Find the value of sin(α - θ)?

α is in the 1st quadrant.
Thus, sin(α) > 0 and cos(α) > 0
cos(α) = 5/6
Since sin²(α) + cos²(α) = 1
then, sin(α) = √[1 - cos²(α)] = √[1 - (5/6)²] = (√11)/6

θ is in the 4th quadrant.
Thus, sin(θ) < 0 and cos(θ) > 0
sin(θ) = -3/4
Since sin²(θ) + cos²(θ) = 1
then, cos(θ) = √[1 - sin²(α)] = √[1 - (-3/4)²] = (√7)/4

sin(α - θ)
= sin(α)cos(θ) - cos(α)sin(θ)
= [(√11)/6] × [(√7)/4] - (5/6) × (-3/4)
= [(√77)/24] + (15/24)
= (15 + √77)/24

i) Since α is in 1st quadrant, all trigonometric relations of α are > 0
Hence sin(α) = √(1 - cos²α) = (√11)/6

ii) θ is in 4th quadrant; so cos(θ) > 0
Hence cos(θ) = √(1 - sin²θ) = (√7)/4

iii) Your formula for sin(α - θ) = sin(α)cos(θ) - cos(α)sin(θ) is correct;
now you need to substitute the values and evaluate.
==> sin(α - θ) = {(√11)/6}*{(√7)/4} - {(5/6)}*{-(3/4)} = √77/4 + 15/24 = (√77 + 15)/24 = 0.9906

cos(α) = 5/6 → α is in the quadrant I → sin(α) > 0

sin(θ) = - 3/4 → θ is in the quadrant IV → cos(θ) > 0



cos²(α) + sin²(α) = 1

sin²(α) = 1 - cos²(α)

sin²(α) = 1 - (5/6)²

sin²(α) = (36/36) - (25/36)

sin²(α) = 11/36

sin²(α) = [± (√11)/6]²

sin(α) = ± (√11)/6 → recall: sin(α) > 0

sin(α) = (√11)/6


cos²(θ) + sin²(θ) = 1

cos²(θ) = 1 - sin²(θ)

cos²(θ) = 1 - (- 3/4)²

cos²(θ) = (16/16) - (9/16)

cos²(θ) = 7/16

cos²(θ) = [± (√7)/4]²

cos(θ) = ± (√7)/4 → recall: cos(θ) > 0

cos(θ) = (√7)/4



Do you know this identity?

sin(a - b) = sin(a).cos(b) - cos(a).sin(b) → adapt it to your case: sin(α - θ)

sin(α - θ) = sin(α).cos(θ) - cos(α).sin(θ)

sin(α - θ) = [(√11)/6].[(√7)/4] - (5/6).(- 3/4)

sin(α - θ) = [(√77)/24] + (15/24)

sin(α - θ) = (√77 + 15)/24