Lec11 Chp4 M2001 Summer26 Evaluating Trig Functions

Overview

Shahera Jannah teaches how to evaluate exact trigonometric functions using reference angles, quadrant signs, special triangles, and unit-circle coordinates. Examples include reducing 840° and 945°, evaluating functions at angles such as 225° and 11π/6, and finding all six functions from coordinate data or a right triangle.

Key takeaways

Reference angles reduce angles in any quadrant to a positive acute angle: use 180° − θ in Quadrant II, θ − 180° in Quadrant III, and 360° − θ in Quadrant IV.
For large angles, first find a coterminal angle by repeatedly subtracting 360° or 2π; for example, 945° reduces to 225°, whose reference angle is 45°.
The exact side ratios needed for common angles are 1:√3:2 for a 30-60-90 triangle and 1:1:√2 for a 45-45-90 triangle.
Quadrant signs follow the unit-circle coordinates: sine uses y, cosine uses x, and tangent uses y/x; reciprocal functions inherit the corresponding signs and are undefined when their denominator is zero.
For a point (x, y), calculate the radius with r = √(x² + y²), then use sin θ = y/r, cos θ = x/r, and tan θ = y/x; the example (−20/29, 21/29) has r = 1.
At quadrantal angles, use the unit-circle axis coordinates directly instead of inventing a reference triangle, which quickly identifies both exact values and undefined functions.

Chapters

0:00 Class Schedule and Review of a Quadrant II Triangle

The quiz is postponed from tomorrow to Monday, and a chapter exam is planned for the following week.
For the review, sin θ = 8/10 in Quadrant II gives y = 8 and hypotenuse r = 10.
The Pythagorean theorem gives |x| = √(10² − 8²) = 6, and Quadrant II makes x = −6.

1:38 Find All Six Trigonometric Functions from sin θ = 8/10

Using x = −6, y = 8, and r = 10, cos θ = −6/10 and tan θ = 8/(−6).
The reciprocal functions are csc θ = 10/8, sec θ = 10/(−6), and cot θ = (−6)/8.
The exercise reinforces reducing each fraction and using the quadrant to determine signs.

8:12 Reference Angle Rules for the Four Quadrants

A reference angle is positive, less than 90°, and formed between the terminal side and the x-axis.
In Quadrant I, the angle is its own reference angle; in Quadrant II, subtract it from 180°.
In Quadrant III, subtract 180° from the angle; in Quadrant IV, subtract the angle from 360°.
For negative degree angles, add 360° first to find a positive coterminal angle.

10:35 Degree Reference-Angle Examples: 20°, 135°, 240°, and 295°

20° is already a positive angle below 90°, so its reference angle is 20°.
For 135° in Quadrant II, 180° − 135° = 45°.
For 240° in Quadrant III, 240° − 180° = 60°.
For 295° in Quadrant IV, 360° − 295° = 65°.

16:06 Reference Angles in Radians and Coterminal-Angle Shortcuts

A radian angle can be converted to degrees, evaluated, and converted back to radians for the answer.
For negative radian angles, adding 2π produces a positive coterminal angle that is easier to place.
Examples use π/3, −5π/4, 11π/6, and −2π/3 to illustrate finding the nearby multiple of π or 2π.

22:08 Reduce Large Angles Before Finding Their Reference Angles

For 840°, subtract 360° twice to get the coterminal angle 120°.
Since 120° is in Quadrant II, its reference angle is 180° − 120° = 60°.
For 945°, subtract 360° twice to get 225°; its Quadrant III reference angle is 225° − 180° = 45°.
For large radian angles, use the same process by repeatedly adding or subtracting 2π.

26:30 Why Reference Angles Make Exact Trig Values Possible

The reference angle for 945° is 45°, so its triangle uses the 45-45-90 side ratio 1:1:√2.
Because 945° is coterminal with 225°, it lies in Quadrant III, where both x and y are negative.
Thus sin 945° = −1/√2, equivalent to −√2/2.
The same reference-angle method reduces unfamiliar angles to known special-triangle values.

36:27 Formal Radian Reference-Angle Method and Special-Angle Equivalences

The formal method first finds a positive coterminal angle, then compares it with π or 2π to determine the reference angle.
Useful degree-radian equivalences include 30° = π/6, 45° = π/4, and 60° = π/3.
The instructor recommends converting radians to degrees when that makes the angle's quadrant and position easier to see.

42:18 Evaluate sin 225° with a 45-45-90 Triangle

Subtract 180° from 225° to get a 45° reference angle in Quadrant III.
The 45-45-90 triangle has legs 1 and hypotenuse √2.
Sine is opposite over hypotenuse, and the Quadrant III sign makes sin 225° = −1/√2 = −√2/2.

47:11 Evaluate cos(π/6) by Converting to 30 Degrees

The angle π/6 equals 30°, and a 30-60-90 triangle has side lengths 1, √3, and 2.
For the illustrated fourth-quadrant angle, x is positive and y is negative.
Cosine is adjacent over hypotenuse, giving cos(π/6) = √3/2 when the angle is in Quadrant I.

55:01 Memorize the 30-60-90 and 45-45-90 Side Ratios

The 30-60-90 triangle uses side lengths 1, √3, and 2, with 1 opposite 30° and 2 opposite 90°.
The 45-45-90 triangle uses equal legs of 1 and hypotenuse √2.
These ratios supply exact values for common reference angles without a calculator.

1:01:29 Evaluate Functions at 11π/6 Using Its 30-Degree Reference Angle

Subtract 11π/6 from 2π to get the reference angle π/6, equivalent to 30°.
The angle 11π/6 lies in Quadrant IV, so x is positive and y is negative.
The 30-60-90 triangle gives side magnitudes 1, √3, and 2, allowing exact function values to be read from the sides.
An alternate method converts 11π/6 to 330° before drawing the triangle.

1:03:38 Independent Practice with Reference Angles and Exact Trig Values

Students work through a set of five additional trigonometric evaluations with neighboring classmates.
They choose between converting radians to degrees first or finding a radian reference angle directly.
The practice focuses on identifying the quadrant, selecting a special triangle when appropriate, and applying the correct sign.

1:16:38 Quadrantal Angles Use Unit-Circle Coordinates, Not Triangles

The key quadrantal angles are 0°, 90°, 180°, 270°, and 360°, corresponding to 0, π/2, π, 3π/2, and 2π.
Their unit-circle coordinates are the axis points (1, 0), (0, 1), (−1, 0), and (0, −1).
For quadrantal angles, use x and y directly rather than constructing a 30-60-90 or 45-45-90 triangle.

1:22:32 Use Reciprocal and Quotient Identities at Axis Angles

At 180°, the unit-circle point is (−1, 0), so sin 180° = 0 and cos 180° = −1.
Since csc θ = 1/sin θ, csc 180° is undefined because its denominator is zero.
Cotangent is cos θ/sin θ, so cot 180° is also undefined; tangent is sin θ/cos θ and equals 0.

1:24:24 Evaluate csc(3π/2) from the Unit Circle

The angle 3π/2 equals 270°, whose unit-circle coordinates are (0, −1).
Cosecant is 1/sin θ, and the y-coordinate gives sin(3π/2) = −1.
Therefore, csc(3π/2) = −1.

1:28:36 Evaluate sin(−135°) with a Quadrant III Reference Triangle

A clockwise rotation of 135° ends in Quadrant III, coterminal with 225°.
The reference angle is 45°, so the triangle has legs 1 and hypotenuse √2.
Both coordinates are negative in Quadrant III, giving sin(−135°) = −1/√2 = −√2/2.

1:32:30 Evaluate cos 360° and tan 180° from Axis Coordinates

At 360°, the unit-circle point is (1, 0), so cos 360° = 1.
At 180°, the point is (−1, 0), and tan θ = y/x gives tan 180° = 0.
Because cot θ = x/y, cot 180° is undefined when y = 0.

1:36:14 Evaluate sec(7π/6) with a 30-60-90 Triangle

Converting 7π/6 to degrees gives 210°, which is in Quadrant III.
The reference angle is 210° − 180° = 30°, so the adjacent-side magnitude is √3 and the hypotenuse is 2.
Cosine is negative in Quadrant III, so sec(7π/6) = −2/√3, equivalent to −2√3/3.

1:42:15 Find a Missing Radius from Coordinates (−20/29, 21/29)

The ordered pair gives x = −20/29 and y = 21/29, placing the point in Quadrant II.
Use r = √(x² + y²) to obtain r = √(400/841 + 441/841) = 1.
A unit-circle point has sine equal to y and cosine equal to x.

1:49:09 Compute All Six Functions from a Unit-Circle Coordinate

For (x, y) = (−20/29, 21/29), sin θ = 21/29 and cos θ = −20/29.
The quotient identity gives tan θ = y/x = −21/20.
Taking reciprocals gives csc θ = 29/21, sec θ = −29/20, and cot θ = −20/21.

1:55:45 Set Up an Exam-Style Triangle Problem

The next assessment-style exercise asks students to find all six trigonometric functions from a right triangle.
The first step is to identify the known sides and use the Pythagorean theorem to find the missing side.
The class is reminded to complete the homework for additional practice applying the day's methods.

1:59:20 Find All Six Functions for a Triangle with Hypotenuse 10

With hypotenuse 10 and one leg 3, the other leg is √(10² − 3²) = √91.
For the angle used in the exercise, sine is √91/10, cosine is 3/10, and tangent is √91/3.
The reciprocal values are csc = 10/√91, sec = 10/3, and cot = 3/√91.

Transcript excerpt

0:05 Good morning everyone. How are you guys today? >> Good. Um I was thinking about quiz tomorrow and then I decided well I'm not going to give you quiz tomorrow. I'm going to give you on Monday instead of tomorrow. Okay. So no quiz tomorrow. It will be Monday and uh next week test day you're going to have your exam too. Uh any other comments? So chapter three, we doing chapter

0:53 three, four, right? So chapter 3, we must do Monday midnight. Um, and I posted the notes just now. Half of chapter 4. I'm going to give you half of chapter 4 notes later, another day. Um, yeah, no quiz tomorrow. in study children Monday. >> Hello. >> Okay. Any question before we start? Any comments, concern, complaint? Okay. Then let's refresh our memory with one the last question we did in our last class. It says find the exact uh value of five

1:41 trig functions for the angle that satisfy the given two condition. The first condition is sign is8 and it lies on quadrant 2. So let's draw our angle on quadrant 2. Angle always start from the origin. It has to touch either positive or negative x-axis. So here we have an angle. So we have a triangle and in this triangle um we have sign is8 looks weird right? What can we do with this 8? Great we can rewrite this.

2:29 Sign of the angle is 8 / 10. Now sign is what about what in terms of triangle? Opposite over hypotenus. But this is opposite from this angle 8 and this is hypotenus here 10. Now we have two sides of our triangle. Can you find the missing side of the triangle? How can you find it? H by the Pythagorean theorem. So either

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Related notes

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Shahera Jannah's Math World! 😁😎.