How do you verify: #sqrt(1+sin x)-sqrt(1-sin x) =2*sin(x/2)# ?

#sqrt(1+sin x)-sqrt(1-sin x) =2*sin(x/2)#

2 Answers
Apr 14, 2018

One cannot verify #sqrt(1+sin(x))-sqrt(1-sin(x)) =2*sin(x/2)# because it is not an identity.

Explanation:

I shall show that #sqrt(1+sin(x))-sqrt(1-sin(x)) =2*sin(x/2)# is not an identity by substituting the real value #x= (3pi)/4#:

#sqrt(1+sin((3pi)/4))-sqrt(1-sin((3pi)/4)) =2*sin((3pi)/8)#

#0.756!=1.848#

Both sides of an identity must be equal for all real values of x; this is clearly NOT an identity.

Apr 14, 2018

This is valid for #x in [0,pi/2]# - i.e. in the first quadrant

Explanation:

We know #sin x = 2sin(x/2) cos(x/2)#

Thus

#1+sin x = sin^2(x/2)+cos^2(x/2) +2sin(x/2)cos(x/2)#
#qquad qquad qquad quad = (sin(x/2)+cos(x/2))^2#

and

#1-sin x = sin^2(x/2)+cos^2(x/2) -2sin(x/2)cos(x/2)#
#qquad qquad qquad quad = (sin(x/2)-cos(x/2))^2#

The subtlety here is that when we take the square root, we have to take the positive square root. When #x# is in the first quadrant, #0 <= x <= pi/2#, we have

#sin(x/2)+cos(x/2) > 0#

so that

#sqrt(1+ sin x) = sin(x/2)+cos(x/2)#

For the other square root, note that for #0 <= x <= pi/2#, we have
#0 <= x/2 <= pi/4# so that

#0<= tan(x/2) <=1 implies sin(x/2)<= cos(x/2)#

Thus

#sqrt(1-sin x) = cos(x/2)-sin(x/2)#

So, finally

#sqrt(1+sin x)-sqrt(1-sin x) #
#= (sin(x/2)+cos(x/2))-(cos(x/2)-sin(x/2) )#
#= 2*sin(x/2)#