Question

If `cos(t)=-7/8` with `pi<=t<=3pi/2`, how do you find the value of `cos(t/2)`?

Answer 1

`cos(t/2)=-1/4`

Explanation:

`"using the "color(blue)"half angle formula"`

`•color(white)(x)cos(t/2)=+-sqrt((1+cost)/2)`

`pi<=t<=(3pi)/2rArrpi/2<=t/2<=(3pi)/4`

`rArrcos(t/2)=-sqrt((1-7/8)/2`

`color(white)(rArrcos(t/2))=-sqrt(1/16)=-1/4`

Answer 2

`cos(t/2)=-1/4`

Explanation:

Identity:

`color(red)bb(cos(x/2)=sqrt(1/2(1+cosx)`

`cos(t/2)=sqrt(1/2(1+(-7/8))`

`cos(t/2)=sqrt(1/2(1/8))`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=sqrt(1/16)=+-1/4`

If:

`pi <= t <= (3pi)/2`

Then:

`t/2`

`pi/2 <=t/2 <= (3pi)/4`

This is in the II quadrant.

Where the cosine is negative:

`cos(t/2)=-1/4`

Answer 3

The use of the some trigonometric identities is the key to solving this question.

Explanation:

First let's look at the double angle formula:
`cos(2x) = cos^2(x) - sin^2(x)`

This cannot be used in its current form since `sin^2(x)` is present. But, rearranging the pythagorean identity allows for the `sin^2(x)` to be replaced in terms of `cos(x)` :
`sin^2(x) + cos^2(x) = 1`
`sin^2(x) = 1 - cos^2(x)`

Substitution of `sin^2(x)` into double angle formula:
`cos(2x) = cos^2(x) - (1 - cos^2(x))`
`cos(2x) = 2cos^2(x) - 1`

Now, looking at the question, if we let `t = 2x` and `t/2 = x` it changes the double angle formula to a form which can be directly used:
`cos(t) = 2cos^2(t/2) - 1`

Finally, substituting in the value of `cos(t)` and rearranging the equation yields an answer:
`-7/8 = 2cos^2(t/2) - 1`
`1/8 = 2cos^2(t/2)`
`1/16 = cos^2(t/2)`
`1/4 = cos(t/2)`

Hence, the value of `cos(t/2)` is `+-1/4` .
The question specifies an answer for the third quadrant and so
`cos(t/2) = -1/4`