cosx= -8/17, 180<x<270
find
sin x/2=
cos x/2=
tan x/2=
To find sin(x/2), cos(x/2), and tan(x/2), we can use the double-angle identities:
sin(x/2) = ±√[(1 - cos(x))/2]
cos(x/2) = ±√[(1 + cos(x))/2]
tan(x/2) = sin(x/2) / cos(x/2)
Let's first find the value of cos(x). We are given that cos(x) = -8/17.
To find sin(x), we can use the Pythagorean identity: sin^2(x) = 1 - cos^2(x).
Since we already know cos(x), we can substitute it into the equation:
sin^2(x) = 1 - (-8/17)^2
sin^2(x) = 1 - 64/289
sin^2(x) = (289 - 64)/289
sin^2(x) = 225/289
sin(x) = ±√(225/289)
But since 180° < x < 270°, the sine function is negative in the third quadrant. Therefore, sin(x) = -15/17.
Now, we can substitute sin(x) and cos(x) into the double-angle identities to find sin(x/2), cos(x/2), and tan(x/2):
sin(x/2) = ±√[(1 - cos(x))/2]
sin(x/2) = ±√[(1 - (-8/17))/2]
sin(x/2) = ±√[(25/17)/2]
sin(x/2) = ±√[25/34]
sin(x/2) = ±5/√34 (rationalize the denominator)
cos(x/2) = ±√[(1 + cos(x))/2]
cos(x/2) = ±√[(1 + (-8/17))/2]
cos(x/2) = ±√[(9/17)/2]
cos(x/2) = ±√[9/34]
cos(x/2) = ±3/√34 (rationalize the denominator)
tan(x/2) = sin(x/2) / cos(x/2)
tan(x/2) = (±5/√34) / (±3/√34)
tan(x/2) = ±5/3
Since we are in the third quadrant, where both sin(x/2) and cos(x/2) are negative, the final answers are:
sin(x/2) = -5/√34
cos(x/2) = -3/√34
tan(x/2) = -5/3