Answer:
cos(A + B) = [tex]\frac{-3\sqrt{5}+1}{8}[/tex]
Step-by-step explanation:
Let us revise some rule of trigonometry
The sign of the trigonometry functions in the four quadrants
In the given question
→ Angle A is in the 4th quadrant
∵ sin(A) = -1/2
→ Use the 1st rule above to find cos(A)
∵ (-1/2)² + cos²(A) = 1
∴ 1/4 + cos²(A) = 1
→ Subtract 1/4 from both sides
∴ cos²(A) = 3/4
→ Take square root for both sides
∴ cos(A) = ±√(3/4)
→ In the 4th quadrant cos is a positive value
∴ cos(A) = (√3)/2
→ Angle B is in the 2nd quadrant
∵ sin(B) = 1/4
→ Use the 1st rule above to find cos(B)
∵ (1/4)² + cos²(B) = 1
∴ 1/16 + cos²(B) = 1
→ Subtract 1/16 from both sides
∴ cos²(B) = 15/16
→ Take square root for both sides
∴ cos(B) = ±√(15/16)
→ In the 2nd quadrant cos is a negative value
∴ cos(B) = (-√15)/4
Let us find the exact value of cos(A + B)
→ By using the 2nd rule above
∵ cos(A + B) = cos(A) cos(B) - sin(A) sin(B)
∴ cos(A + B) = [tex](\frac{\sqrt{3}}{2})(-\frac{\sqrt{15}}{4})-(-\frac{1}{2})(\frac{1}{4}) =-\frac{3\sqrt{5}}{8}+\frac{1}{8}[/tex]
∴ cos(A + B) = [tex]\frac{-3\sqrt{5}+1}{8}[/tex]
