Question

How do you find the value of `sin(theta/2)` given `tantheta=2` and `0<theta<pi/2`?

Answer 1

`sin(theta/2)=sqrt((sqrt5-1)/(2sqrt5))`

Explanation:

It is observed that in range `0<theta<pi/2` all trigonometric functions are positive. Further as `theta` is in first quadrant, so is `theta/2` hence trigonometric ratios of `teta/2` too are all positive.

Now as `tantheta=2`, `sectheta=sqrt(1+tan^2theta)`

= `sqrt(1+2^2)=sqrt5`

and hence `costheta=1/sqrt5`

Now, `sin(theta/2)=sqrt((1-costheta)/2)`

= `sqrt((1-1/sqrt5)/2)=sqrt((sqrt5-1)/(2sqrt5))`

Answer 2

`:.sin(theta/2)={(sqrt5-1)sqrt(50+10sqrt5)}/20`

Explanation:

Let us use the Identity`: tantheta=(2tan(theta/2))/(1-tan^2(theta/2)`

Letting, `tan(theta/2)=t, tantheta=(2t)/(1-t^2).............(1)`.

Since, `tantheta=2`, we have, by `(1)`,

`2=(2t)/(1-t^2)rArr1-t^2=t, or, t^2+t-1+0`

`t=(-1+-sqrt(1^2-4*1*(-1)))/2=(-1+-sqrt5)/2`

Let us note that, by data, `0<,theta<,pi/2rArr 0<,theta/2<,pi/4`

`:. theta/2` lies in the First Quadrant, in which, `tan(theta/2)>0,`

and, also, `sin(theta/2)>0`

This means that `tan(theta/2)!=(-1-sqrt5)/2`

`:. tan(theta/2)=(sqrt5-1)/2rArr cot(theta/2)=2/(sqrt5-1)`

Now, `csc^2(theta/2)=1+cot^2(theta/2)=1+4/(sqrt5-1)^2`

`:.csc^2(theta/2)={(5-2sqrt5+1)+4}/(sqrt5-1)^2=(10-2sqrt5)/(sqrt5-1)^2`

`:.sin(theta/2)=+(sqrt5-1)/sqrt(10-2sqrt5)={(sqrt5-1)sqrt(10+2sqrt5)}/{sqrt(10-2sqrt5)sqrt(10+2sqrt5)}`

`={(sqrt5-1)sqrt(10+2sqrt5)}/sqrt(100-20)={(sqrt5-1)sqrt(10+2sqrt5)}/(4sqrt5)`

`={(sqrt5-1)(sqrt5)sqrt(10+2sqrt5)}/(4*5)={(sqrt5-1)sqrt(50+10sqrt5)}/20`