > 0.
  • 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi: In Quadrant IV, x>0x > 0 and y<0y < 0: the opposite of PP's signs.

5. What is sin⁡(θ+π)\sin(\theta + \pi)?

  1. −45-\frac{4}{5}
  2. 45\frac{4}{5}
  3. 35\frac{3}{5}
  4. −35-\frac{3}{5}
Show the answer

Turning another π\pi (half a turn) sends PP to the opposite point on the circle, (35,−45)\left(\frac{3}{5}, -\frac{4}{5}\right). Its yy-coordinate is −45-\frac{4}{5}.

  • Correct: −45-\frac{4}{5}: Half a turn flips both coordinates' signs, so the height becomes −45-\frac{4}{5}.
  • 45\frac{4}{5}: 45\frac{4}{5} would be sin⁡(θ+2π)\sin(\theta + 2\pi): a full turn, back to PP.
  • 35\frac{3}{5}: 35\frac{3}{5} is the new xx-coordinate, cos⁡(θ+π)\cos(\theta + \pi).
  • −35-\frac{3}{5}: This is cos⁡θ\cos\theta, the xx-coordinate of PP itself.

6. What is 150∘150^\circ in radians?

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

Show the answer

150⋅π180=5π6150 \cdot \frac{\pi}{180} = \frac{5\pi}{6}.

  • Answer: 5π6\frac{5\,\pi}{6} ≈ 2.618

7. Which angles have the same terminal ray as π3\frac{\pi}{3}? Select all that apply.

  1. 7π3\frac{7\pi}{3}
  2. −5π3-\frac{5\pi}{3}
  3. 4π3\frac{4\pi}{3}
  4. 2π3\frac{2\pi}{3}
Show the answer

Angles that differ by a multiple of 2π2\pi share a terminal ray: π3+2π=7π3\frac{\pi}{3} + 2\pi = \frac{7\pi}{3} and π3−2π=−5π3\frac{\pi}{3} - 2\pi = -\frac{5\pi}{3}.

  • Correct: 7π3\frac{7\pi}{3}: One extra full turn: π3+2π\frac{\pi}{3} + 2\pi.
  • Correct: −5π3-\frac{5\pi}{3}: One full turn clockwise: π3−2π\frac{\pi}{3} - 2\pi.
  • 4π3\frac{4\pi}{3}: This differs by π\pi, half a turn, so its ray points the opposite way.
  • 2π3\frac{2\pi}{3}: This is the mirror image across the yy-axis, in Quadrant II.

Part 9 · Summary

Summary

An angle in radians is arc length over radius, so a full turn is 2π2\pi. Place the angle θ\theta in standard position; its terminal ray meets the unit circle at (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), and tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} is the ray's slope. These definitions work for every angle, and the quadrant sets the signs.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic