Lesson Notes By Weeks and Term v5 - Grade 10
Trigonometric functions – Week 6 focus
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Trigonometric functions – Week 6 focus
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Subject: Mathematics
Class: Grade 10
Term: 2nd Term
Week: 6
Theme: General lesson support
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Trigonometry is a fundamental branch of mathematics that explores the relationships between angles and sides of triangles. In Grade 10, we begin a more formal study of trigonometric functions beyond right-angled triangles. This week, we will delve into the sine, cosine, and tangent functions for angles beyond 0° to 90°. This is crucial because angles aren't limited to just acute angles in real-world scenarios. Imagine surveyors determining land boundaries, or civil engineers designing bridges; they need trigonometry to work with various angles and spatial relationships.
The Cartesian Plane and Trigonometric Functions: We extend the definition of trigonometric functions beyond the confines of a right-angled triangle by placing the triangle within the Cartesian plane. Consider a point P(x; y) on a circle with radius r centred at the origin. Let θ be the angle formed between the positive x-axis and the line O
P. We define the trigonometric functions as follows: Sine (sin θ): sin θ = y/r Cosine (cos θ): cos θ = x/r Tangent (tan θ): tan θ = y/x (where x ≠ 0) Remember that r is always positive (since it represents a radius), but x and y can be positive or negative depending on the quadrant in which the point P lies. Signs of Trigonometric Functions in Each Quadrant: The signs of x and y in each quadrant determine the signs of sin θ, cos θ, and tan θ. *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 (Tangent positive) *Quadrant IV (270° 0, y 0, tan θ < 0 (Cosine positive) A helpful mnemonic to remember this is "All Students Take Coffee" (ASTC) or "All Stations To Cape Town": All (Quadrant I): All functions are positive.
Students (Quadrant II): Sine is positive.
Take (Quadrant III): Tangent is positive.
Coffee (Quadrant IV): Cosine is positive.
Special Angles: It's crucial to know the trigonometric ratios for special angles: 0°, 30°, 45°, 60°, 90°, 180°, 270°, and 360°. These values are often used in problem-solving and can be derived from geometric considerations of equilateral triangles and squares. Here's a table summarizing the values (important to memorize or be able to derive quickly): | Angle (θ) | sin θ | cos θ | tan θ | |---|---|---|---| | 0° | 0 | 1 | 0 | | 30° | 1/2 | √3/2 | 1/√3 or √3/3 | | 45° | √2/2 | √2/2 | 1 | | 60° | √3/2 | 1/2 | √3 | | 90° | 1 | 0 | Undefined | | 180° | 0 | -1 | 0 | | 270° | -1 | 0 | Undefined | | 360° | 0 | 1 | 0 | Reduction Formulae: Reduction formulae allow us to express trigonometric functions of angles greater than 90° in terms of acute angles (angles between 0° and 90°). These formulae are based on the symmetry properties of the unit circle and the signs of the functions in each quadrant.
Quadrant II: sin (180° - θ) = sin θ cos (180° - θ) = -cos θ tan (180° - θ) = -tan θ Quadrant III: sin (180° + θ) = -sin θ cos (180° + θ) = -cos θ tan (180° + θ) = tan θ Quadrant IV: sin (360° - θ) = -sin θ cos (360° - θ) = cos θ tan (360° - θ) = -tan θ Additionally, angles beyond 360° are co-terminal, meaning 360°+θ is the same angle as θ sin (360° + θ) = sin θ cos (360° + θ) = cos θ tan (360° + θ) = tan θ Solving Trigonometric Equations: Solving trigonometric equations involves finding the angle(s) that satisfy a given equation. We typically seek solutions within a specified range (e.g., 0° ≤ θ ≤ 360°).
Example 1: Finding Trigonometric Ratios Given point P(-3, 4). Find sin θ, cos θ, and tan θ.
Step 1: Find r. r = √((-3)^2 + 4^2) = √(9 + 16) = √25 = 5 Step 2: Apply the definitions. sin θ = y/r = 4/5 cos θ = x/r = -3/5 tan θ = y/x = 4/-3 = -4/3 Example 2: Using Reduction Formulae Simplify sin 210°.
Step 1: Identify the quadrant. 210° lies in Quadrant II
I. Step 2: Apply the reduction formula. sin 210° = sin (180° + 30°) = -sin 30° Step 3: Evaluate. -sin 30° = -1/2 Example 3: Solving a Trigonometric Equation Solve for θ: cos θ = -√3/2, 0° ≤ θ ≤ 360°.
Step 1: Determine the reference angle. The reference angle is the acute angle whose cosine is √3/
2. This is 30°.
Step 2: Identify the quadrants where cosine is negative. Cosine is negative in Quadrants II and II
I. Step 3: Find the solutions in each quadrant.
Quadrant II: θ = 180° - 30° = 150° Quadrant III: θ = 180° + 30° = 210° Solution: θ = 150° or 210° Guided Practice (With Solutions)
Question 1: Determine the sign of tan 135°.
Solution: 135° lies in Quadrant I
I. In Quadrant II, only sine is positive; therefore, tangent is negative.
Answer: Negative.
Question 2: Evaluate cos 300° without a calculator.
Solution: 300° lies in Quadrant I
V. Using the reduction formula: cos 300° = cos (360° - 60°) = cos 60° cos 60° = 1/2 Answer: 1/2 Question 3: Simplify the expression: sin(180° + x) * cos(360° - x)
Solution: sin(180° + x) = -sin x (Quadrant III) cos(360° - x) = cos x (Quadrant IV) Therefore, sin(180° + x) cos(360° - x) = (-sin x) * (cos x) = -sin x cos x Answer: -sin x cos x Question 4: If sin θ = -0.6 and 180° < θ < 270°, find cos θ.
Solution: We know sin θ = y/r, and cos θ = x/r, and r is always positive. Since sin θ is negative, y = -0.6r We need to find x.
Using the Pythagorean theorem: x² + y² = r² x² + (-0.6r)² = r² x² + 0.36r² = r² x² = 0.64r² x = ±0.8r Since 180° < θ < 270° (Quadrant III), x is negative. So, x = -0.8r Therefore, cos θ = x/r = -0.8r/r = -0.8 Answer: -0.8 Independent Practice (Questions Only) State the quadrant in which the following angles lie: 225°, 310°, 100°, 180°, 390°. Determine the sign of sin θ, cos θ, and tan θ for θ = 200°. Evaluate sin 150°, cos 225°, and tan 330° without using a calculator.