Prove frac{sin(2x)}{sqrt{2-2cos(2x)}} = sin(x)
\[\frac{\sin(2x)}{\sqrt{2-2\cos(2x)}} = \sin(x) \]
u know the formula for cos 2x ?
and find any restrictions on x
i actually just need that part .. the restriction
ok, the denominator has square root sign and the expression under it cannot be negative. Also , denominator cannot be 0. make sense till here ?
yes
so\( 1-cos 2x \ne 0\) so \(cos2x \ne 1\) so 2x cannot equal 0,2pi,4pi,... so x cannot equal 0,pi,2pi,3pi...... ok?
oh yeah .. cant remember properly but the test said the identity holds for 0<x<180 degrees.. is that true ?? idn if supposed to find that
seems correct, because cos 2x will be negative in 3rd quadrant, then denominator can no longer be 2cos x....so yeah.
how can i get that restriction ??
u need cos 2x to be negative, right ? only then u could write denominator as 2cos x now, cos 2x negative implies, 2x is between 90 to 270 , right? so actually here i am getting x is between 45 to 135.... 45<x<135 as restriction.... does this makes sense?
yeah i get you
so overall restrictions are 45<x<135 and x \(\ne\) 0,180,360,......
can i solve 2-2cos(2x)>0 "?
if u solve that what u get?
x>0,pi,2pi,3pi
?
oh,nopes... when we take cos 2x as negative, we already took care of expression under square root to be positive....so need of that.
oh so you just solved 2-2cos(2x)=0 and 2-2cos(2x)<0
so is it 0<x<180 ??
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