Suppose x is an angle in the second quadrant such that cosx=-8/17, what is the value of tan(3x/2)?
first get you results to sine and cosine of x
since you are in II, from your diagram,
sinx = 15/17
tanx = -15/8
use the formula for tan(3x) you just found in your other post.
then apply the formula for tan(A/2)
actually that is a nice one:
tan (A/2) = (1 - cosA)/sinA
good luck
To find the value of tan(3x/2), we first need to determine the value of x. We are given that cos(x) = -8/17 and x lies in the second quadrant.
In the second quadrant, both sine and tangent values are positive. Since cosine is negative in the second quadrant, we can use the Pythagorean identity to find the value of sin(x).
Using the Pythagorean identity:
sin^2(x) + cos^2(x) = 1
sin^2(x) + (-8/17)^2 = 1
sin^2(x) + 64/289 = 1
sin^2(x) = 225/289
sin(x) = sqrt(225/289)
sin(x) = 15/17
Now that we have the values of sin(x) and cos(x), we can use them to find the value of tan(x):
tan(x) = sin(x)/cos(x)
tan(x) = (15/17)/(-8/17)
tan(x) = -15/8
Now, we need to find the value of 3x/2. Multiplying tan(x) by 3/2:
tan(3x/2) = (3/2) * (-15/8)
tan(3x/2) = -45/16
Therefore, the value of tan(3x/2) is -45/16.