Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

This page supports the lesson note with a companion video and a short classroom-ready summary.

For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.

Performance objectives

• Define sine, cosine, and tangent for angles between 0° and 360°.
• Determine the sign (+ or -) of trig functions in each quadrant.
• Calculate trig ratios for any angle using reference angles.
• Solve problems involving trig ratios for angles beyond 90°.
• Use a calculator to find trig ratios and angles.

Lesson summary

This week, we delve into the exciting world of trigonometric functions, specifically focusing on defining and understanding sine, cosine, and tangent for angles beyond acute angles (0° to 90°). We will extend our knowledge from right-angled triangles to the Cartesian plane, allowing us to work with angles of any magnitude. This is a crucial step in understanding periodic phenomena, which are prevalent in many aspects of our lives, from sound waves and light to electrical circuits and even tidal patterns along the South African coastline. Understanding these functions enables us to model and predict these real-world phenomena.

Lesson notes

2.1 Extending Trigonometric Ratios to the Cartesian Plane Previously, we defined sine, cosine, and tangent using the sides of a right-angled triangle (SOH CAH TOA). Now, we extend this to the Cartesian plane. Consider a point P(x, y) on the terminal arm of an angle θ in standard position (vertex at the origin, initial arm along the positive x-axis). Let 'r' be the distance from the origin (0, 0) to point P. We can think of 'r' as the hypotenuse of a right-angled triangle. Importantly, 'r' is always positive.

Then we define: Sine (sin θ) = y / r Cosine (cos θ) = x / r Tangent (tan θ) = y / x (where x ≠ 0) Notice that if θ is an acute angle (between 0° and 90°), these definitions are consistent with our previous SOH CAH TOA definitions. (y is opposite, x is adjacent, and r is hypotenuse).

However, these new definitions allow us to define trigonometric ratios for any angle. 2.2 Quadrants and Signs of Trigonometric Ratios The Cartesian plane is divided into four quadrants. The sign of x and y coordinates varies in each quadrant, and therefore the sign of sin θ, cos θ, and tan θ also varies. Remember that 'r' is always positive. *Quadrant I (0° 0, y >

0. Therefore, sin θ > 0, cos θ > 0, tan θ > 0 (All positive) *Quadrant II (90°

0. Therefore, sin θ > 0, cos θ 0 (Only Tangent positive) *Quadrant IV (270° 0, y 0, tan θ R ) is the acute angle formed between the terminal arm of the angle θ and the x-axis. Using reference angles simplifies the process of finding trigonometric ratios for angles beyond the first quadrant. Here's how to find the reference angle for different quadrants: Quadrant I: θ R = θ Quadrant II: θ R = 180° - θ Quadrant III: θ R = θ - 180° Quadrant IV: θ R = 360° - θ To find the trigonometric ratios of any angle θ: Determine the quadrant in which θ lies. Find the reference angle θ R . Determine the sign of the trigonometric ratio based on the quadrant (using ASTC). Evaluate the trigonometric ratio of the reference angle θ R . The sign is already determined in step 3. 2.4 Worked Examples Example 1: Find sin 150° 150° lies in Quadrant I

I. Reference angle: θ R = 180° - 150° = 30° In Quadrant II, sine is positive. sin 150° = sin 30° = 1/2 Example 2: Find cos 240° 240° lies in Quadrant II

I. Reference angle: θ R = 240° - 180° = 60° In Quadrant III, cosine is negative. cos 240° = - cos 60° = -1/2 Example 3: Find tan 315° 315° lies in Quadrant I

V. Reference angle: θ R = 360° - 315° = 45° In Quadrant IV, tangent is negative. tan 315° = - tan 45° = -1 Example 4: If cos θ = -0.6 and 180° 2 = x 2 + y 2 , so 5 2 = (-3) 2 + y 2 .

Therefore, y 2 = 25 - 9 = 16, and y = ±

4. Since θ is in Quadrant III, y is negative, so y = -4. sin θ = y/r = -4/5 = -0.8 tan θ = y/x = -4/-3 = 4/3 = 1.33 (approx) Guided Practice (With Solutions)

Question 1: Determine the quadrant in which the angle 210° lies and state the sign of sin 210°.

Solution: 210° lies between 180° and 270°, so it is in Quadrant II

I. In Quadrant III, only tangent is positive.

Therefore, sin 210° is negative.

Question 2: Find the reference angle for 135°.

Solution: 135° lies in Quadrant I

I. The reference angle is θ R = 180° - 135° = 45°.

Question 3: Find the value of cos 300° without using a calculator.

Solution: 300° lies in Quadrant IV. The reference angle is θ R = 360° - 300° = 60°. In Quadrant IV, cosine is positive.

Therefore, cos 300° = cos 60° = 1/

2. Question 4: Given that sin θ = 0.8 and θ is an acute angle, find cos θ and tan θ.

Solution: Since θ is an acute angle, we can use right-angled triangle definitions. sin θ = y/r = 0.8 = 8/10 = 4/

5. So y = 4 and r =

5. Using Pythagoras, x 2 + y 2 = r 2 => x 2 + 4 2 = 5 2 => x 2 = 9 => x = 3 (since x is positive for acute angles) cos θ = x/r = 3/5 = 0.6 tan θ = y/x = 4/3 = 1.33 (approx.) Independent Practice (Questions Only) Determine the quadrant in which each of the following angles lies: 75°, 190°, 340°, 100°, 280°. Find the reference angle for each of the following angles: 120°, 225°, 330°, 95°, 200°. Without using a calculator, find the values of: sin 225°, cos 135°, tan 330°. If cos θ = -√3/2 and 90° < θ < 180°, find sin θ and tan θ. If tan θ = -1 and 270° < θ < 360°, find sin θ and cos θ.

Calculate: 2sin 150° + 3cos 225° - tan 315°. If sin θ = -0.6 and 180° < θ < 270°, calculate the value of cos θ. What are the possible values of θ if cos θ = 0.5 and 0° ≤ θ ≤ 360°? What are the possible values of θ if sin θ = -1 and 0° ≤ θ ≤ 360°? Determine the values of x for 0° ≤ x ≤ 360° that satisfies 2cosx + 1 = 0