Question

How do you find the exact values of `sin(u/2), cos(u/2), tan(u/2)` using the half angle formulas given `tanu=-5/8, (3pi)/2<u<2pi`?

Answer 1

`sin(u/2)=(sqrt89-8)/sqrt(178-16sqrt89)`
`cos(u/2)=-5/sqrt((178-16sqrt89)`
`tan(u/2)=-(sqrt89-8)/5`

Explanation:

As `tanu=(2tan(u/2))/(1-tan^2(u/2)`

`(2tan(u/2))/((1-tan^2(u/2)))=-5/8`

or `16tan(u/2))=-5+5tan^2(u/2)`

or `5tan^2(u/2)-16tan(u/2)-5=0`

and using quadratic formula

`tan(u/2)=(-(-16)+-sqrt((-16)^2-4xx5xx(-5)))/10`

= `(16+-sqrt(256+100))/10=(16+-sqrt356)/10=(8+-sqrt89)/5`

As `(3pi)/2 < u < 2pi`, we have `(3pi)/4 < u/2 < pi` i.e. `u/2` lies in Q2

and hence `tan(u/2)` is negative i.e. `tan(u/2)=-(sqrt89-8)/5`

and `sec(u/2)=-sqrt(1+(sqrt89-8)^2/25)`

= `-sqrt(1+(89+64-16sqrt89)/25)`

= `-sqrt((178-16sqrt89)/25)`

and `cos(u/2)=-5/sqrt((178-16sqrt89)`

and `sin(u/2)=cos(u/2)tan(u/2)=-5/sqrt((178-16sqrt89))xx(-(sqrt89-8)/5)`

= `(sqrt89-8)/sqrt(178-16sqrt89)`