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}#