Question

how do I finish this

If π < t < 3π/2 then π/2 < t/2 < 3π/4; for such values of t, cos(t/2) < 0 and sin(t/2) > 0

From this and other trig identities,
cos(t/2) = -√[(1 + cos(t))/2]
sin(t/2) = √[(1 - cos(t))/2]

how do I finish this

If π < t < 3π/2 then π/2 < t/2 < 3π/4; for such values of t, cos(t/2) < 0 and sin(t/2) > 0

From this and other trig identities,
cos(t/2) = -√[(1 + cos(t))/2]
sin(t/2) = √[(1 - cos(t))/2]

@Mathematics

Answer 1

So what are you suppose to do on this??
Im kinda lost

Answer 2

this is the original question

Answer 3

If cos(t)=−3/4 where π<t<3π/2, find the values of the following trigonometric functions?
cos(2t) =
sin(2t)=
cos(t/2)=
sin(t/2)=

Answer 4

i have found the first two to be

cos(2t) = 2 cos²(t) - 1 = 2(-¾)² - 1 = 1/8

sin(t) = -√(1 - cos²(t)) = -√(1 - (-¾)²) = -√7 / 4
sin(2t) = 2 sin(t) cos(t) = 2(-√7/4)(-¾) = 3 √7 / 8

Answer 5

and i started the remaining two but dont know how to finish it

Answer 6

Ok

Answer 7

\[\cos(t/2) = \sqrt{\frac{1+\cos(x)}{2}}\]

Answer 8

that cos(x) should be cos(t) in this case.

Answer 9

\[\cos(t/2) = \sqrt{\frac{1+(-\frac{3}{4})}{2}}\]

Answer 10

ok

Answer 11

one sec

Answer 12

im having extreme difficulties entering this into my homework software lol

Answer 13

you should get \[\frac{\sqrt{8}}{8}\]

Answer 14

it rejected that..uhhh