Trigonometric Identities

I solved it!

 

Given:

α lies in quadrant II, and tan(α) = -12/5.

β lies in quadrant IV, and cos(β) = 3/5.

We need to find sin(α + β).

 

Step 1: Determine sin(α) and cos(α)

 

Use the given value of tan(α): tan(α) = -12/5

Use the identity tan(α) = sin(α)/cos(α): tan(α) = sin(α)/cos(α)

Assume sin(α) = 12k and cos(α) = 5k, where k is a constant.

Apply the Pythagorean identity sin²(α) + cos²(α) = 1: (12k)² + (5k)² = 1 → 144k² + 25k² = 1 → 169k² = 1 → k² = 1/169 → k = 1/13

Substitute k back to find sin(α) and cos(α): sin(α) = 12k = 12/13, cos(α) = 5k = 5/13

Since α is in quadrant II, cos(α) should be negative: cos(α) = -5/13

Step 2: Determine sin(β) from cos(β)

 

Use the given value of cos(β): cos(β) = 3/5

Apply the Pythagorean identity sin²(β) + cos²(β) = 1: sin²(β) + (3/5)² = 1 → sin²(β) + 9/25 = 1 → sin²(β) = 1 - 9/25 = 25/25 - 9/25 = 16/25

Take the square root to find sin(β): sin(β) = ±4/5

Since β is in quadrant IV, sin(β) should be negative: sin(β) = -4/5

Step 3: Use the angle sum formula for sine

 

Use the angle sum formula: sin(α + β) = sin(α) cos(β) + cos(α) sin(β)

Substitute the known values of sin(α), cos(α), sin(β), and cos(β): sin(α + β) = (12/13)(3/5) + (-5/13)(-4/5)

Simplify each term: sin(α + β) = (12 * 3) / (13 * 5) + (5 * 4) / (13 * 5) = 36/65 + 20/65

Add the fractions: sin(α + β) = (36 + 20) / 65 = 56/65

Therefore, the exact value of sin(α + β) is: 56/65