If cos(t)=(-7/8) where pi<t<(3pi/2), find the values of the following trig functions.

cos(2t)=?
sin(2t)=?
cos(t/2)=?
sin(t/2)=?

if cos t = -7/8, then sin t = -√15/8 , t is in III

cos 2t = cos^2t - sin^2t = 49/64 - 15/64 = 34/64
sin 2t = 2sintcost = 2(-7/8)(-√15/8) = 14√15/64

cos t = 2cos^2 (t/2) - 1
-7/8 + 1 = 2cos^2 t
cos^2(t/2) = 1/16
cos (t/2) = -1/4 , if t is in III

cos t = 1 - 2sin^2 (t/2)
2sin^2 (t/2) = 1 + 7/8
sin^2 (t/2) = 15/16
sin (t/2) = √15/4 , for t in III

To find the values of the trigonometric functions, we'll first need to find the values of cosine and sine for the original angle t, which is given as cos(t) = -7/8.

1. Finding cos(t):
Since cos(t) = -7/8, and t is in the range (π, 3π/2), we know that t is in the third quadrant. In the third quadrant, cosine is negative, so cos(t) = -7/8.

2. Finding sin(t):
To find sin(t), we can use the Pythagorean Identity: sin^2(t) + cos^2(t) = 1. Since we already know cos(t) = -7/8, we can substitute it into the equation and solve for sin(t).

sin^2(t) + (-7/8)^2 = 1
sin^2(t) + 49/64 = 1
sin^2(t) = 1 - 49/64
sin^2(t) = 15/64

Taking the square root of both sides, we get:
sin(t) = ±√(15/64) = ±(√15 / 8)

Since t is in the third quadrant, sin(t) is negative:
sin(t) = -(√15 / 8)

3. Finding cos(2t):
To find cos(2t), we can use the double-angle identity for cosine:

cos(2t) = cos^2(t) - sin^2(t)

Substituting the values we found earlier:
cos(2t) = (-7/8)^2 - (-(√15 / 8))^2
cos(2t) = 49/64 - 15/64
cos(2t) = 34/64
cos(2t) = 17/32

So, cos(2t) = 17/32.

4. Finding sin(2t):
Similarly, using the double-angle identity for sine:

sin(2t) = 2sin(t)cos(t)

Substituting the values we found earlier:
sin(2t) = 2*(-(√15 / 8)) * (-7/8)
sin(2t) = (14√15 / 64)

So, sin(2t) = 14√15 / 64.

5. Finding cos(t/2):
To find cos(t/2), we can use the half-angle identity for cosine:

cos(t/2) = ±√((1 + cos(t)) / 2)

Substituting the values we found earlier:
cos(t/2) = ±√((1 + (-7/8)) / 2)
cos(t/2) = ±√((1 - 7/8) / 2)
cos(t/2) = ±√((1/8) / 2)
cos(t/2) = ±√(1/16)
cos(t/2) = ±(1/4)

So, cos(t/2) = ±1/4.

6. Finding sin(t/2):
Similarly, using the half-angle identity for sine:

sin(t/2) = ±√((1 - cos(t)) / 2)

Substituting the values we found earlier:
sin(t/2) = ±√((1 - (-7/8)) / 2)
sin(t/2) = ±√((1 + 7/8) / 2)
sin(t/2) = ±√((15/8) / 2)
sin(t/2) = ±√(15/16)
sin(t/2) = ±√(15) / 4

So, sin(t/2) = ±√(15) / 4.

In summary:
cos(2t) = 17/32
sin(2t) = 14√15 / 64
cos(t/2) = ±1/4
sin(t/2) = ±√(15) / 4

To find the values of the trigonometric functions, we need to use the double angle and half angle identities.

Given that cos(t) = -7/8 where pi < t < (3pi/2):

1. cos(2t) = 2cos^2(t) - 1 (double angle identity for cosine)
Plugging in the value of cos(t) = -7/8:
cos(2t) = 2(-7/8)^2 - 1
cos(2t) = 2(49/64) - 1
cos(2t) = 98/64 - 1
cos(2t) = -34/64
Simplifying, we get:
cos(2t) = -17/32

2. sin(2t) = 2sin(t)cos(t) (double angle identity for sine)
Plugging in the value of cos(t) = -7/8:
sin(2t) = 2sin(t)(-7/8)
sin(2t) = -14sin(t)/8
Simplifying, we get:
sin(2t) = -7sin(t)/4

3. cos(t/2) = ±√[(1 + cos(t))/2] (half angle identity for cosine)
Since cos(t) = -7/8 is negative, we need to use the negative sign:
cos(t/2) = -√[(1 + cos(t))/2]
Plugging in the value of cos(t) = -7/8:
cos(t/2) = -√[(1 - 7/8)/2]
cos(t/2) = -√[(1/8)/2]
cos(t/2) = -√(1/16)
Simplifying, we get:
cos(t/2) = -1/4

4. sin(t/2) = ±√[(1 - cos(t))/2] (half angle identity for sine)
Since cos(t) = -7/8 is negative, we need to use the positive sign:
sin(t/2) = √[(1 - cos(t))/2]
Plugging in the value of cos(t) = -7/8:
sin(t/2) = √[(1 + 7/8)/2]
sin(t/2) = √[(15/8)/2]
sin(t/2) = √(15/16)
Simplifying, we get:
sin(t/2) = √15/4