Question

For `x in (0,pi/2) and sinx=3/5` how to find `tan(x/2)`?

Answer 1

`tan (x/2) = 1/3`

Explanation:

The basic trigonometric functions of angles in the range `(0, pi/2)` are described in terms of ratios of the sides of right angled triangles.

I will use `A` instead of `x` because I already have a picture...

`color(white)()`

In our example:

`sin A = "opposite"/"hypotenuse" = 3/5`

`cos A = "adjacent"/"hypotenuse" = 4/5`

`tan A = "opposite"/"adjacent" = 3/4`

The half angle formulas for `cos` and `sin` tell us that:

`cos (A/2) = +-sqrt((1+cos A)/2) = +-sqrt(9/10) = +-sqrt(90)/10`

`sin (A/2) = +-sqrt((1-cos A)/2) = +-sqrt(1/10) = +-sqrt(10)/10`

Note that we can identify that we need the `+` sign here since the angles are all in Q1.

Hence:

`tan (A/2) = (sin (A/2))/(cos (A/2)) = sqrt(10)/sqrt(90) = 1/3`

Answer 2

`tan (x/2) = 1/3`

Explanation:

`sin x = 3/5`. First, find cos x .
`cos^2 x = 1 - sin^2 x = 1 - 9/25 = 16/25`
`cos x = +- 4/5`
x is in Quadrant 1, then, cos x is positive --> `cos x = 4/5`
Find `cos (x/2)` by using trig identity:
`2cos^2 x = 1 + cos 2x`.
In this case:
`2cos^2 (x/2) = 1 + cos x = 1 + 4/5 = 9/5`
`cos^2 (x/2) = 9/10` --> `cos (x/2) = +- 3/sqrt10`
`x/2` is in Quadrant 1, then `cos x/2` is positive.
To find `sin (x/2)` use trig identity:
`sin x = 2sin (x/2)cos (x/2)`
`sin (x/2) = (sin x)/(2cos (x/2)) = (3/5)(sqrt10/6) = sqrt10/10`
`tan (x/2) = sin (x/2)/(cos (x/2)) = (sqrt10/10)(sqrt10/3) = 10/30 = 1/3`