Question

Given that tan a=-(√5)/2 where a is an obtuse angle, can u please help me to find the value of sin(a/2) without using a calculator?

Answer 1

`sin (a/2)= sqrt(5/6)`

Explanation:

Given `tan a = -sqrt5 /2`, `sin a` would be =`sqrt5/(sqrt((sqrt5)^2+2^2))=sqrt5/sqrt(5+4)=sqrt 5/3` and `cos a` would be = `sina/tana=(-sqrt5/3)/(sqrt5/2)=-2/3`.

Further, as an obtuse angle is in `Q2`, where `sina` would be positive and `cosa` would be negative. Also note that `a/2` would be in `Q1` and its all trigonometric ratios would be positive.

Now use formula `cos a= 1- 2sin^2 (a/2)`

Thus `sin^2 (a/2)=1/2 (1-cos a)`

=`1/2 (1+2/3)=5/6`

`sin (a/2)=sqrt(5/6)`

Answer 2

`tana=-sqrt5/2`

`=>(2tan(a/2))/(1-tan^2(a/2))=-sqrt5/2`

`=>4tan(a/2))=-sqrt5+sqrt5tan^2(a/2)`

`=>sqrt5tan^2(a/2)-4tan(a/2)-sqrt5=0`

`=>sqrt5tan^2(a/2)-5tan(a/2)+tan(a/2)-sqrt5=0`

`=>sqrt5tan(a/2)(tan(a/2)-sqrt5)+(tan(a/2)-sqrt5)=0`

`=>(sqrt5tan(a/2)+1)(tan(a/2)-sqrt5)=0`

so

`tan(a/2)=-1/sqrt5->` rejected as a being obtuse `a/2` is acute and `tan(a/2)` should be positive.

Hence `tan(a/2)=sqrt5` accepted

Now `sin(a/2)=1/csc(a/2)`

`=>sin(a/2)=1/sqrt(1+cot^2(a/2))`

`=>sin(a/2)=1/sqrt(1+(1/sqrt5)^2)=sqrt(5/6)`