Question

How do you use the half angle formula to solve for sin(x/2) when given sin(x)=(24/25)?

Answer 1

`sin(x/2)=3/5 or sqrt(17)/5`

Explanation:

angle x can be in first quadrant or second quadrant
1. first quadrant (`0 < x < pi/2`)
`cos x = 1 - sin^2(x) = sqrt(25^2-24^2)/25=7/25`
2. second quadrant (`pi/2 < x < pi`)
`cos x = -7/25`

`cos x = 1-2*sin^2(x/2)`
`sin(x/2)=sqrt((1-cos (x))/2)`
(angle `x/2` is in first quadrant. `sin(x/2)>0`)
`sin (x/2)=sqrt(1/2((25+-7)/25)`
`sin(x/2)=3/5 or 4/5`

Answer 2

`sin(x/2)=3/5, or, 4/5`.

Explanation:

We know that, `cosx=1-2sin^2(x/2)..................(star)`.

But, `sinx=24/25`.

`:. cos^2x=1-sin^2x=1-(24/25)^2=(25^2-24^2)/25^2`.

`:. cos^2x={(25+24)(25-24)}/25^2=49/25^2`.

`:. cosx=+-7/25`.

Utilising this value of `cosx` in `(star)`, we get,

`2sin^2(x/2)=1-(+-7/25), i.e.,`

`2sin^2(x/2)=18/25, or, 2sin^2(x/2)=32/25`.

`:. sin^2(x/2)=9/25, or, 16/25`.

Let us note that, `sinx gt 0 :. 0 lt x lt pi :. 0 lt x/2 lt pi/2`.

`:. sin(x/2)" must be "+ve`.

`:. sin(x/2)=+3/5, or, +4/5`.

`color(red)("Enjoy Maths.!")`