If 180<A<270 and sinA= -radical(5)/3, find tan(1/2)A.
A is in QIII, so cosA = -2/3
tan(A/2) = (1-cosA)/sinA = (5/3)/(-√5/3) = -√5
To find tan(1/2)A, let's first find the value of A.
Given that sinA = -√5/3, we can determine the quadrant in which angle A lies by considering the sign of sinA. Since sinA is negative, angle A must be in the third quadrant (180° < A < 270°).
Now, let's solve for A:
Since sinA = -√5/3, we can use the Pythagorean identity to find cosA:
cos^2A = 1 - sin^2A
cos^2A = 1 - (-√5/3)^2
cos^2A = 1 - 5/9
cos^2A = 9/9 - 5/9
cos^2A = 4/9
cosA = ± 2/3
Since A lies in the third quadrant, cosA must be negative:
cosA = -2/3
Therefore, A = arccos(-2/3) ≈ 143.13° (rounded to two decimal places).
Now, let's find 1/2A:
1/2A = 1/2 * 143.13° ≈ 71.57° (rounded to two decimal places).
Finally, we can determine tan(1/2)A:
Since tan(1/2)A = sin(1/2)A / cos(1/2)A, we need to find the values of sin(1/2)A and cos(1/2)A.
Using the half-angle identities for sine and cosine, we have:
sin(1/2)A = √((1 - cosA) / 2)
sin(1/2)A = √((1 - (-2/3)) / 2)
sin(1/2)A = √((3/3 + 2/3) / 2)
sin(1/2)A = √(5/6)
cos(1/2)A = √((1 + cosA) / 2)
cos(1/2)A = √((1 + (-2/3)) / 2)
cos(1/2)A = √((3/3 - 2/3) / 2)
cos(1/2)A = √(1/6)
Now, we can find tan(1/2)A:
tan(1/2)A = sin(1/2)A / cos(1/2)A
tan(1/2)A = (√(5/6)) / (√(1/6))
tan(1/2)A = (√5) / (√1) or √5
Therefore, tan(1/2)A = √5.