Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• State and prove the compound angle identities.
• Use identities to find exact values for non-standard angles (e.g., 15°, 75°).
• Apply identities to simplify complex trigonometric expressions.
• Solve trigonometric equations using compound angle identities.
• Solve problems in 2D and 3D.

Lesson summary

This week, we delve into the fascinating world of compound angle identities in trigonometry. These identities allow us to express trigonometric functions of sums or differences of angles in terms of trigonometric functions of the individual angles. Mastering these identities is crucial not only for success in Grade 12 Mathematics but also for various fields like engineering, physics, and even surveying where angle calculations are essential. Imagine engineers designing bridges or architects planning buildings – they constantly use these principles.

Lesson notes

What are Compound Angle Identities? Compound angle identities (also called angle sum and difference identities) are a set of trigonometric identities that express trigonometric functions of sums or differences of angles in terms of trigonometric functions of the individual angles.

The core identities are: sin(A + B) = sinA cosB + cosA sinB sin(A - B) = sinA cosB - cosA sinB cos(A + B) = cosA cosB - sinA sinB cos(A - B) = cosA cosB + sinA sinB tan(A + B) = (tanA + tanB) / (1 - tanA tanB) tan(A - B) = (tanA - tanB) / (1 + tanA tanB) Derivation of sin(A + B) and cos(A + B) (Conceptual Understanding): While a rigorous geometric proof is beyond the scope of this lesson note, it is crucial to have a conceptual understanding of where these identities come from. Consider two angles, A and B, constructed within a unit circle. By considering the coordinates of points on the circle corresponding to angles A, B, and A+B, and using distance formulas and trigonometric ratios, one can derive these identities. It involves breaking down the larger angle (A+B) into smaller, manageable components using known trigonometric relationships in right-angled triangles.

Example 1: Finding the exact value of sin 75° We can express 75° as the sum of two standard angles: 75° = 45° + 30°. Using the identity sin(A + B) = sinA cosB + cosA sinB: sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (√2 / 2) (√3 / 2) + (√2 / 2) (1 / 2) = (√6 / 4) + (√2 / 4) = (√6 + √2) / 4 Therefore, sin 75° = (√6 + √2) / 4 Example 2: Finding the exact value of cos 15° We can express 15° as the difference of two standard angles: 15° = 45° - 30°. Using the identity cos(A - B) = cosA cosB + sinA sinB: cos 15° = cos (45° - 30°) = cos 45° cos 30° + sin 45° sin 30° = (√2 / 2) (√3 / 2) + (√2 / 2) (1 / 2) = (√6 / 4) + (√2 / 4) = (√6 + √2) / 4 Therefore, cos 15° = (√6 + √2) / 4 Example 3: Simplifying a Trigonometric Expression Simplify: cos(x + y) + cos(x - y)

Using the compound angle identities: cos(x + y) = cos x cos y - sin x sin y cos(x - y) = cos x cos y + sin x sin y Adding the two expressions: cos(x + y) + cos(x - y) = (cos x cos y - sin x sin y) + (cos x cos y + sin x sin y) = 2 cos x cos y Therefore, cos(x + y) + cos(x - y) simplifies to 2 cos x cos y Example 4: Solving a Trigonometric Equation Solve for x: sin(x + 30°) = cos x, for 0° ≤ x ≤ 360° Using the identity sin(A + B) = sinA cosB + cosA sinB: sin(x + 30°) = sin x cos 30° + cos x sin 30° = sin x (√3 / 2) + cos x (1 / 2) Therefore, the equation becomes: sin x (√3 / 2) + cos x (1 / 2) = cos x sin x (√3 / 2) = cos x - cos x (1 / 2) sin x (√3 / 2) = cos x (1 / 2) sin x / cos x = (1 / 2) / (√3 / 2) tan x = 1 / √3 tan x = √3 / 3 The reference angle is 30°. Since tan x is positive in the 1st and 3rd quadrants: x = 30° or x = 180° + 30° = 210° Therefore, the solutions are x = 30° and x = 210°. Guided Practice (With Solutions)

Question 1: Expand and simplify: sin(x - 45°)

Solution: Using the identity sin(A - B) = sinA cosB - cosA sinB: sin(x - 45°) = sin x cos 45° - cos x sin 45° = sin x (√2 / 2) - cos x (√2 / 2) = (√2 / 2)(sin x - cos x)

Commentary: The key here is correctly identifying the 'A' and 'B' and substituting them into the correct compound angle identity. Remember the values of sin 45° and cos 45°.

Question 2: Determine the exact value of cos 105° Solution: We can express 105° as the sum of two standard angles: 105° = 60° + 45°. Using the identity cos(A + B) = cosA cosB - sinA sinB: cos 105° = cos (60° + 45°) = cos 60° cos 45° - sin 60° sin 45° = (1 / 2) (√2 / 2) - (√3 / 2) (√2 / 2) = (√2 / 4) - (√6 / 4) = (√2 - √6) / 4 Therefore, cos 105° = (√2 - √6) / 4

Commentary: Choose standard angles that add or subtract to give the target angle. Pay close attention to the signs in the compound angle formula, especially for cosine.

Question 3: If sin A = 3/5 and cos B = -5/13, where A is in the first quadrant and B is in the second quadrant, find the value of sin(A + B).

Solution: We need to find cos A and sin B first. Since A is in the first quadrant, cos A is positive.

Using the Pythagorean identity: cos²A = 1 - sin²A cos²A = 1 - (3/5)² = 1 - 9/25 = 16/25 cos A = √(16/25) = 4/5 Since B is in the second quadrant, sin B is positive.

Using the Pythagorean identity: sin²B = 1 - cos²B sin²B = 1 - (-5/13)² = 1 - 25/169 = 144/169 sin B = √(144/169) = 12/13 Now, use the identity sin(A + B) = sinA cosB + cosA sinB: sin(A + B) = (3/5) (-5/13) + (4/5) (12/13) = -15/65 + 48/65 = 33/65 Therefore, sin(A + B) = 33/65

Commentary: Remember to determine the correct signs of cos A and sin B based on the quadrants in which angles A and B lie. Always use the Pythagorean identity to find the missing trigonometric ratios.