Question

How do you find the value?

`cos` `alpha/2` if `tan alpha= 40/9, (180 < alpha<270)`

Answer 1

`cos(alpha/2)=-4/sqrt41`

Explanation:

Given:
`(180ltalpha<270)`
indicating that `cosalpha` is negative

`tanalpha=40/9`
opposite side is 40
adjacent side is 9
hypotenuse will be 41,
because
`sqrt(9^2+40^2)=sqrt(81+1600)=sqrt1681=41`
`cosalpha="(adjacent side)/(hypotenuse)"`
`cosalpha=-9/41`
`cos(alpha/2)=+-sqrt((1+cosalpha)/2)=+-sqrt((1-9/41)/2`
`=+-sqrt((41-9)/(41xx2))=+-sqrt(32/2xx1/41)=+-sqrt(16)/sqrt(41)`

`cos(alpha/2)=+-4/sqrt41`

As mentioned, `180^@ltalphalt270^@,`
it follows that
`180/2ltalpha/2<270/2`

`90ltalpha/2<135`, where `cos(alpha/2)` is negative

Hence,
`cos(alpha/2)=-4/sqrt41`

Answer 2

Start with the identity:

`tan^2(alpha) +1 = sec^2(alpha)`

Substitute `sec^2(alpha)= 1/cos^2(alpha)`

`tan^2(alpha)+ 1= 1/cos^2(alpha)`

Multiply both sides by `cos^2(alpha)/(tan^2(alpha) + 1)`:

`cos^2(alpha) = 1/(tan^2(alpha) + 1)`

Use the square root operation on both sides:

`cos(alpha) = +-sqrt(1/(tan^2(alpha) + 1))`

We are told that `180^@ < alpha < 270^@`, therefore we choose the negative value:

`cos(alpha) = -sqrt(1/(tan^2(alpha) + 1))`

Add 1 to both sides:

`1+ cos(alpha) = 1-sqrt(1/(tan^2(alpha) + 1))`

Multiply both sides by `1/2`:

`(1+ cos(alpha))/2 = (1-sqrt(1/(tan^2(alpha) + 1)))/2`

Use the square root operation on both sides:

`+-sqrt((1+ cos(alpha))/2) = +-sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)`

Substitute `+-sqrt((1+ cos(alpha))/2)= cos(alpha/2)`

`cos(alpha/2) = +-sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)`

From `180^@ < alpha < 270^@` we derive `90^@ < alpha/2 < 135^@` and conclude that the cosine function is negative within the specified domain:

`cos(alpha/2) = -sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)`

Substitute `tan^2(alpha) = (40/9)^2`:

`cos(alpha/2) = -sqrt((1-sqrt(1/((40/9)^2 + 1)))/2)`

I used WolframAlpha to simplify the above into an exact form:

`cos(alpha/2) = -(4sqrt41)/41`