Question

Use the information about the angle θ to find the exact value of?

a. `sin(θ/2)`
b. `cos(θ/2)`
c. `tan(θ/2)`

`cot θ = 3/4`, `pi` < θ < `(3pi)/2`

Answer 1

`sin(theta/2) = sqrt(10)/5`

`cos(theta/2) = sqrt(10)/10`

`tan(theta/2) = 2`

Explanation:

Given: `cot(theta) = 3/4, pi < theta < (3pi)/2`

We need `cos(theta)` for the formulas of the half-angles :

`cos(theta) = cos(cot^-1(x))`

`cos(theta) = +-x/sqrt(x^2+1)`

Substitute `x = 3/4`:

`cos(theta) = +-(3/4)/sqrt((3/4)^2+1)`

`cos(theta) = +-(3/4)/sqrt((3/4)^2+1)`

`cos(theta) = +-3/sqrt((3^2+4^2)`

`cos(theta) = +-3/5`

The cosine function is negative in the third quadrant:

`cos(theta) = -3/5" [1]"`

For the half-angle formulas, please observe that:

`pi < theta < (3pi)/2 to pi/2 < theta/2 < (3pi)/4`

This means that sine and cosine of the half-angles are positive.

`sin(theta/2) = sqrt(1-cos(theta))/2`

Substitute equation [1]:

`sin(theta/2) = sqrt(1+(3/5))/2`

`sin(theta/2) = sqrt(10)/5`

`cos(theta/2) = sqrt(1+cos(theta))/2`

Substitute equation [1]:

`cos(theta/2) = sqrt(1-3/5)/2`

`cos(theta/2) = sqrt(10)/10`

`tan(theta/2) = sin(theta/2)/cos(theta/2)`

`tan(theta/2) = (sqrt(10)/5)/(sqrt(10)/10)`

`tan(theta/2) = 2`

Answer 2

`sin (t/2) = (2sqrt2)/(sqrt10)`
`cos (t/2) = - sqrt2/(sqrt10)`
`tan (t/2) = - 2`

Explanation:

`cot t = 3/4` --> `tan t = 4/3`
First find cos t.
`cos^2 t = 1/(1 +tan^2 t) = 1/(1 + 16/9) = 9/25`
`cos t = +- 3/5`.
Since t is in Quadrant 3, then, cos t is negative
`cos t = - 3/5`.
To find `sin (t/2)` and `cos (t/2)` use these identities:
`2cos^2 (t/2) = 1 + cos t`, and
`2sin^2 (t/2) = 1 - cos t`.
In this case:
a. `2sin^2 (t/2) = 1 - cos t = 1 + 3/5 = 8/5`
`sin^2 (t/2) = 8/10`
`sin (t/2) = +- 2sqrt2/sqrt10`
Because t lies in Q.3, then, `(t/2)` lies in Q.2, and `sin (t/2)` is positive
b. `2cos^2 (t/2) = 1 + cos t = 1 - 3/5 = 2/5`
`cos^2 (t/2) = 2/10`
`cos (t/2) = +- sqrt2/(sqrt10)`
Because `(t/2)` lies in Q. 2, therefor, `cos t/2` is negative.
`cos (t/2) = - sqrt2/(sqrt10)`
c. `tan (t/2) = sin/(cos) = ((2sqrt2)/(sqrt10))(sqrt10/-sqrt2)`.
`tan (t/2) = - 2`