Question #c5d00

Let us draw a figure to depict given and what is to calculated

`sin A=sin(40)=X`
`=>"opposite"/ "hypotenuse"=X/1`
This is shown in the figure above.
From Pythagoras theorem
`AB=sqrt(1-X^2)` .....(1)
To calculate `sin(20)` we need to consider `DeltaADB`
In this triangle to find `AD and DB` we proceed as below

If `AD` is the angular bisector of `angle CAB`, by angle bisector theorem we have
`CD:DB :: AC:AB`
`=>(CD)/(DB):: (AC)/(AB)`, From (1)
`=>1/(DB):: 1/sqrt(1-X^2)`
From ratio and proportion
`DB=X xxsqrt(1-X^2)/(1+sqrt(1-X^2))` .......(2)

Also using Pythagoras theorem
`AD=sqrt(DB^2+AB^2)`
from (1) and (2)
`AD=sqrt((X xxsqrt(1-X^2)/(1+sqrt(1-X^2)))^2+(sqrt(1-X^2))^2)`
`=>AD=sqrt(1-X^2)sqrt((X/(1+sqrt(1-X^2)))^2+1)`
`=>AD=sqrt(1-X^2)sqrt((X^2+(1+sqrt(1-X^2))^2)/(1+sqrt(1-X^2))^2)`
`=>AD=sqrt(1-X^2)/(1+sqrt(1-X^2))sqrt(X^2+1+2sqrt(1-X^2)+1-X^2)`
`=>AD=sqrt 2sqrt(1-X^2)/(1+sqrt(1-X^2))sqrt(1+sqrt(1-X^2))`
`=>AD=sqrt 2sqrt(1-X^2)/sqrt(1+sqrt(1-X^2))` .......(3)
Now `sin(20)=(DB)/(AD)`
Inserting the values from (2) and (3) we obtain

`sin(20)=(X xxsqrt(1-X^2)/(1+sqrt(1-X^2)))/(sqrt 2sqrt(1-X^2)/sqrt(1+sqrt(1-X^2)))`
`sin(20)=X /(sqrt 2 sqrt(1+sqrt(1-X^2)))`
Rationalizing the denominator by multiplying and dividing RHS with the conjugate of denominator we get
RHS`=X /(sqrt 2 sqrt(1+sqrt(1-X^2)))xxsqrt(1-sqrt(1-X^2))/sqrt(1-sqrt(1-X^2))`
`=>`RHS`=X sqrt(1-sqrt(1-X^2))/(sqrt 2 sqrt(1-(1-X^2)))`
`=>`RHS`= sqrt((1-sqrt(1-X^2))/ 2 )`

Let `sin(20)=t`, so that `cos(20)=sqrt(1-t^2).`

Now given that, `sin(40)=X`

`:. sin(20+20)=X.`

Now to expand `LHS`, we use the identity `sin(A+B)=sin(A)cos(B)+cos(A)sin(B).`

`:. sin(20)cos(20)+cos(20)sin(20)=X.`

`:. 2sin(20)cos(20)=X.`

Putting the values of `sin(20)` & `cos(20),`

`:.2t*sqrt(1-t^2)=X.`

Squaring, `4t^2*(1-t^2)=X^2.`

`:. 4t^2-4t^4=X^2.`

`:. 4t^4-4t^2=-X^2.`

Completing square,
`4t^4-4t^2+1=1-X^2.`

`:. (2t^2-1)^2=1-X^2.`

`:. 2t^2-1=+-sqrt(1-X^2),` giving,

`t=sin(20)=sqrt{(1+-sqrt(1-X^2))/2}`

The `+` solution is not good since `sin 20^o` should be less than `sin 30^o=1/2` (function `y=sin x` is monotonically increasing in the 1st quadrant).

So the final solution is
`t=sin(20)=sqrt{(1-sqrt(1-X^2))/2}`