How do you prove: #(4sin^2x)/(2sinx+1)=2sinx-1#?

1 Answer
Mar 7, 2017

The problem is that this is not a valid trig identity

Explanation:

For example, if #x=0# then
LS
#color(white)("XXX")=(4sin^2(x))/(2sin(x)+1)=(4xx1^2)/(2xx1+1)=4/3#
and
RS
#color(white)("XXX")=2sin(x)-1=2xx1-1=1#

Clearly #LS!=RS#

Perhaps the numerator (LS) should have been #4sin^2(x)-1#
(in which case the solution is fairly simple since #4sin^2(x)-1# can be factored as #(2sin(x)-1)(2sin(x)+1)#).