Sine, Cosine, and Reference Angles

• The goal is to solve for the sine and cosine of 45°, 30°, and 60°.
• Also, you will learn what reference angles are and fill out more of the unit circle sheet.
• Challenge Problem 1 has been graded.
• People proving the Triangle Postulate did well, but proving the Pythagorean Theorem was trickier.
• When proving a statement, the statement cannot be used in the proof.
• The Pythagorean Theorem ONLY applies to right triangles, unlike the triangle inequality, which applies to all triangles.
• If last week's Wednesday class was missed, the Challenge Problem can be made up in office hours.
• The WeBWorks from Week 1 is due this week.
• Extensions can be requested if needed.
• If stuck on a problem, the "Email Instructor" button can be used.
• The Canvas Quiz is due tonight.

Exact Values of Trigonometric Functions

• If you're fine with a decimal approximation, plug the value into your calculator.
• sin(28.6°) = 0.47869...
• An exact value can be found for some angles.
• Exact values have been computed already for sin(0°), cos(0°), sin(90°), cos(90°), etc.

Sine and Cosine of 45°

• Find the side lengths of a right triangle with a hypotenuse of 1, where one of the angles is 45°.
• Since 90° + 45° + 45° = 180°, the non-right angles are the same: 45°.
• The two legs opposite the non-right angles have the same length, l.
• By the Pythagorean theorem, l² + l² = 1. Solving for l, is √½ = √2 / 2.
• Thus, cos(45°) = √2 / 2 and sin(45°) = √2 / 2.

Sine and Cosine of 30°

• Find the side lengths of a right triangle with a hypotenuse of 1, where one of the angles is 30°.
• Flip the triangle around its longer (adjacent to 30°) side ℓ to make a bigger triangle.
• The triangle’s angles are 180° − 90° − 30° = 60°, 60°, and 2(30°)60°.
• The triangle’s side lengths are the same.
• Because 2h = 1 means that h = ½, by the Pythagorean theorem, ℓ = √3 / 2.
• Thus, cos(30°) = √3 / 2 and sin(30°) = ½.

Sine and Cosine of 60°

• Find the side lengths of a right triangle of hypotenuse 1 where one of the angles is 60°.
• The shorter side is adjacent to 60° that is ½.
• The longer side is opposite 60° that is √3 / 2.
• Therefore, cos(60°) = ½ and sin(60°) = √3 / 2.
• Other angles, such as 15° and 18°, can be computed to find the exact values of sin and cos, but for now, worry about 0°, 30°, 45°, 60°, and 90°.
• sin(15°) = (√6 - √2) / 4 and cos(15°) = (√6 + √2) / 4.
• sin(18°) = (√5 - 1) / 4 and cos(18°) = √(10 + 2√5) / 4.

Reference Angles

• Assume that θ is an angle in the unit circle.
• The reference triangle of that angle is the right triangle with vertices (0,0), (cos(θ), 0), and (cos(θ), sin(θ)).
• In other words, the reference triangle of θ is the right triangle "aligned with" the x-axis that contains both the origin and θ's reference point.
• The reference angle of that angle is the interior angle of the reference triangle that is formed at the vertex (0,0).
• α is equal to "how far away" θ is from either 0° = 0 or 180° = π.
• α will always be an angle in the first quadrant.
• |cos(θ)| = cos(α) and |sin(θ)| = sin(α).
• To get the signs of cos(θ) and sin(θ), check θ's quadrant.
• The reference angle is 30° because 180° + 30° = 210°.
• |cos(210°)| = √3 / 2 and |sin(210°)| = ½.

Sine and Cosine

• The sin and cos of angles between 0° and 90° are intimately linked with right triangles of hypotenuse 1.
• The sin and cos of an angle anywhere else on the unit circle can be expressed in terms of the sin or cos between 0° and 90° (namely, its reference angle).
• Triangles along the y-axis are useful to draw – i.e., with coordinates (0,0), (0, sin(θ)), and (cos(θ), sin(θ)).
• if ß is the interior angle of that triangle along the vertex then |cos(θ)| = sin(ß) and |sin(θ)| = cos(ß).
• sin(θ) = cos(90° – θ) and cos(θ) = sin(90° – θ).

Angles and Reference Points

• Rotating 360° = 2π is the same as not having rotated at all concerning our reference points.
• sin(θ) = sin(θ + 2π) and cos(θ) = cos(θ + 2π).
• The sin and cos functions are periodic, with a period of 360° = 2π.
• If you add or subtract 360° = 2π from an angle, the angle’s sine and cosine stay the same.
• There exists some angle θ₀ ∈ [0, 2π) and some integer n ∈ {..., -2, -1, 0, 1, 2, ... } such that θ = θ₀ + 2πn, meaning that:
  • sin(θ) = sin(θ₀ + 2πn) = sin(θ₀)
  • cos(θ) = cos(θ₀ + 2πn) = cos(θ₀)
• Formulas also work if you replace 2π with 360°.
• sin(765°) = sin(2 * 360° + 45°) = sin(45°) = √2 / 2.
• sin(-π / 4) = sin(-1 * 2π + 7π / 4) = sin(7π / 4) = -√2 / 2.